MATH 121 - COLLEGE ALGEBRA -
Rational Functions
Question Bank - Set 2
Liberty University
Question 1
Question
Let f(x) = 3x2−4x−4
x2+2x−3. Find the vertical asymptotes, horizontal asymptotes, and
holes (if any) of the rational function f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to 0 and
solve for x.x2+ 2x−3 = 0
(x+ 3)(x−1) = 0
So, x=−3and x= 1 are the vertical asymptotes.
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and denominator. Since the degrees are the same, divide the leading
coefficients.
Horizontal asymptote :y=3
1= 3
Step 3: To find the holes, factor the rational function and simplify.
f(x) = 3x2−4x−4
x2+ 2x−3
=(x−2)(3x+ 2)
(x+ 3)(x−1)
Since the factor (x−2) cancel out from the numerator and denominator, x= 2
is a hole in the graph of the function.
Therefore, the vertical asymptotes are x=−3and x= 1, the horizontal
asymptote is y= 3, and there is a hole at x= 2 in the graph of the rational
function f(x).
Question 2
Question
Consider the rational function f(x) = 2x2−7x−30
x2−5x−6. Find the domain of the
function f(x).
Solution
Step 1: To find the domain of a rational function, we need to identify any
values of xthat would make the denominator zero, as these values would result
in division by zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−5x−6 = 0
Step 3: Factor the quadratic equation:
(x−6)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x:
x−6 = 0 =⇒x= 6
x+ 1 = 0 =⇒x=−1
Step 5: Therefore, the values x= 6 and x=−1make the denominator equal
to zero, so they must be excluded from the domain.
Step 6: The domain of the rational function f(x) = 2x2−7x−30
x2−5x−6is all real num-
bers except x= 6 and x=−1, so the domain is (−∞,−1) ∪(−1,6) ∪(6,∞).
Question 3
Question
Given the rational function f(x) = 2x2+5x−3
x2−4x−5, find the horizontal asymptotes (if
any) of the function.
Solution
Step 1: To find the horizontal asymptotes of a rational function, we compare
the degrees of the numerator and denominator polynomials. The degree of the
numerator is 2 and the degree of the denominator is 2, so we must consider the
leading coefficients of both polynomials.
Step 2: The leading coefficient of the numerator is 2, and the leading coeffi-
cient of the denominator is 1. Since the degrees of the two polynomials are equal,
we can find the horizontal asymptote by comparing their leading coefficients.
Step 3: The horizontal asymptote is given by the ratio of the leading coeffi-
cients. Therefore, the horizontal asymptote of the function is y=2
1= 2.
2
Question 4
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes of the ra-
tional function:
f(x) = x2−3x
x2−4
Solution
To find the vertical asymptotes of a rational function, we need to identify the
values of xthat make the denominator equal to zero, but not the numerator.
This will result in division by zero.
Step 1: Find Vertical Asymptotes Setting the denominator equal to
zero and solving for x:
x2−4 = 0
(x+ 2)(x−2) = 0
x=−2or x= 2
So, there are vertical asymptotes at x=−2and x= 2.
Step 2: Find the Horizontal Asymptotes To find horizontal asymp-
totes, we compare the degrees of the numerator and denominator of the function.
Since the degrees of the numerator and denominator are the same (2), the
horizontal asymptote is the ratio of the leading coefficients. The horizontal
asymptote is y=1
1= 1.
Step 3: Find any Holes To find any holes in the graph, we simplify the
function by canceling any common factors between the numerator and denomi-
nator.
f(x) = x(x−3)
(x+ 2)(x−2)
Notice that there is a common factor of xin the numerator and x−2in the
denominator, which gives us a hole at x= 0.
Therefore, the function f(x)has vertical asymptotes at x=−2and x= 2,
a horizontal asymptote at y= 1, and a hole at x= 0.
Question 5
Question
Simplify the following rational expression:
6x3−12x2+ 6x
2x2−4x
3
Solution
Step 1: Factor out the common terms in the numerator and denominator.
6x(x2−2x+ 1)
2x(x−2)
Step 2: Simplify the expression by canceling out the common factors.
6(x2−2x+ 1)
2(x−2)
Step 3: Further simplify the expression by factoring the quadratic in the
numerator. 6(x−1)2
2(x−2)
Step 4: Simplify by canceling out the common factors.
6(x−1)
2
Step 5: Finally, simplify the expression to obtain the simplified rational
expression.
3(x−1)
Question 6
Question
Find the domain of the rational function given by:
f(x) = x2−4
x2−3x−4
Solution
Step 1: To find the domain of the rational function, we need to determine the
values of xfor which the function is defined. The function is defined as long as
the denominator is not equal to zero, since division by zero is undefined.
Step 2: We start by factoring the denominator, x2−3x−4, to determine its
zeros:
x2−3x−4 = (x−4)(x+ 1)
Setting this expression equal to zero and solving for x, we get:
x−4 = 0 or x+ 1 = 0
x= 4 or x=−1
4
Step 3: Therefore, the function f(x)is undefined at x= 4 and x=−1since
these are the values that make the denominator zero.
Step 4: The domain of the function is all real numbers except x= 4 and
x=−1. So, the domain can be expressed in interval notation as:
(−∞,−1) ∪(−1,4) ∪(4,∞)
Question 7
Question
Find the domain of the rational function:
f(x) = 3x−1
x2−4x+ 3
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that make the denominator equal to zero, since division by zero is unde-
fined.
Step 2: To find the domain of f(x), we need to determine the values of x
that make the denominator x2−4x+ 3 equal to zero.
Step 3: We will solve the equation x2−4x+ 3 = 0 to find these values. This
can be factored as (x−3)(x−1) = 0.
Step 4: Setting each factor to zero gives us x−3 = 0 and x−1 = 0. Solving
these equations gives x= 3 and x= 1.
Step 5: Therefore, the values x= 1 and x= 3 make the denominator of the
function equal to zero and must be excluded from the domain.
Step 6: In interval notation, the domain of the function f(x) = 3x−1
x2−4x+3 is
(−∞,1) ∪(1,3) ∪(3,∞).
Question 8
Question
Find the domain of the rational function:
f(x) = 2x+ 1
x2−9
Solution
Step 1: The domain of a rational function consists of all real numbers except
for the values of xthat make the denominator equal to zero since division by
zero is undefined. So, we need to find the values of xthat make x2−9 = 0.
5
Step 2: We can factor the denominator x2−9using the difference of squares
formula: a2−b2= (a+b)(a−b). So, x2−9 = (x+ 3)(x−3).
Step 3: Setting x2−9 = 0 gives us:
(x+ 3)(x−3) = 0
Step 4: Solving for x, we get:
x+ 3 = 0 or x−3 = 0
x=−3or x= 3
Step 5: Therefore, the domain of the function f(x) = 2x+1
x2−9is all real numbers
except x=−3and x= 3, so the domain is:
(−∞,−3) ∪(−3,3) ∪(3,∞)
Question 9
Question
Let f(x) = 4x2+5x−6
2x2−3x−2. Find the vertical asymptotes, horizontal asymptotes, and
the x-intercepts of the function f(x).
Solution
Step 1: Find Vertical Asymptotes Vertical asymptotes occur when the de-
nominator of the rational function is equal to zero. So, to find the vertical
asymptotes, we set the denominator 2x2−3x−2equal to zero and solve for x.
2x2−3x−2 = 0
Factor the quadratic:
(x−2)(2x+ 1) = 0
Solve for x:x−2 = 0 =⇒x= 2 or 2x+ 1 = 0 =⇒x=−1
2
Therefore, the vertical asymptotes are x= 2 and x=−1
2.
Step 2: Find Horizontal Asymptotes To find the horizontal asymptotes,
we need to compare the degrees of the numerator and denominator. Since the
degrees are the same (both are 2), we look at the leading terms of the numerator
and denominator. The horizontal asymptote can be found by dividing the lead-
ing coefficient of the numerator by the leading coefficient of the denominator.
y=4
2= 2
Therefore, the horizontal asymptote is y= 2.
6
Step 3: Find X-Intercepts To find the x-intercepts, we set the numerator
4x2+ 5x−6equal to zero and solve for x.
4x2+ 5x−6 = 0
This can be factored into:
(4x−3)(x+ 2) = 0
Solve for x:4x−3 = 0 =⇒x=3
4or x+ 2 = 0 =⇒x=−2
Therefore, the x-intercepts are (3
4,0)and (−2,0).
Question 10
Question
Let f(x) = 3x2−2x−8
2x2−7x−4. Find the vertical asymptotes of the function f(x).
Solution
Step 1: To find the vertical asymptotes of f(x), we need to look for the values of
xthat make the denominator of f(x)equal to 0, since division by 0 is undefined.
Step 2: Set the denominator 2x2−7x−4equal to 0 and solve for x:
2x2−7x−4 = 0
Step 3: This is a quadratic equation that can be factored or solved using the
quadratic formula. Factoring gives:
(2x+ 1)(x−4) = 0
which gives solutions x=−1/2and x= 4.
Step 4: Therefore, the vertical asymptotes of the function f(x)are x=−1/2
and x= 4. These are the values of xfor which the function f(x)is not defined
due to division by zero in the denominator.
Question 11
Question
Simplify the rational expression:
3x3+x2−4x
x2−4
7
Solution
Step 1: Factor out any common factors in the numerator and denominator.
3x3+x2−4x
x2−4=x(3x2+x−4)
(x+ 2)(x−2)
Step 2: Factor the quadratic expression in the numerator.
3x2+x−4 = 3x2+ 4x−3x−4 = x(3x+ 4) −1(3x+ 4) = (x−1)(3x+ 4)
Step 3: Replace the factored numerator into the expression.
x(3x2+x−4)
(x+ 2)(x−2) =x(x−1)(3x+ 4)
(x+ 2)(x−2)
Thus, the simplified form of the rational expression is x(x−1)(3x+4)
(x+2)(x−2) .
Question 12
Question
Simplify the rational function:
3x2−9x
6x2−18
Solution
Step 1: Factor out the greatest common factor in the numerator and the de-
nominator. Step 2: Simplify by canceling out common factors.
Step 1: Factor out the greatest common factor in the numerator and the
denominator: 3x2−9x
6x2−18 =3x(x−3)
6(x2−3)
Step 2: Simplify by canceling out common factors:
3x(x−3)
6(x2−3) =(x−3)
2(x2−3)
Therefore, the simplified form of the rational function is x−3
2(x2−3) .
Question 13
Question
Find the domain of the rational function:
f(x) = x2−4
x2−5x+ 6
8
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the function is defined. The function is defined for all x
values except where the denominator is equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−5x+ 6 = 0
Step 3: Factor the quadratic equation:
(x−2)(x−3) = 0
Step 4: Solve for x:
x−2 = 0 ⇒x= 2
x−3 = 0 ⇒x= 3
Step 5: The domain of the function f(x)is all real numbers except x= 2
and x= 3. Therefore, the domain is:
x∈R, x = 2, x = 3
Question 14
Question
Simplify the rational function:
f(x) = 3x2−x−2
x2−3x−4
Solution
Step 1: Factor the numerator and the denominator.
We have:
f(x) = 3x2−x−2
x2−3x−4
Factoring the numerator:
3x2−x−2 = (3x+ 2)(x−1)
Factoring the denominator:
x2−3x−4 = (x−4)(x+ 1)
Step 2: Rewrite the rational function with the factored numerator and de-
nominator.
9
Therefore,
f(x) = (3x+ 2)(x−1)
(x−4)(x+ 1)
Step 3: Simplify the expression as far as possible.
Thus, the simplified form of the rational function is:
f(x) = 3x+ 2
x−4
Question 15
Question
Let f(x) = 4x2+7x+2
x2−3x−10 . Find the equations of the horizontal and vertical asymp-
totes of the function f(x).
Solution
Step 1: Determine the vertical asymptotes by finding the values of xthat make
the denominator equal to zero. Set x2−3x−10 = 0 and solve for x.
x2−3x−10 = 0
(x−5)(x+ 2) = 0
x= 5,−2
Therefore, the vertical asymptotes are x= 5 and x=−2.
Step 2: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator. Since the degrees are the same, the horizontal asymptote
is the ratio of the leading coefficients. Divide the leading coefficients of the nu-
merator and denominator.
lim
x→±∞
4x2+ 7x+ 2
x2−3x−10 = lim
x→±∞
4
1= 4
Therefore, the equation of the horizontal asymptote is y= 4.
In conclusion, the function f(x)has vertical asymptotes at x= 5 and x=
−2, and a horizontal asymptote at y= 4.
Question 16
Question
Simplify the following rational expression:
4x3−8x2+ 4x
2x2−2x
10
Solution
Step 1: Factor out any common terms in the numerator and the denominator:
4x3−8x2+ 4x
2x2−2x=4x(x2−2x+ 1)
2x(x−1)
Step 2: Simplify the expression by canceling out common factors:
4x(x2−2x+ 1)
2x(x−1) =4(x2−2x+ 1)
2(x−1) =4(x−1)2
2(x−1)
Step 3: Further simplify by canceling out common factors:
4(x−1)2
2(x−1) =4(x−1)(x−1)
2(x−1) =4(x−1)
2= 2(x−1)
Therefore, the simplified form of the given rational expression is 2(x−1).
Question 17
Question
Solve the following rational inequality for x:
2x2−3x−2
x−1>x−2
x+ 3
Solution
Step 1: Identify the domain of the rational inequality by finding the values of x
that make the denominators equal to zero.
For the given rational inequality, the denominators are x−1and x+ 3.
Setting each denominator equal to zero gives us:
x−1 = 0 ⇒x= 1 and x+ 3 = 0 ⇒x=−3
So, the domain of the rational inequality is all real numbers except x= 1
and x=−3.
Step 2: Determine the critical points of the rational inequality by finding
the values of xthat make the numerator and denominators equal to zero.
The numerator and denominators of the rational inequality are 2x2−3x−2,
x−1, and x+ 3. Setting each expression equal to zero gives us:
2x2−3x−2 = 0 ⇒(2x+ 1)(x−2) = 0
x−1 = 0 ⇒x= 1
x+ 3 = 0 ⇒x=−3
The critical points are x=−3,x=−1
2, and x= 1.
Step 3: Create a sign chart to test the intervals determined by the critical
points.
11
| | −3|−1
2| 1 | +∞| | |—|——|———–|—–|———|—-| | | | | | | | | | | | | | |
|||||||
Step 4: Plug in a test value from each interval into the rational inequality
and determine the sign of the expression.
For x < −3, pick x=−4:
2(−4)2−3(−4) −2
−4−1>−4−2
−4+3
Simplifying, we get:
42
−5>−6
This is false, so x < −3is not part of the solution.
Step 5: Continue testing the other intervals until you have determined the
solution set.
Can you continue from here to determine the solution set for the rational
inequality?
Question 18
Question
Simplify the following rational expression:
3x2−2x−5
4x2+ 3x−7∇ · 2x2+ 3x+ 1
2x2−5x−3
Solution
To simplify the given rational expression, we will follow these steps:
Step 1: Write the division as a multiplication by taking the reciprocal of
the divisor.
3x2−2x−5
4x2+ 3x−7∇ · 2x2+ 3x+ 1
2x2−5x−3=3x2−2x−5
4x2+ 3x−7·2x2−5x−3
2x2+ 3x+ 1
Step 2: Factor all quadratic expressions.
3x2−2x−5 = (3x+ 1)(x−5)
4x2+ 3x−7 = (4x−7)(x+ 1)
2x2+ 3x+ 1 = (2x+ 1)(x+ 1)
2x2−5x−3 = (2x+ 1)(x−3)
12
Step 3: Substitute the factored expressions back into the initial expression
and simplify.
(3x+ 1)(x−5)
(4x−7)(x+ 1) ·(2x−5)(x−3)
(2x+ 1)(x+ 1) =(3x+ 1)(x−5)(2x−5)(x−3)
(4x−7)(x+ 1)(2x+ 1)(x+ 1)
=(3x2−14x−5)(2x2−11x+ 15)
(4x2−3x−7)(2x2+ 3x+ 1)
Therefore, the simplified form of the given rational expression is (3x2−14x−5)(2x2−11x+15)
(4x2−3x−7)(2x2+3x+1) .
Question 19
Question
Simplify the following rational expression:
5x2−4x−5
x2−3x−4∇ · 3x2−5x−2
x2−5x+ 6
Solution
To simplify the given rational expression, we will follow these steps:
Step 1: Factor the numerators and denominators of the fractions.
5x2−4x−5
x2−3x−4∇ · 3x2−5x−2
x2−5x+ 6 =(5x+ 1)(x−5)
(x−4)(x+ 1) ∇ · (3x+ 1)(x−2)
(x−3)(x−2)
Step 2: Rewrite the division as multiplication by the reciprocal of the second
fraction.
(5x+ 1)(x−5)
(x−4)(x+ 1) ×(x−3)(x−2)
(3x+ 1)(x−2) =(5x+ 1)(x−5)
(x−4)(x+ 1) ×(x−3)
(3x+ 1)
Step 3: Multiply the fractions by multiplying the numerators together and
the denominators together.
(5x+ 1)(x−5)(x−3)
(x−4)(x+ 1)(3x+ 1) =(5x2−25x+x−5)(x−3)
(x−4)(x+ 1)(3x+ 1)
=(5x2−24x−5)(x−3)
(x−4)(x+ 1)(3x+ 1)
Step 4: Expand and simplify the numerator.
(5x2−24x−5)(x−3) = 5x3−24x2−5x−15x2+ 72x+ 15
= 5x3−39x2+ 67x+ 15
Therefore, the simplified expression is:
5x3−39x2+ 67x+ 15
(x−4)(x+ 1)(3x+ 1)
13
Question 20
Question
Solve the following rational inequality and express your answer in interval no-
tation:
(x−3)(x+ 2) ≥0
Solution
Step 1: Find the critical points by setting the expression equal to zero and
solving for x.
(x−3)(x+ 2) = 0
This gives x= 3 and x=−2as critical points.
Step 2: Create intervals using the critical points (x= 3 and x=−2). We
have three intervals to consider: (−∞,−2),(−2,3), and (3,∞).
Step 3: Test a value in each interval to determine the sign of (x−3)(x+ 2).
• For x=−3:(−3−3)(−3 + 2) = (−6)(−1) = 6 >0, so this interval is
positive.
• For x= 0:(0 −3)(0 + 2) = (−3)(2) = −6<0, so this interval is negative.
• For x= 4:(4 −3)(4 + 2) = (1)(6) = 6 >0, so this interval is positive.
Step 4: Identify the solution by considering where the inequality is satisfied,
i.e., where the expression is greater than or equal to zero. The solution is the
combination of intervals where (x−3)(x+ 2) ≥0:
(−∞,−2] ∪[3,∞).
Question 21
Question
Given the rational function f(x) = 2x2+5x−3
x2−4x−5, find the vertical asymptotes of
f(x).
Solution
Step 1: To find the vertical asymptotes of f(x), we need to determine the values
of xfor which the denominator of the rational function is equal to 0.
Step 2: Set the denominator x2−4x−5equal to 0and solve for x:
x2−4x−5 = 0
(x−5)(x+ 1) = 0
14
Step 3: Setting each factor to zero gives the solutions:
x−5 = 0 or x+ 1 = 0
x= 5 or x=−1
Step 4: Therefore, the vertical asymptotes of the function f(x)are x= 5
and x=−1.
Question 22
Question
Find the domain of the rational function:
f(x) = 2x+ 5
x2−9
Solution
Step 1: The domain of a rational function is all real numbers except for the
values of xthat make the denominator equal to zero since division by zero is
undefined. Step 2: Set the denominator of f(x)equal to zero and solve for x.
x2−9 = 0
x2= 9
x=±3
Step 3: The values of xthat make the denominator equal to zero are x= 3 and
x=−3. Step 4: Therefore, the domain of the function f(x) = 2x+ 5
x2−9is all
real numbers except for x= 3 and x=−3. In interval notation, the domain is
(−∞,−3) ∪(−3,3) ∪(3,∞).
Question 23
Question
Let f(x) = 5x2−10x
x2+3x−4. Determine the equations of the vertical asymptotes, hor-
izontal asymptotes, and oblique asymptotes (if any) of the rational function
f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator of the rational
function equal to zero and solve for x.
Denominator: x2+ 3x−4 = 0
(x+ 4)(x−1) = 0
15
Step 2: The vertical asymptotes occur at the values of xwhere the denom-
inator is zero. Thus, the equations of the vertical asymptotes are x=−4and
x= 1.
Step 3: To determine the horizontal asymptote, compare the degrees of the
numerator and denominator. Since the degree of the numerator is 2 and the
degree of the denominator is also 2, we look at the leading coefficients. The
equation of the horizontal asymptote is the ratio of the leading coefficients:
y=5
1= 5
Step 4: To check for an oblique asymptote, divide the numerator by the
denominator using long division or synthetic division.
x+ 4 5x2−10x5x
5x2+ 20x
−30x
−30x+ 0
Step 5: The quotient is 5and the remainder is −30x. Thus, the oblique
asymptote is the line y= 5x−30.
Therefore, the equations of the asymptotes for the rational function f(x)
are: Vertical asymptotes: x=−4and x= 1 Horizontal asymptote: y= 5
Oblique asymptote: y= 5x−30
Question 24
Question
Find the domain of the rational function defined by
f(x) = 2x2−5x−3
x2−4x+ 3
.
Solution
Step 1: Set the denominator equal to zero and solve for xto find the values that
make the function undefined.
x2−4x+ 3 = 0
(x−1)(x−3) = 0
x= 1 or x= 3
Step 2: The function is undefined for x= 1 and x= 3, so the domain of
f(x)is all real numbers except x= 1 and x= 3. Therefore, the domain of the
function is (−∞,1) ∪(1,3) ∪(3,∞).
16
Question 25
Question
For the rational function
f(x) = 2x2+x−3
x2+ 2x−8,
find the vertical asymptotes, horizontal asymptotes, and any holes in the graph.
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero
and solving for x.
x2+ 2x−8 = 0
(x+ 4)(x−2) = 0
This gives us x=−4and x= 2 as the vertical asymptotes.
Step 2: Find the horizontal asymptote by comparing the degrees of the
numerator and denominator. Since the degree of the numerator and denomi-
nator are the same, we compare the leading coefficients to find the horizontal
asymptote.
lim
x→∞
2x2+x−3
x2+ 2x−8= lim
x→∞
2 + 1
x−3
x2
1 + 2
x−8
x2
=2
1
= 2
Therefore, the horizontal asymptote is y= 2.
Step 3: Find any holes in the graph by factoring and canceling out common
factors in the numerator and denominator. Since 2x2+x−3does not factor
nicely, we can use polynomial long division or synthetic division to divide it by
x2+ 2x−8. Performing long division, we get:
2x+5 −3
x2+ 2x−8 2x2+x−3
2x2+4x−16
(−) (−3x) (+24)
x8
Therefore, we have f(x) = 2x2+x−3
x2+2x−8= 2x+5+ x+8
x2+2x−8. The hole in the graph
occurs at the point (2,9).
Question 26
Question
Let f(x) = 4x2−3x−4
x2−2x−3. Find the domain of f(x).
17
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that make the denominator equal to zero, as division by zero is undefined.
Therefore, we need to find the values of xthat make the denominator x2−2x−3
equal to zero.
Step 2: To find these values, we solve the equation x2−2x−3 = 0.
Step 3: Factoring the quadratic equation, we get (x−3)(x+ 1) = 0.
Step 4: Setting each factor to zero gives x−3 = 0 or x+ 1 = 0.
Step 5: Solving x−3 = 0 yields x= 3.
Step 6: Solving x+ 1 = 0 yields x=−1.
Step 7: Therefore, the values x= 3 and x=−1make the denominator of
f(x)equal to zero, and they are not in the domain of f(x).
Step 8: Thus, the domain of the function f(x)is all real numbers except
x= 3 and x=−1, which can be expressed as (−∞,−1) ∪(−1,3) ∪(3,∞).
Question 27
Question
Let f(x) = 2x2−x−1
x2−4x+3 . Determine the equations of the vertical and horizontal
asymptotes of the graph of y=f(x).
Solution
Step 1: To find the vertical asymptotes, we need to determine where the denomi-
nator of the function, x2−4x+3, becomes zero. This occurs when x2−4x+3 = 0.
Factoring the quadratic, we get (x−1)(x−3) = 0. Therefore, the vertical
asymptotes are x= 1 and x= 3.
Step 2: To find the horizontal asymptote, we compare the degrees of the
numerator and denominator of the function. Since the degrees are the same,
we look at the leading coefficients. The horizontal asymptote is the ratio of
the leading coefficients, which is 2/1=2. Therefore, y= 2 is the horizontal
asymptote.
Question 28
Question
Find the domain of the rational function
f(x) = x2+ 5x+ 6
x2−4x−12.
18
Solution
Step 1: We first need to find the restrictions on the domain of the function.
The function f(x)is defined for all real numbers except where the denominator
is equal to zero, since division by zero is undefined. Step 2: We find where the
denominator x2−4x−12 is equal to zero by solving the equation:
x2−4x−12 = 0.
Step 3: Factor the quadratic equation:
(x−6)(x+ 2) = 0.
Step 4: Set each factor to zero and solve for x:
x−6 = 0 or x+ 2 = 0.
Step 5: Solving x−6 = 0 gives x= 6 and solving x+ 2 = 0 gives x=−2. Step
6: Therefore, the restrictions on the domain are x= 6 and x=−2. Step 7:
The domain of the rational function f(x)is all real numbers except x= 6 and
x=−2, so the domain is (−∞,−2) ∪(−2,6) ∪(6,∞).
Question 29
Question
Solve the following rational equation for x:
1
x−1−2
x+ 2 =3
x2+x−2
Solution
Step 1: To begin, let’s find a common denominator for all the fractions on both
sides of the equation. The common denominator could be (x−1)(x+ 2)(x+ 1),
but since the third fraction on the right already has x2+x−2in its denominator,
we can simplify to (x−1)(x+ 2). Let’s rewrite the equation with this common
denominator: (x+ 2) −2(x−1)
(x−1)(x+ 2) =3
x2+x−2
Step 2: Simplify the left side of the equation:
x+ 2 −2x+ 2
(x−1)(x+ 2) =3
x2+x−2
−x+ 4
(x−1)(x+ 2) =3
x2+x−2
19
Step 3: Now, express each side as a single fraction:
−x+ 4
(x−1)(x+ 2) =3
(x−1)(x+ 2)
Step 4: Cross multiply to eliminate the denominators:
(−x+ 4)(x2+x−2) = 3(x−1)(x+ 2)
Step 5: Expand both sides of the equation:
−x3+x2+ 2x+ 4x2−4x−8 = 3(x2+x−2)
−x3+ 5x2−2x−8 = 3x2+ 3x−6
Step 6: Combine like terms on both sides:
−x3+ 5x2−2x−8 = 3x2+ 3x−6
−x3+ 2x2−5x−2 = 0
Step 7: Now we have a cubic equation that we need to solve. While there
are algebraic methods for finding the roots of cubic equations, in this case
the roots are not nice. Therefore, we will leave the answer as the equation
−x3+ 2x2−5x−2 = 0.
Question 30
Question
Simplify the rational function:
f(x) = x2−4
x2−5x+ 6
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify by canceling
out common factors if possible. If not, proceed with the factored form.
Step 1: We need to factor the numerator and denominator of the rational
function f(x):
The numerator x2−4can be factored as a difference of squares: x2−4 =
(x+ 2)(x−2).
The denominator x2−5x+6 can be factored as: x2−5x+6 = (x−2)(x−3).
Therefore, the rational function f(x)can be rewritten as:
f(x) = (x+ 2)(x−2)
(x−2)(x−3)
20
Step 2: Now we simplify the function by canceling out the common factor
(x−2) in the numerator and denominator.
f(x) = x+ 2
x−3
So, the simplified form of the rational function f(x)is x+2
x−3.
Question 31
Question
Find the domain of the rational function f(x) = x2−5x+ 6
x2−4x−5.
Solution
Step 1: To find the domain of the rational function, we need to identify the
values of xthat would make the denominator equal to zero since division by
zero is undefined.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x:
x−5 = 0 =
⇒x= 5
x+ 1 = 0 =⇒x=−1
Step 5: The values x= 5 and x=−1would make the denominator equal
to zero. Therefore, the domain of the rational function f(x)is all real numbers
except x= 5 and x=−1.
Step 6: Thus, the domain of the function f(x) = x2−5x+ 6
x2−4x−5is (−∞,−1) ∪
(−1,5) ∪(5,∞).
Question 32
Question
Simplify the following rational expression:
3x2−7x−6
2x2+ 5x−3
21
Solution
Step 1: Factor both the numerator and denominator. Step 2: Find the common
factors to simplify the expression.
Step 1:
Factor the numerator 3x2−7x−6: The factors of 3×(−6) = −18 that add up
to −7are −9and 2. So, we can write 3x2−7x−6as (3x+ 2)(x−3).
Factor the denominator 2x2+ 5x−3: The factors of 2×(−3) = −6that add
up to 5are 6and −1. So, we can write 2x2+ 5x−3as (2x−1)(x+ 3).
Thus, the expression becomes:
(3x+ 2)(x−3)
(2x−1)(x+ 3)
Step 2:
Now, simplify by cancelling out the common factors:
(3x+ 2)(x−3)
(2x−1)(x+ 3)
Therefore, the simplified form of the given rational expression is:
x−3
2x−1
Question 33
Question
Solve the rational equation: 4
x+3 −2
x−1=5
(x+3)(x−1) .
Solution
Step 1: Find a common denominator for all terms in the equation. Step 2:
Multiply every term by the common denominator to eliminate the fractions.
Step 3: Simplify the equation and solve for x. Step 4: Check for extraneous
solutions.
Step 1: Find a common denominator The common denominator for all
terms in the equation is (x+ 3)(x−1).
Step 2: Multiply every term by the common denominator Multiplying every
term by the common denominator, we get:
(x+ 3)(x−1) (4
x+ 3 −2
x−1)= (x+ 3)(x−1) (5
(x+ 3)(x−1))
4(x−1) −2(x+ 3) = 5
22
Step 3: Simplify the equation and solve for xSolving the simplified equa-
tion:
4x−4−2x−6 = 5
2x−10 = 5
2x= 15
x=15
2
Step 4: Check for extraneous solutions Check if x=15
2leads to any de-
nominator being equal to zero. Checking x=15
2in the original equation, we
find that x=15
2does not make any denominator zero.
Therefore, the solution to the equation is x=15
2.
Question 34
Question
Solve the rational inequality: 5
x−3≥3.
Solution
Step 1: Begin by setting up the inequality:
5
x−3≥3
Step 2: Multiply both sides of the inequality by (x−3) to remove the fraction:
5≥3(x−3)
Step 3: Distribute on the right side of the inequality:
5≥3x−9
Step 4: Add 9 to both sides of the inequality:
14 ≥3x
Step 5: Divide by 3 to solve for x:
14
3≥x
Step 6: Thus, the solution to the rational inequality is x≤14
3.
23
Question 35
Question
Simplify the rational function:
f(x) = 3x2−5x−2
2x2+ 7x−4
Solution
Step 1: Factor the numerator and the denominator. Step 2: Simplify by can-
celing out common factors.
Step 1: Factor the numerator and denominator of the rational function:
f(x) = 3x2−5x−2
2x2+ 7x−4
Numerator:
3x2−5x−2 = (3x+ 1)(x−2)
Denominator:
2x2+ 7x−4 = (2x−1)(x+ 4)
Therefore, the rational function can be expressed as:
f(x) = (3x+ 1)(x−2)
(2x−1)(x+ 4)
Step 2: Simplify the rational function by canceling out common factors:
f(x) = (3x+ 1)(x−2)
(2x−1)(x+ 4) =3x+ 1
2x−1·x−2
x+ 4
Thus, the simplified form of the rational function f(x)is:
f(x) = 3x+ 1
2x−1·x−2
x+ 4
24
Question 2
Question
Consider the rational function f(x) = 2x2−7x−30
x2−5x−6. Find the domain of the
function f(x).
Solution
Step 1: To find the domain of a rational function, we need to identify any
values of xthat would make the denominator zero, as these values would result
in division by zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−5x−6 = 0
Step 3: Factor the quadratic equation:
(x−6)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x:
x−6 = 0 =⇒x= 6
x+ 1 = 0 =⇒x=−1
Step 5: Therefore, the values x= 6 and x=−1make the denominator equal
to zero, so they must be excluded from the domain.
Step 6: The domain of the rational function f(x) = 2x2−7x−30
x2−5x−6is all real num-
bers except x= 6 and x=−1, so the domain is (−∞,−1) ∪(−1,6) ∪(6,∞).
Question 3
Question
Given the rational function f(x) = 2x2+5x−3
x2−4x−5, find the horizontal asymptotes (if
any) of the function.
Solution
Step 1: To find the horizontal asymptotes of a rational function, we compare
the degrees of the numerator and denominator polynomials. The degree of the
numerator is 2 and the degree of the denominator is 2, so we must consider the
leading coefficients of both polynomials.
Step 2: The leading coefficient of the numerator is 2, and the leading coeffi-
cient of the denominator is 1. Since the degrees of the two polynomials are equal,
we can find the horizontal asymptote by comparing their leading coefficients.
Step 3: The horizontal asymptote is given by the ratio of the leading coeffi-
cients. Therefore, the horizontal asymptote of the function is y=2
1= 2.
2
Question 4
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes of the ra-
tional function:
f(x) = x2−3x
x2−4
Solution
To find the vertical asymptotes of a rational function, we need to identify the
values of xthat make the denominator equal to zero, but not the numerator.
This will result in division by zero.
Step 1: Find Vertical Asymptotes Setting the denominator equal to
zero and solving for x:
x2−4 = 0
(x+ 2)(x−2) = 0
x=−2or x= 2
So, there are vertical asymptotes at x=−2and x= 2.
Step 2: Find the Horizontal Asymptotes To find horizontal asymp-
totes, we compare the degrees of the numerator and denominator of the function.
Since the degrees of the numerator and denominator are the same (2), the
horizontal asymptote is the ratio of the leading coefficients. The horizontal
asymptote is y=1
1= 1.
Step 3: Find any Holes To find any holes in the graph, we simplify the
function by canceling any common factors between the numerator and denomi-
nator.
f(x) = x(x−3)
(x+ 2)(x−2)
Notice that there is a common factor of xin the numerator and x−2in the
denominator, which gives us a hole at x= 0.
Therefore, the function f(x)has vertical asymptotes at x=−2and x= 2,
a horizontal asymptote at y= 1, and a hole at x= 0.
Question 5
Question
Simplify the following rational expression:
6x3−12x2+ 6x
2x2−4x
3
Solution
Step 1: Factor out the common terms in the numerator and denominator.
6x(x2−2x+ 1)
2x(x−2)
Step 2: Simplify the expression by canceling out the common factors.
6(x2−2x+ 1)
2(x−2)
Step 3: Further simplify the expression by factoring the quadratic in the
numerator. 6(x−1)2
2(x−2)
Step 4: Simplify by canceling out the common factors.
6(x−1)
2
Step 5: Finally, simplify the expression to obtain the simplified rational
expression.
3(x−1)
Question 6
Question
Find the domain of the rational function given by:
f(x) = x2−4
x2−3x−4
Solution
Step 1: To find the domain of the rational function, we need to determine the
values of xfor which the function is defined. The function is defined as long as
the denominator is not equal to zero, since division by zero is undefined.
Step 2: We start by factoring the denominator, x2−3x−4, to determine its
zeros:
x2−3x−4 = (x−4)(x+ 1)
Setting this expression equal to zero and solving for x, we get:
x−4 = 0 or x+ 1 = 0
x= 4 or x=−1
4
Step 3: Therefore, the function f(x)is undefined at x= 4 and x=−1since
these are the values that make the denominator zero.
Step 4: The domain of the function is all real numbers except x= 4 and
x=−1. So, the domain can be expressed in interval notation as:
(−∞,−1) ∪(−1,4) ∪(4,∞)
Question 7
Question
Find the domain of the rational function:
f(x) = 3x−1
x2−4x+ 3
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that make the denominator equal to zero, since division by zero is unde-
fined.
Step 2: To find the domain of f(x), we need to determine the values of x
that make the denominator x2−4x+ 3 equal to zero.
Step 3: We will solve the equation x2−4x+ 3 = 0 to find these values. This
can be factored as (x−3)(x−1) = 0.
Step 4: Setting each factor to zero gives us x−3 = 0 and x−1 = 0. Solving
these equations gives x= 3 and x= 1.
Step 5: Therefore, the values x= 1 and x= 3 make the denominator of the
function equal to zero and must be excluded from the domain.
Step 6: In interval notation, the domain of the function f(x) = 3x−1
x2−4x+3 is
(−∞,1) ∪(1,3) ∪(3,∞).
Question 8
Question
Find the domain of the rational function:
f(x) = 2x+ 1
x2−9
Solution
Step 1: The domain of a rational function consists of all real numbers except
for the values of xthat make the denominator equal to zero since division by
zero is undefined. So, we need to find the values of xthat make x2−9 = 0.
5
Step 2: We can factor the denominator x2−9using the difference of squares
formula: a2−b2= (a+b)(a−b). So, x2−9 = (x+ 3)(x−3).
Step 3: Setting x2−9 = 0 gives us:
(x+ 3)(x−3) = 0
Step 4: Solving for x, we get:
x+ 3 = 0 or x−3 = 0
x=−3or x= 3
Step 5: Therefore, the domain of the function f(x) = 2x+1
x2−9is all real numbers
except x=−3and x= 3, so the domain is:
(−∞,−3) ∪(−3,3) ∪(3,∞)
Question 9
Question
Let f(x) = 4x2+5x−6
2x2−3x−2. Find the vertical asymptotes, horizontal asymptotes, and
the x-intercepts of the function f(x).
Solution
Step 1: Find Vertical Asymptotes Vertical asymptotes occur when the de-
nominator of the rational function is equal to zero. So, to find the vertical
asymptotes, we set the denominator 2x2−3x−2equal to zero and solve for x.
2x2−3x−2 = 0
Factor the quadratic:
(x−2)(2x+ 1) = 0
Solve for x:x−2 = 0 =⇒x= 2 or 2x+ 1 = 0 =⇒x=−1
2
Therefore, the vertical asymptotes are x= 2 and x=−1
2.
Step 2: Find Horizontal Asymptotes To find the horizontal asymptotes,
we need to compare the degrees of the numerator and denominator. Since the
degrees are the same (both are 2), we look at the leading terms of the numerator
and denominator. The horizontal asymptote can be found by dividing the lead-
ing coefficient of the numerator by the leading coefficient of the denominator.
y=4
2= 2
Therefore, the horizontal asymptote is y= 2.
6
Step 3: Find X-Intercepts To find the x-intercepts, we set the numerator
4x2+ 5x−6equal to zero and solve for x.
4x2+ 5x−6 = 0
This can be factored into:
(4x−3)(x+ 2) = 0
Solve for x:4x−3 = 0 =⇒x=3
4or x+ 2 = 0 =⇒x=−2
Therefore, the x-intercepts are (3
4,0)and (−2,0).
Question 10
Question
Let f(x) = 3x2−2x−8
2x2−7x−4. Find the vertical asymptotes of the function f(x).
Solution
Step 1: To find the vertical asymptotes of f(x), we need to look for the values of
xthat make the denominator of f(x)equal to 0, since division by 0 is undefined.
Step 2: Set the denominator 2x2−7x−4equal to 0 and solve for x:
2x2−7x−4 = 0
Step 3: This is a quadratic equation that can be factored or solved using the
quadratic formula. Factoring gives:
(2x+ 1)(x−4) = 0
which gives solutions x=−1/2and x= 4.
Step 4: Therefore, the vertical asymptotes of the function f(x)are x=−1/2
and x= 4. These are the values of xfor which the function f(x)is not defined
due to division by zero in the denominator.
Question 11
Question
Simplify the rational expression:
3x3+x2−4x
x2−4
7
Solution
Step 1: Factor out any common factors in the numerator and denominator.
3x3+x2−4x
x2−4=x(3x2+x−4)
(x+ 2)(x−2)
Step 2: Factor the quadratic expression in the numerator.
3x2+x−4 = 3x2+ 4x−3x−4 = x(3x+ 4) −1(3x+ 4) = (x−1)(3x+ 4)
Step 3: Replace the factored numerator into the expression.
x(3x2+x−4)
(x+ 2)(x−2) =x(x−1)(3x+ 4)
(x+ 2)(x−2)
Thus, the simplified form of the rational expression is x(x−1)(3x+4)
(x+2)(x−2) .
Question 12
Question
Simplify the rational function:
3x2−9x
6x2−18
Solution
Step 1: Factor out the greatest common factor in the numerator and the de-
nominator. Step 2: Simplify by canceling out common factors.
Step 1: Factor out the greatest common factor in the numerator and the
denominator: 3x2−9x
6x2−18 =3x(x−3)
6(x2−3)
Step 2: Simplify by canceling out common factors:
3x(x−3)
6(x2−3) =(x−3)
2(x2−3)
Therefore, the simplified form of the rational function is x−3
2(x2−3) .
Question 13
Question
Find the domain of the rational function:
f(x) = x2−4
x2−5x+ 6
8
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the function is defined. The function is defined for all x
values except where the denominator is equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−5x+ 6 = 0
Step 3: Factor the quadratic equation:
(x−2)(x−3) = 0
Step 4: Solve for x:
x−2 = 0 ⇒x= 2
x−3 = 0 ⇒x= 3
Step 5: The domain of the function f(x)is all real numbers except x= 2
and x= 3. Therefore, the domain is:
x∈R, x = 2, x = 3
Question 14
Question
Simplify the rational function:
f(x) = 3x2−x−2
x2−3x−4
Solution
Step 1: Factor the numerator and the denominator.
We have:
f(x) = 3x2−x−2
x2−3x−4
Factoring the numerator:
3x2−x−2 = (3x+ 2)(x−1)
Factoring the denominator:
x2−3x−4 = (x−4)(x+ 1)
Step 2: Rewrite the rational function with the factored numerator and de-
nominator.
9
Therefore,
f(x) = (3x+ 2)(x−1)
(x−4)(x+ 1)
Step 3: Simplify the expression as far as possible.
Thus, the simplified form of the rational function is:
f(x) = 3x+ 2
x−4
Question 15
Question
Let f(x) = 4x2+7x+2
x2−3x−10 . Find the equations of the horizontal and vertical asymp-
totes of the function f(x).
Solution
Step 1: Determine the vertical asymptotes by finding the values of xthat make
the denominator equal to zero. Set x2−3x−10 = 0 and solve for x.
x2−3x−10 = 0
(x−5)(x+ 2) = 0
x= 5,−2
Therefore, the vertical asymptotes are x= 5 and x=−2.
Step 2: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator. Since the degrees are the same, the horizontal asymptote
is the ratio of the leading coefficients. Divide the leading coefficients of the nu-
merator and denominator.
lim
x→±∞
4x2+ 7x+ 2
x2−3x−10 = lim
x→±∞
4
1= 4
Therefore, the equation of the horizontal asymptote is y= 4.
In conclusion, the function f(x)has vertical asymptotes at x= 5 and x=
−2, and a horizontal asymptote at y= 4.
Question 16
Question
Simplify the following rational expression:
4x3−8x2+ 4x
2x2−2x
10
Solution
Step 1: Factor out any common terms in the numerator and the denominator:
4x3−8x2+ 4x
2x2−2x=4x(x2−2x+ 1)
2x(x−1)
Step 2: Simplify the expression by canceling out common factors:
4x(x2−2x+ 1)
2x(x−1) =4(x2−2x+ 1)
2(x−1) =4(x−1)2
2(x−1)
Step 3: Further simplify by canceling out common factors:
4(x−1)2
2(x−1) =4(x−1)(x−1)
2(x−1) =4(x−1)
2= 2(x−1)
Therefore, the simplified form of the given rational expression is 2(x−1).
Question 17
Question
Solve the following rational inequality for x:
2x2−3x−2
x−1>x−2
x+ 3
Solution
Step 1: Identify the domain of the rational inequality by finding the values of x
that make the denominators equal to zero.
For the given rational inequality, the denominators are x−1and x+ 3.
Setting each denominator equal to zero gives us:
x−1 = 0 ⇒x= 1 and x+ 3 = 0 ⇒x=−3
So, the domain of the rational inequality is all real numbers except x= 1
and x=−3.
Step 2: Determine the critical points of the rational inequality by finding
the values of xthat make the numerator and denominators equal to zero.
The numerator and denominators of the rational inequality are 2x2−3x−2,
x−1, and x+ 3. Setting each expression equal to zero gives us:
2x2−3x−2 = 0 ⇒(2x+ 1)(x−2) = 0
x−1 = 0 ⇒x= 1
x+ 3 = 0 ⇒x=−3
The critical points are x=−3,x=−1
2, and x= 1.
Step 3: Create a sign chart to test the intervals determined by the critical
points.
11
| | −3|−1
2| 1 | +∞| | |—|——|———–|—–|———|—-| | | | | | | | | | | | | | |
|||||||
Step 4: Plug in a test value from each interval into the rational inequality
and determine the sign of the expression.
For x < −3, pick x=−4:
2(−4)2−3(−4) −2
−4−1>−4−2
−4+3
Simplifying, we get:
42
−5>−6
This is false, so x < −3is not part of the solution.
Step 5: Continue testing the other intervals until you have determined the
solution set.
Can you continue from here to determine the solution set for the rational
inequality?
Question 18
Question
Simplify the following rational expression:
3x2−2x−5
4x2+ 3x−7∇ · 2x2+ 3x+ 1
2x2−5x−3
Solution
To simplify the given rational expression, we will follow these steps:
Step 1: Write the division as a multiplication by taking the reciprocal of
the divisor.
3x2−2x−5
4x2+ 3x−7∇ · 2x2+ 3x+ 1
2x2−5x−3=3x2−2x−5
4x2+ 3x−7·2x2−5x−3
2x2+ 3x+ 1
Step 2: Factor all quadratic expressions.
3x2−2x−5 = (3x+ 1)(x−5)
4x2+ 3x−7 = (4x−7)(x+ 1)
2x2+ 3x+ 1 = (2x+ 1)(x+ 1)
2x2−5x−3 = (2x+ 1)(x−3)
12
Step 3: Substitute the factored expressions back into the initial expression
and simplify.
(3x+ 1)(x−5)
(4x−7)(x+ 1) ·(2x−5)(x−3)
(2x+ 1)(x+ 1) =(3x+ 1)(x−5)(2x−5)(x−3)
(4x−7)(x+ 1)(2x+ 1)(x+ 1)
=(3x2−14x−5)(2x2−11x+ 15)
(4x2−3x−7)(2x2+ 3x+ 1)
Therefore, the simplified form of the given rational expression is (3x2−14x−5)(2x2−11x+15)
(4x2−3x−7)(2x2+3x+1) .
Question 19
Question
Simplify the following rational expression:
5x2−4x−5
x2−3x−4∇ · 3x2−5x−2
x2−5x+ 6
Solution
To simplify the given rational expression, we will follow these steps:
Step 1: Factor the numerators and denominators of the fractions.
5x2−4x−5
x2−3x−4∇ · 3x2−5x−2
x2−5x+ 6 =(5x+ 1)(x−5)
(x−4)(x+ 1) ∇ · (3x+ 1)(x−2)
(x−3)(x−2)
Step 2: Rewrite the division as multiplication by the reciprocal of the second
fraction.
(5x+ 1)(x−5)
(x−4)(x+ 1) ×(x−3)(x−2)
(3x+ 1)(x−2) =(5x+ 1)(x−5)
(x−4)(x+ 1) ×(x−3)
(3x+ 1)
Step 3: Multiply the fractions by multiplying the numerators together and
the denominators together.
(5x+ 1)(x−5)(x−3)
(x−4)(x+ 1)(3x+ 1) =(5x2−25x+x−5)(x−3)
(x−4)(x+ 1)(3x+ 1)
=(5x2−24x−5)(x−3)
(x−4)(x+ 1)(3x+ 1)
Step 4: Expand and simplify the numerator.
(5x2−24x−5)(x−3) = 5x3−24x2−5x−15x2+ 72x+ 15
= 5x3−39x2+ 67x+ 15
Therefore, the simplified expression is:
5x3−39x2+ 67x+ 15
(x−4)(x+ 1)(3x+ 1)
13
Question 20
Question
Solve the following rational inequality and express your answer in interval no-
tation:
(x−3)(x+ 2) ≥0
Solution
Step 1: Find the critical points by setting the expression equal to zero and
solving for x.
(x−3)(x+ 2) = 0
This gives x= 3 and x=−2as critical points.
Step 2: Create intervals using the critical points (x= 3 and x=−2). We
have three intervals to consider: (−∞,−2),(−2,3), and (3,∞).
Step 3: Test a value in each interval to determine the sign of (x−3)(x+ 2).
• For x=−3:(−3−3)(−3 + 2) = (−6)(−1) = 6 >0, so this interval is
positive.
• For x= 0:(0 −3)(0 + 2) = (−3)(2) = −6<0, so this interval is negative.
• For x= 4:(4 −3)(4 + 2) = (1)(6) = 6 >0, so this interval is positive.
Step 4: Identify the solution by considering where the inequality is satisfied,
i.e., where the expression is greater than or equal to zero. The solution is the
combination of intervals where (x−3)(x+ 2) ≥0:
(−∞,−2] ∪[3,∞).
Question 21
Question
Given the rational function f(x) = 2x2+5x−3
x2−4x−5, find the vertical asymptotes of
f(x).
Solution
Step 1: To find the vertical asymptotes of f(x), we need to determine the values
of xfor which the denominator of the rational function is equal to 0.
Step 2: Set the denominator x2−4x−5equal to 0and solve for x:
x2−4x−5 = 0
(x−5)(x+ 1) = 0
14
Step 3: Setting each factor to zero gives the solutions:
x−5 = 0 or x+ 1 = 0
x= 5 or x=−1
Step 4: Therefore, the vertical asymptotes of the function f(x)are x= 5
and x=−1.
Question 22
Question
Find the domain of the rational function:
f(x) = 2x+ 5
x2−9
Solution
Step 1: The domain of a rational function is all real numbers except for the
values of xthat make the denominator equal to zero since division by zero is
undefined. Step 2: Set the denominator of f(x)equal to zero and solve for x.
x2−9 = 0
x2= 9
x=±3
Step 3: The values of xthat make the denominator equal to zero are x= 3 and
x=−3. Step 4: Therefore, the domain of the function f(x) = 2x+ 5
x2−9is all
real numbers except for x= 3 and x=−3. In interval notation, the domain is
(−∞,−3) ∪(−3,3) ∪(3,∞).
Question 23
Question
Let f(x) = 5x2−10x
x2+3x−4. Determine the equations of the vertical asymptotes, hor-
izontal asymptotes, and oblique asymptotes (if any) of the rational function
f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator of the rational
function equal to zero and solve for x.
Denominator: x2+ 3x−4 = 0
(x+ 4)(x−1) = 0
15
Step 2: The vertical asymptotes occur at the values of xwhere the denom-
inator is zero. Thus, the equations of the vertical asymptotes are x=−4and
x= 1.
Step 3: To determine the horizontal asymptote, compare the degrees of the
numerator and denominator. Since the degree of the numerator is 2 and the
degree of the denominator is also 2, we look at the leading coefficients. The
equation of the horizontal asymptote is the ratio of the leading coefficients:
y=5
1= 5
Step 4: To check for an oblique asymptote, divide the numerator by the
denominator using long division or synthetic division.
x+ 4 5x2−10x5x
5x2+ 20x
−30x
−30x+ 0
Step 5: The quotient is 5and the remainder is −30x. Thus, the oblique
asymptote is the line y= 5x−30.
Therefore, the equations of the asymptotes for the rational function f(x)
are: Vertical asymptotes: x=−4and x= 1 Horizontal asymptote: y= 5
Oblique asymptote: y= 5x−30
Question 24
Question
Find the domain of the rational function defined by
f(x) = 2x2−5x−3
x2−4x+ 3
.
Solution
Step 1: Set the denominator equal to zero and solve for xto find the values that
make the function undefined.
x2−4x+ 3 = 0
(x−1)(x−3) = 0
x= 1 or x= 3
Step 2: The function is undefined for x= 1 and x= 3, so the domain of
f(x)is all real numbers except x= 1 and x= 3. Therefore, the domain of the
function is (−∞,1) ∪(1,3) ∪(3,∞).
16
Question 25
Question
For the rational function
f(x) = 2x2+x−3
x2+ 2x−8,
find the vertical asymptotes, horizontal asymptotes, and any holes in the graph.
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero
and solving for x.
x2+ 2x−8 = 0
(x+ 4)(x−2) = 0
This gives us x=−4and x= 2 as the vertical asymptotes.
Step 2: Find the horizontal asymptote by comparing the degrees of the
numerator and denominator. Since the degree of the numerator and denomi-
nator are the same, we compare the leading coefficients to find the horizontal
asymptote.
lim
x→∞
2x2+x−3
x2+ 2x−8= lim
x→∞
2 + 1
x−3
x2
1 + 2
x−8
x2
=2
1
= 2
Therefore, the horizontal asymptote is y= 2.
Step 3: Find any holes in the graph by factoring and canceling out common
factors in the numerator and denominator. Since 2x2+x−3does not factor
nicely, we can use polynomial long division or synthetic division to divide it by
x2+ 2x−8. Performing long division, we get:
2x+5 −3
x2+ 2x−8 2x2+x−3
2x2+4x−16
(−) (−3x) (+24)
x8
Therefore, we have f(x) = 2x2+x−3
x2+2x−8= 2x+5+ x+8
x2+2x−8. The hole in the graph
occurs at the point (2,9).
Question 26
Question
Let f(x) = 4x2−3x−4
x2−2x−3. Find the domain of f(x).
17
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that make the denominator equal to zero, as division by zero is undefined.
Therefore, we need to find the values of xthat make the denominator x2−2x−3
equal to zero.
Step 2: To find these values, we solve the equation x2−2x−3 = 0.
Step 3: Factoring the quadratic equation, we get (x−3)(x+ 1) = 0.
Step 4: Setting each factor to zero gives x−3 = 0 or x+ 1 = 0.
Step 5: Solving x−3 = 0 yields x= 3.
Step 6: Solving x+ 1 = 0 yields x=−1.
Step 7: Therefore, the values x= 3 and x=−1make the denominator of
f(x)equal to zero, and they are not in the domain of f(x).
Step 8: Thus, the domain of the function f(x)is all real numbers except
x= 3 and x=−1, which can be expressed as (−∞,−1) ∪(−1,3) ∪(3,∞).
Question 27
Question
Let f(x) = 2x2−x−1
x2−4x+3 . Determine the equations of the vertical and horizontal
asymptotes of the graph of y=f(x).
Solution
Step 1: To find the vertical asymptotes, we need to determine where the denomi-
nator of the function, x2−4x+3, becomes zero. This occurs when x2−4x+3 = 0.
Factoring the quadratic, we get (x−1)(x−3) = 0. Therefore, the vertical
asymptotes are x= 1 and x= 3.
Step 2: To find the horizontal asymptote, we compare the degrees of the
numerator and denominator of the function. Since the degrees are the same,
we look at the leading coefficients. The horizontal asymptote is the ratio of
the leading coefficients, which is 2/1=2. Therefore, y= 2 is the horizontal
asymptote.
Question 28
Question
Find the domain of the rational function
f(x) = x2+ 5x+ 6
x2−4x−12.
18
Solution
Step 1: We first need to find the restrictions on the domain of the function.
The function f(x)is defined for all real numbers except where the denominator
is equal to zero, since division by zero is undefined. Step 2: We find where the
denominator x2−4x−12 is equal to zero by solving the equation:
x2−4x−12 = 0.
Step 3: Factor the quadratic equation:
(x−6)(x+ 2) = 0.
Step 4: Set each factor to zero and solve for x:
x−6 = 0 or x+ 2 = 0.
Step 5: Solving x−6 = 0 gives x= 6 and solving x+ 2 = 0 gives x=−2. Step
6: Therefore, the restrictions on the domain are x= 6 and x=−2. Step 7:
The domain of the rational function f(x)is all real numbers except x= 6 and
x=−2, so the domain is (−∞,−2) ∪(−2,6) ∪(6,∞).
Question 29
Question
Solve the following rational equation for x:
1
x−1−2
x+ 2 =3
x2+x−2
Solution
Step 1: To begin, let’s find a common denominator for all the fractions on both
sides of the equation. The common denominator could be (x−1)(x+ 2)(x+ 1),
but since the third fraction on the right already has x2+x−2in its denominator,
we can simplify to (x−1)(x+ 2). Let’s rewrite the equation with this common
denominator: (x+ 2) −2(x−1)
(x−1)(x+ 2) =3
x2+x−2
Step 2: Simplify the left side of the equation:
x+ 2 −2x+ 2
(x−1)(x+ 2) =3
x2+x−2
−x+ 4
(x−1)(x+ 2) =3
x2+x−2
19
Step 3: Now, express each side as a single fraction:
−x+ 4
(x−1)(x+ 2) =3
(x−1)(x+ 2)
Step 4: Cross multiply to eliminate the denominators:
(−x+ 4)(x2+x−2) = 3(x−1)(x+ 2)
Step 5: Expand both sides of the equation:
−x3+x2+ 2x+ 4x2−4x−8 = 3(x2+x−2)
−x3+ 5x2−2x−8 = 3x2+ 3x−6
Step 6: Combine like terms on both sides:
−x3+ 5x2−2x−8 = 3x2+ 3x−6
−x3+ 2x2−5x−2 = 0
Step 7: Now we have a cubic equation that we need to solve. While there
are algebraic methods for finding the roots of cubic equations, in this case
the roots are not nice. Therefore, we will leave the answer as the equation
−x3+ 2x2−5x−2 = 0.
Question 30
Question
Simplify the rational function:
f(x) = x2−4
x2−5x+ 6
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify by canceling
out common factors if possible. If not, proceed with the factored form.
Step 1: We need to factor the numerator and denominator of the rational
function f(x):
The numerator x2−4can be factored as a difference of squares: x2−4 =
(x+ 2)(x−2).
The denominator x2−5x+6 can be factored as: x2−5x+6 = (x−2)(x−3).
Therefore, the rational function f(x)can be rewritten as:
f(x) = (x+ 2)(x−2)
(x−2)(x−3)
20
Step 2: Now we simplify the function by canceling out the common factor
(x−2) in the numerator and denominator.
f(x) = x+ 2
x−3
So, the simplified form of the rational function f(x)is x+2
x−3.
Question 31
Question
Find the domain of the rational function f(x) = x2−5x+ 6
x2−4x−5.
Solution
Step 1: To find the domain of the rational function, we need to identify the
values of xthat would make the denominator equal to zero since division by
zero is undefined.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x:
x−5 = 0 =
⇒x= 5
x+ 1 = 0 =⇒x=−1
Step 5: The values x= 5 and x=−1would make the denominator equal
to zero. Therefore, the domain of the rational function f(x)is all real numbers
except x= 5 and x=−1.
Step 6: Thus, the domain of the function f(x) = x2−5x+ 6
x2−4x−5is (−∞,−1) ∪
(−1,5) ∪(5,∞).
Question 32
Question
Simplify the following rational expression:
3x2−7x−6
2x2+ 5x−3
21
Solution
Step 1: Factor both the numerator and denominator. Step 2: Find the common
factors to simplify the expression.
Step 1:
Factor the numerator 3x2−7x−6: The factors of 3×(−6) = −18 that add up
to −7are −9and 2. So, we can write 3x2−7x−6as (3x+ 2)(x−3).
Factor the denominator 2x2+ 5x−3: The factors of 2×(−3) = −6that add
up to 5are 6and −1. So, we can write 2x2+ 5x−3as (2x−1)(x+ 3).
Thus, the expression becomes:
(3x+ 2)(x−3)
(2x−1)(x+ 3)
Step 2:
Now, simplify by cancelling out the common factors:
(3x+ 2)(x−3)
(2x−1)(x+ 3)
Therefore, the simplified form of the given rational expression is:
x−3
2x−1
Question 33
Question
Solve the rational equation: 4
x+3 −2
x−1=5
(x+3)(x−1) .
Solution
Step 1: Find a common denominator for all terms in the equation. Step 2:
Multiply every term by the common denominator to eliminate the fractions.
Step 3: Simplify the equation and solve for x. Step 4: Check for extraneous
solutions.
Step 1: Find a common denominator The common denominator for all
terms in the equation is (x+ 3)(x−1).
Step 2: Multiply every term by the common denominator Multiplying every
term by the common denominator, we get:
(x+ 3)(x−1) (4
x+ 3 −2
x−1)= (x+ 3)(x−1) (5
(x+ 3)(x−1))
4(x−1) −2(x+ 3) = 5
22
Step 3: Simplify the equation and solve for xSolving the simplified equa-
tion:
4x−4−2x−6 = 5
2x−10 = 5
2x= 15
x=15
2
Step 4: Check for extraneous solutions Check if x=15
2leads to any de-
nominator being equal to zero. Checking x=15
2in the original equation, we
find that x=15
2does not make any denominator zero.
Therefore, the solution to the equation is x=15
2.
Question 34
Question
Solve the rational inequality: 5
x−3≥3.
Solution
Step 1: Begin by setting up the inequality:
5
x−3≥3
Step 2: Multiply both sides of the inequality by (x−3) to remove the fraction:
5≥3(x−3)
Step 3: Distribute on the right side of the inequality:
5≥3x−9
Step 4: Add 9 to both sides of the inequality:
14 ≥3x
Step 5: Divide by 3 to solve for x:
14
3≥x
Step 6: Thus, the solution to the rational inequality is x≤14
3.
23
Question 35
Question
Simplify the rational function:
f(x) = 3x2−5x−2
2x2+ 7x−4
Solution
Step 1: Factor the numerator and the denominator. Step 2: Simplify by can-
celing out common factors.
Step 1: Factor the numerator and denominator of the rational function:
f(x) = 3x2−5x−2
2x2+ 7x−4
Numerator:
3x2−5x−2 = (3x+ 1)(x−2)
Denominator:
2x2+ 7x−4 = (2x−1)(x+ 4)
Therefore, the rational function can be expressed as:
f(x) = (3x+ 1)(x−2)
(2x−1)(x+ 4)
Step 2: Simplify the rational function by canceling out common factors:
f(x) = (3x+ 1)(x−2)
(2x−1)(x+ 4) =3x+ 1
2x−1·x−2
x+ 4
Thus, the simplified form of the rational function f(x)is:
f(x) = 3x+ 1
2x−1·x−2
x+ 4
24
Question 2
Question
Consider the rational function f(x) = 2x2−7x−30
x2−5x−6. Find the domain of the
function f(x).
Solution
Step 1: To find the domain of a rational function, we need to identify any
values of xthat would make the denominator zero, as these values would result
in division by zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−5x−6 = 0
Step 3: Factor the quadratic equation:
(x−6)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x:
x−6 = 0 =⇒x= 6
x+ 1 = 0 =⇒x=−1
Step 5: Therefore, the values x= 6 and x=−1make the denominator equal
to zero, so they must be excluded from the domain.
Step 6: The domain of the rational function f(x) = 2x2−7x−30
x2−5x−6is all real num-
bers except x= 6 and x=−1, so the domain is (−∞,−1) ∪(−1,6) ∪(6,∞).
Question 3
Question
Given the rational function f(x) = 2x2+5x−3
x2−4x−5, find the horizontal asymptotes (if
any) of the function.
Solution
Step 1: To find the horizontal asymptotes of a rational function, we compare
the degrees of the numerator and denominator polynomials. The degree of the
numerator is 2 and the degree of the denominator is 2, so we must consider the
leading coefficients of both polynomials.
Step 2: The leading coefficient of the numerator is 2, and the leading coeffi-
cient of the denominator is 1. Since the degrees of the two polynomials are equal,
we can find the horizontal asymptote by comparing their leading coefficients.
Step 3: The horizontal asymptote is given by the ratio of the leading coeffi-
cients. Therefore, the horizontal asymptote of the function is y=2
1= 2.
2
Question 4
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes of the ra-
tional function:
f(x) = x2−3x
x2−4
Solution
To find the vertical asymptotes of a rational function, we need to identify the
values of xthat make the denominator equal to zero, but not the numerator.
This will result in division by zero.
Step 1: Find Vertical Asymptotes Setting the denominator equal to
zero and solving for x:
x2−4 = 0
(x+ 2)(x−2) = 0
x=−2or x= 2
So, there are vertical asymptotes at x=−2and x= 2.
Step 2: Find the Horizontal Asymptotes To find horizontal asymp-
totes, we compare the degrees of the numerator and denominator of the function.
Since the degrees of the numerator and denominator are the same (2), the
horizontal asymptote is the ratio of the leading coefficients. The horizontal
asymptote is y=1
1= 1.
Step 3: Find any Holes To find any holes in the graph, we simplify the
function by canceling any common factors between the numerator and denomi-
nator.
f(x) = x(x−3)
(x+ 2)(x−2)
Notice that there is a common factor of xin the numerator and x−2in the
denominator, which gives us a hole at x= 0.
Therefore, the function f(x)has vertical asymptotes at x=−2and x= 2,
a horizontal asymptote at y= 1, and a hole at x= 0.
Question 5
Question
Simplify the following rational expression:
6x3−12x2+ 6x
2x2−4x
3
Solution
Step 1: Factor out the common terms in the numerator and denominator.
6x(x2−2x+ 1)
2x(x−2)
Step 2: Simplify the expression by canceling out the common factors.
6(x2−2x+ 1)
2(x−2)
Step 3: Further simplify the expression by factoring the quadratic in the
numerator. 6(x−1)2
2(x−2)
Step 4: Simplify by canceling out the common factors.
6(x−1)
2
Step 5: Finally, simplify the expression to obtain the simplified rational
expression.
3(x−1)
Question 6
Question
Find the domain of the rational function given by:
f(x) = x2−4
x2−3x−4
Solution
Step 1: To find the domain of the rational function, we need to determine the
values of xfor which the function is defined. The function is defined as long as
the denominator is not equal to zero, since division by zero is undefined.
Step 2: We start by factoring the denominator, x2−3x−4, to determine its
zeros:
x2−3x−4 = (x−4)(x+ 1)
Setting this expression equal to zero and solving for x, we get:
x−4 = 0 or x+ 1 = 0
x= 4 or x=−1
4
Step 3: Therefore, the function f(x)is undefined at x= 4 and x=−1since
these are the values that make the denominator zero.
Step 4: The domain of the function is all real numbers except x= 4 and
x=−1. So, the domain can be expressed in interval notation as:
(−∞,−1) ∪(−1,4) ∪(4,∞)
Question 7
Question
Find the domain of the rational function:
f(x) = 3x−1
x2−4x+ 3
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that make the denominator equal to zero, since division by zero is unde-
fined.
Step 2: To find the domain of f(x), we need to determine the values of x
that make the denominator x2−4x+ 3 equal to zero.
Step 3: We will solve the equation x2−4x+ 3 = 0 to find these values. This
can be factored as (x−3)(x−1) = 0.
Step 4: Setting each factor to zero gives us x−3 = 0 and x−1 = 0. Solving
these equations gives x= 3 and x= 1.
Step 5: Therefore, the values x= 1 and x= 3 make the denominator of the
function equal to zero and must be excluded from the domain.
Step 6: In interval notation, the domain of the function f(x) = 3x−1
x2−4x+3 is
(−∞,1) ∪(1,3) ∪(3,∞).
Question 8
Question
Find the domain of the rational function:
f(x) = 2x+ 1
x2−9
Solution
Step 1: The domain of a rational function consists of all real numbers except
for the values of xthat make the denominator equal to zero since division by
zero is undefined. So, we need to find the values of xthat make x2−9 = 0.
5
Step 2: We can factor the denominator x2−9using the difference of squares
formula: a2−b2= (a+b)(a−b). So, x2−9 = (x+ 3)(x−3).
Step 3: Setting x2−9 = 0 gives us:
(x+ 3)(x−3) = 0
Step 4: Solving for x, we get:
x+ 3 = 0 or x−3 = 0
x=−3or x= 3
Step 5: Therefore, the domain of the function f(x) = 2x+1
x2−9is all real numbers
except x=−3and x= 3, so the domain is:
(−∞,−3) ∪(−3,3) ∪(3,∞)
Question 9
Question
Let f(x) = 4x2+5x−6
2x2−3x−2. Find the vertical asymptotes, horizontal asymptotes, and
the x-intercepts of the function f(x).
Solution
Step 1: Find Vertical Asymptotes Vertical asymptotes occur when the de-
nominator of the rational function is equal to zero. So, to find the vertical
asymptotes, we set the denominator 2x2−3x−2equal to zero and solve for x.
2x2−3x−2 = 0
Factor the quadratic:
(x−2)(2x+ 1) = 0
Solve for x:x−2 = 0 =⇒x= 2 or 2x+ 1 = 0 =⇒x=−1
2
Therefore, the vertical asymptotes are x= 2 and x=−1
2.
Step 2: Find Horizontal Asymptotes To find the horizontal asymptotes,
we need to compare the degrees of the numerator and denominator. Since the
degrees are the same (both are 2), we look at the leading terms of the numerator
and denominator. The horizontal asymptote can be found by dividing the lead-
ing coefficient of the numerator by the leading coefficient of the denominator.
y=4
2= 2
Therefore, the horizontal asymptote is y= 2.
6
Step 3: Find X-Intercepts To find the x-intercepts, we set the numerator
4x2+ 5x−6equal to zero and solve for x.
4x2+ 5x−6 = 0
This can be factored into:
(4x−3)(x+ 2) = 0
Solve for x:4x−3 = 0 =⇒x=3
4or x+ 2 = 0 =⇒x=−2
Therefore, the x-intercepts are (3
4,0)and (−2,0).
Question 10
Question
Let f(x) = 3x2−2x−8
2x2−7x−4. Find the vertical asymptotes of the function f(x).
Solution
Step 1: To find the vertical asymptotes of f(x), we need to look for the values of
xthat make the denominator of f(x)equal to 0, since division by 0 is undefined.
Step 2: Set the denominator 2x2−7x−4equal to 0 and solve for x:
2x2−7x−4 = 0
Step 3: This is a quadratic equation that can be factored or solved using the
quadratic formula. Factoring gives:
(2x+ 1)(x−4) = 0
which gives solutions x=−1/2and x= 4.
Step 4: Therefore, the vertical asymptotes of the function f(x)are x=−1/2
and x= 4. These are the values of xfor which the function f(x)is not defined
due to division by zero in the denominator.
Question 11
Question
Simplify the rational expression:
3x3+x2−4x
x2−4
7
Solution
Step 1: Factor out any common factors in the numerator and denominator.
3x3+x2−4x
x2−4=x(3x2+x−4)
(x+ 2)(x−2)
Step 2: Factor the quadratic expression in the numerator.
3x2+x−4 = 3x2+ 4x−3x−4 = x(3x+ 4) −1(3x+ 4) = (x−1)(3x+ 4)
Step 3: Replace the factored numerator into the expression.
x(3x2+x−4)
(x+ 2)(x−2) =x(x−1)(3x+ 4)
(x+ 2)(x−2)
Thus, the simplified form of the rational expression is x(x−1)(3x+4)
(x+2)(x−2) .
Question 12
Question
Simplify the rational function:
3x2−9x
6x2−18
Solution
Step 1: Factor out the greatest common factor in the numerator and the de-
nominator. Step 2: Simplify by canceling out common factors.
Step 1: Factor out the greatest common factor in the numerator and the
denominator: 3x2−9x
6x2−18 =3x(x−3)
6(x2−3)
Step 2: Simplify by canceling out common factors:
3x(x−3)
6(x2−3) =(x−3)
2(x2−3)
Therefore, the simplified form of the rational function is x−3
2(x2−3) .
Question 13
Question
Find the domain of the rational function:
f(x) = x2−4
x2−5x+ 6
8
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the function is defined. The function is defined for all x
values except where the denominator is equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−5x+ 6 = 0
Step 3: Factor the quadratic equation:
(x−2)(x−3) = 0
Step 4: Solve for x:
x−2 = 0 ⇒x= 2
x−3 = 0 ⇒x= 3
Step 5: The domain of the function f(x)is all real numbers except x= 2
and x= 3. Therefore, the domain is:
x∈R, x = 2, x = 3
Question 14
Question
Simplify the rational function:
f(x) = 3x2−x−2
x2−3x−4
Solution
Step 1: Factor the numerator and the denominator.
We have:
f(x) = 3x2−x−2
x2−3x−4
Factoring the numerator:
3x2−x−2 = (3x+ 2)(x−1)
Factoring the denominator:
x2−3x−4 = (x−4)(x+ 1)
Step 2: Rewrite the rational function with the factored numerator and de-
nominator.
9
Therefore,
f(x) = (3x+ 2)(x−1)
(x−4)(x+ 1)
Step 3: Simplify the expression as far as possible.
Thus, the simplified form of the rational function is:
f(x) = 3x+ 2
x−4
Question 15
Question
Let f(x) = 4x2+7x+2
x2−3x−10 . Find the equations of the horizontal and vertical asymp-
totes of the function f(x).
Solution
Step 1: Determine the vertical asymptotes by finding the values of xthat make
the denominator equal to zero. Set x2−3x−10 = 0 and solve for x.
x2−3x−10 = 0
(x−5)(x+ 2) = 0
x= 5,−2
Therefore, the vertical asymptotes are x= 5 and x=−2.
Step 2: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator. Since the degrees are the same, the horizontal asymptote
is the ratio of the leading coefficients. Divide the leading coefficients of the nu-
merator and denominator.
lim
x→±∞
4x2+ 7x+ 2
x2−3x−10 = lim
x→±∞
4
1= 4
Therefore, the equation of the horizontal asymptote is y= 4.
In conclusion, the function f(x)has vertical asymptotes at x= 5 and x=
−2, and a horizontal asymptote at y= 4.
Question 16
Question
Simplify the following rational expression:
4x3−8x2+ 4x
2x2−2x
10
Solution
Step 1: Factor out any common terms in the numerator and the denominator:
4x3−8x2+ 4x
2x2−2x=4x(x2−2x+ 1)
2x(x−1)
Step 2: Simplify the expression by canceling out common factors:
4x(x2−2x+ 1)
2x(x−1) =4(x2−2x+ 1)
2(x−1) =4(x−1)2
2(x−1)
Step 3: Further simplify by canceling out common factors:
4(x−1)2
2(x−1) =4(x−1)(x−1)
2(x−1) =4(x−1)
2= 2(x−1)
Therefore, the simplified form of the given rational expression is 2(x−1).
Question 17
Question
Solve the following rational inequality for x:
2x2−3x−2
x−1>x−2
x+ 3
Solution
Step 1: Identify the domain of the rational inequality by finding the values of x
that make the denominators equal to zero.
For the given rational inequality, the denominators are x−1and x+ 3.
Setting each denominator equal to zero gives us:
x−1 = 0 ⇒x= 1 and x+ 3 = 0 ⇒x=−3
So, the domain of the rational inequality is all real numbers except x= 1
and x=−3.
Step 2: Determine the critical points of the rational inequality by finding
the values of xthat make the numerator and denominators equal to zero.
The numerator and denominators of the rational inequality are 2x2−3x−2,
x−1, and x+ 3. Setting each expression equal to zero gives us:
2x2−3x−2 = 0 ⇒(2x+ 1)(x−2) = 0
x−1 = 0 ⇒x= 1
x+ 3 = 0 ⇒x=−3
The critical points are x=−3,x=−1
2, and x= 1.
Step 3: Create a sign chart to test the intervals determined by the critical
points.
11
| | −3|−1
2| 1 | +∞| | |—|——|———–|—–|———|—-| | | | | | | | | | | | | | |
|||||||
Step 4: Plug in a test value from each interval into the rational inequality
and determine the sign of the expression.
For x < −3, pick x=−4:
2(−4)2−3(−4) −2
−4−1>−4−2
−4+3
Simplifying, we get:
42
−5>−6
This is false, so x < −3is not part of the solution.
Step 5: Continue testing the other intervals until you have determined the
solution set.
Can you continue from here to determine the solution set for the rational
inequality?
Question 18
Question
Simplify the following rational expression:
3x2−2x−5
4x2+ 3x−7∇ · 2x2+ 3x+ 1
2x2−5x−3
Solution
To simplify the given rational expression, we will follow these steps:
Step 1: Write the division as a multiplication by taking the reciprocal of
the divisor.
3x2−2x−5
4x2+ 3x−7∇ · 2x2+ 3x+ 1
2x2−5x−3=3x2−2x−5
4x2+ 3x−7·2x2−5x−3
2x2+ 3x+ 1
Step 2: Factor all quadratic expressions.
3x2−2x−5 = (3x+ 1)(x−5)
4x2+ 3x−7 = (4x−7)(x+ 1)
2x2+ 3x+ 1 = (2x+ 1)(x+ 1)
2x2−5x−3 = (2x+ 1)(x−3)
12
Step 3: Substitute the factored expressions back into the initial expression
and simplify.
(3x+ 1)(x−5)
(4x−7)(x+ 1) ·(2x−5)(x−3)
(2x+ 1)(x+ 1) =(3x+ 1)(x−5)(2x−5)(x−3)
(4x−7)(x+ 1)(2x+ 1)(x+ 1)
=(3x2−14x−5)(2x2−11x+ 15)
(4x2−3x−7)(2x2+ 3x+ 1)
Therefore, the simplified form of the given rational expression is (3x2−14x−5)(2x2−11x+15)
(4x2−3x−7)(2x2+3x+1) .
Question 19
Question
Simplify the following rational expression:
5x2−4x−5
x2−3x−4∇ · 3x2−5x−2
x2−5x+ 6
Solution
To simplify the given rational expression, we will follow these steps:
Step 1: Factor the numerators and denominators of the fractions.
5x2−4x−5
x2−3x−4∇ · 3x2−5x−2
x2−5x+ 6 =(5x+ 1)(x−5)
(x−4)(x+ 1) ∇ · (3x+ 1)(x−2)
(x−3)(x−2)
Step 2: Rewrite the division as multiplication by the reciprocal of the second
fraction.
(5x+ 1)(x−5)
(x−4)(x+ 1) ×(x−3)(x−2)
(3x+ 1)(x−2) =(5x+ 1)(x−5)
(x−4)(x+ 1) ×(x−3)
(3x+ 1)
Step 3: Multiply the fractions by multiplying the numerators together and
the denominators together.
(5x+ 1)(x−5)(x−3)
(x−4)(x+ 1)(3x+ 1) =(5x2−25x+x−5)(x−3)
(x−4)(x+ 1)(3x+ 1)
=(5x2−24x−5)(x−3)
(x−4)(x+ 1)(3x+ 1)
Step 4: Expand and simplify the numerator.
(5x2−24x−5)(x−3) = 5x3−24x2−5x−15x2+ 72x+ 15
= 5x3−39x2+ 67x+ 15
Therefore, the simplified expression is:
5x3−39x2+ 67x+ 15
(x−4)(x+ 1)(3x+ 1)
13
Question 20
Question
Solve the following rational inequality and express your answer in interval no-
tation:
(x−3)(x+ 2) ≥0
Solution
Step 1: Find the critical points by setting the expression equal to zero and
solving for x.
(x−3)(x+ 2) = 0
This gives x= 3 and x=−2as critical points.
Step 2: Create intervals using the critical points (x= 3 and x=−2). We
have three intervals to consider: (−∞,−2),(−2,3), and (3,∞).
Step 3: Test a value in each interval to determine the sign of (x−3)(x+ 2).
• For x=−3:(−3−3)(−3 + 2) = (−6)(−1) = 6 >0, so this interval is
positive.
• For x= 0:(0 −3)(0 + 2) = (−3)(2) = −6<0, so this interval is negative.
• For x= 4:(4 −3)(4 + 2) = (1)(6) = 6 >0, so this interval is positive.
Step 4: Identify the solution by considering where the inequality is satisfied,
i.e., where the expression is greater than or equal to zero. The solution is the
combination of intervals where (x−3)(x+ 2) ≥0:
(−∞,−2] ∪[3,∞).
Question 21
Question
Given the rational function f(x) = 2x2+5x−3
x2−4x−5, find the vertical asymptotes of
f(x).
Solution
Step 1: To find the vertical asymptotes of f(x), we need to determine the values
of xfor which the denominator of the rational function is equal to 0.
Step 2: Set the denominator x2−4x−5equal to 0and solve for x:
x2−4x−5 = 0
(x−5)(x+ 1) = 0
14
Step 3: Setting each factor to zero gives the solutions:
x−5 = 0 or x+ 1 = 0
x= 5 or x=−1
Step 4: Therefore, the vertical asymptotes of the function f(x)are x= 5
and x=−1.
Question 22
Question
Find the domain of the rational function:
f(x) = 2x+ 5
x2−9
Solution
Step 1: The domain of a rational function is all real numbers except for the
values of xthat make the denominator equal to zero since division by zero is
undefined. Step 2: Set the denominator of f(x)equal to zero and solve for x.
x2−9 = 0
x2= 9
x=±3
Step 3: The values of xthat make the denominator equal to zero are x= 3 and
x=−3. Step 4: Therefore, the domain of the function f(x) = 2x+ 5
x2−9is all
real numbers except for x= 3 and x=−3. In interval notation, the domain is
(−∞,−3) ∪(−3,3) ∪(3,∞).
Question 23
Question
Let f(x) = 5x2−10x
x2+3x−4. Determine the equations of the vertical asymptotes, hor-
izontal asymptotes, and oblique asymptotes (if any) of the rational function
f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator of the rational
function equal to zero and solve for x.
Denominator: x2+ 3x−4 = 0
(x+ 4)(x−1) = 0
15
Step 2: The vertical asymptotes occur at the values of xwhere the denom-
inator is zero. Thus, the equations of the vertical asymptotes are x=−4and
x= 1.
Step 3: To determine the horizontal asymptote, compare the degrees of the
numerator and denominator. Since the degree of the numerator is 2 and the
degree of the denominator is also 2, we look at the leading coefficients. The
equation of the horizontal asymptote is the ratio of the leading coefficients:
y=5
1= 5
Step 4: To check for an oblique asymptote, divide the numerator by the
denominator using long division or synthetic division.
x+ 4 5x2−10x5x
5x2+ 20x
−30x
−30x+ 0
Step 5: The quotient is 5and the remainder is −30x. Thus, the oblique
asymptote is the line y= 5x−30.
Therefore, the equations of the asymptotes for the rational function f(x)
are: Vertical asymptotes: x=−4and x= 1 Horizontal asymptote: y= 5
Oblique asymptote: y= 5x−30
Question 24
Question
Find the domain of the rational function defined by
f(x) = 2x2−5x−3
x2−4x+ 3
.
Solution
Step 1: Set the denominator equal to zero and solve for xto find the values that
make the function undefined.
x2−4x+ 3 = 0
(x−1)(x−3) = 0
x= 1 or x= 3
Step 2: The function is undefined for x= 1 and x= 3, so the domain of
f(x)is all real numbers except x= 1 and x= 3. Therefore, the domain of the
function is (−∞,1) ∪(1,3) ∪(3,∞).
16
Question 25
Question
For the rational function
f(x) = 2x2+x−3
x2+ 2x−8,
find the vertical asymptotes, horizontal asymptotes, and any holes in the graph.
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero
and solving for x.
x2+ 2x−8 = 0
(x+ 4)(x−2) = 0
This gives us x=−4and x= 2 as the vertical asymptotes.
Step 2: Find the horizontal asymptote by comparing the degrees of the
numerator and denominator. Since the degree of the numerator and denomi-
nator are the same, we compare the leading coefficients to find the horizontal
asymptote.
lim
x→∞
2x2+x−3
x2+ 2x−8= lim
x→∞
2 + 1
x−3
x2
1 + 2
x−8
x2
=2
1
= 2
Therefore, the horizontal asymptote is y= 2.
Step 3: Find any holes in the graph by factoring and canceling out common
factors in the numerator and denominator. Since 2x2+x−3does not factor
nicely, we can use polynomial long division or synthetic division to divide it by
x2+ 2x−8. Performing long division, we get:
2x+5 −3
x2+ 2x−8 2x2+x−3
2x2+4x−16
(−) (−3x) (+24)
x8
Therefore, we have f(x) = 2x2+x−3
x2+2x−8= 2x+5+ x+8
x2+2x−8. The hole in the graph
occurs at the point (2,9).
Question 26
Question
Let f(x) = 4x2−3x−4
x2−2x−3. Find the domain of f(x).
17
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that make the denominator equal to zero, as division by zero is undefined.
Therefore, we need to find the values of xthat make the denominator x2−2x−3
equal to zero.
Step 2: To find these values, we solve the equation x2−2x−3 = 0.
Step 3: Factoring the quadratic equation, we get (x−3)(x+ 1) = 0.
Step 4: Setting each factor to zero gives x−3 = 0 or x+ 1 = 0.
Step 5: Solving x−3 = 0 yields x= 3.
Step 6: Solving x+ 1 = 0 yields x=−1.
Step 7: Therefore, the values x= 3 and x=−1make the denominator of
f(x)equal to zero, and they are not in the domain of f(x).
Step 8: Thus, the domain of the function f(x)is all real numbers except
x= 3 and x=−1, which can be expressed as (−∞,−1) ∪(−1,3) ∪(3,∞).
Question 27
Question
Let f(x) = 2x2−x−1
x2−4x+3 . Determine the equations of the vertical and horizontal
asymptotes of the graph of y=f(x).
Solution
Step 1: To find the vertical asymptotes, we need to determine where the denomi-
nator of the function, x2−4x+3, becomes zero. This occurs when x2−4x+3 = 0.
Factoring the quadratic, we get (x−1)(x−3) = 0. Therefore, the vertical
asymptotes are x= 1 and x= 3.
Step 2: To find the horizontal asymptote, we compare the degrees of the
numerator and denominator of the function. Since the degrees are the same,
we look at the leading coefficients. The horizontal asymptote is the ratio of
the leading coefficients, which is 2/1=2. Therefore, y= 2 is the horizontal
asymptote.
Question 28
Question
Find the domain of the rational function
f(x) = x2+ 5x+ 6
x2−4x−12.
18
Solution
Step 1: We first need to find the restrictions on the domain of the function.
The function f(x)is defined for all real numbers except where the denominator
is equal to zero, since division by zero is undefined. Step 2: We find where the
denominator x2−4x−12 is equal to zero by solving the equation:
x2−4x−12 = 0.
Step 3: Factor the quadratic equation:
(x−6)(x+ 2) = 0.
Step 4: Set each factor to zero and solve for x:
x−6 = 0 or x+ 2 = 0.
Step 5: Solving x−6 = 0 gives x= 6 and solving x+ 2 = 0 gives x=−2. Step
6: Therefore, the restrictions on the domain are x= 6 and x=−2. Step 7:
The domain of the rational function f(x)is all real numbers except x= 6 and
x=−2, so the domain is (−∞,−2) ∪(−2,6) ∪(6,∞).
Question 29
Question
Solve the following rational equation for x:
1
x−1−2
x+ 2 =3
x2+x−2
Solution
Step 1: To begin, let’s find a common denominator for all the fractions on both
sides of the equation. The common denominator could be (x−1)(x+ 2)(x+ 1),
but since the third fraction on the right already has x2+x−2in its denominator,
we can simplify to (x−1)(x+ 2). Let’s rewrite the equation with this common
denominator: (x+ 2) −2(x−1)
(x−1)(x+ 2) =3
x2+x−2
Step 2: Simplify the left side of the equation:
x+ 2 −2x+ 2
(x−1)(x+ 2) =3
x2+x−2
−x+ 4
(x−1)(x+ 2) =3
x2+x−2
19
Step 3: Now, express each side as a single fraction:
−x+ 4
(x−1)(x+ 2) =3
(x−1)(x+ 2)
Step 4: Cross multiply to eliminate the denominators:
(−x+ 4)(x2+x−2) = 3(x−1)(x+ 2)
Step 5: Expand both sides of the equation:
−x3+x2+ 2x+ 4x2−4x−8 = 3(x2+x−2)
−x3+ 5x2−2x−8 = 3x2+ 3x−6
Step 6: Combine like terms on both sides:
−x3+ 5x2−2x−8 = 3x2+ 3x−6
−x3+ 2x2−5x−2 = 0
Step 7: Now we have a cubic equation that we need to solve. While there
are algebraic methods for finding the roots of cubic equations, in this case
the roots are not nice. Therefore, we will leave the answer as the equation
−x3+ 2x2−5x−2 = 0.
Question 30
Question
Simplify the rational function:
f(x) = x2−4
x2−5x+ 6
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify by canceling
out common factors if possible. If not, proceed with the factored form.
Step 1: We need to factor the numerator and denominator of the rational
function f(x):
The numerator x2−4can be factored as a difference of squares: x2−4 =
(x+ 2)(x−2).
The denominator x2−5x+6 can be factored as: x2−5x+6 = (x−2)(x−3).
Therefore, the rational function f(x)can be rewritten as:
f(x) = (x+ 2)(x−2)
(x−2)(x−3)
20
Step 2: Now we simplify the function by canceling out the common factor
(x−2) in the numerator and denominator.
f(x) = x+ 2
x−3
So, the simplified form of the rational function f(x)is x+2
x−3.
Question 31
Question
Find the domain of the rational function f(x) = x2−5x+ 6
x2−4x−5.
Solution
Step 1: To find the domain of the rational function, we need to identify the
values of xthat would make the denominator equal to zero since division by
zero is undefined.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x:
x−5 = 0 =
⇒x= 5
x+ 1 = 0 =⇒x=−1
Step 5: The values x= 5 and x=−1would make the denominator equal
to zero. Therefore, the domain of the rational function f(x)is all real numbers
except x= 5 and x=−1.
Step 6: Thus, the domain of the function f(x) = x2−5x+ 6
x2−4x−5is (−∞,−1) ∪
(−1,5) ∪(5,∞).
Question 32
Question
Simplify the following rational expression:
3x2−7x−6
2x2+ 5x−3
21
Solution
Step 1: Factor both the numerator and denominator. Step 2: Find the common
factors to simplify the expression.
Step 1:
Factor the numerator 3x2−7x−6: The factors of 3×(−6) = −18 that add up
to −7are −9and 2. So, we can write 3x2−7x−6as (3x+ 2)(x−3).
Factor the denominator 2x2+ 5x−3: The factors of 2×(−3) = −6that add
up to 5are 6and −1. So, we can write 2x2+ 5x−3as (2x−1)(x+ 3).
Thus, the expression becomes:
(3x+ 2)(x−3)
(2x−1)(x+ 3)
Step 2:
Now, simplify by cancelling out the common factors:
(3x+ 2)(x−3)
(2x−1)(x+ 3)
Therefore, the simplified form of the given rational expression is:
x−3
2x−1
Question 33
Question
Solve the rational equation: 4
x+3 −2
x−1=5
(x+3)(x−1) .
Solution
Step 1: Find a common denominator for all terms in the equation. Step 2:
Multiply every term by the common denominator to eliminate the fractions.
Step 3: Simplify the equation and solve for x. Step 4: Check for extraneous
solutions.
Step 1: Find a common denominator The common denominator for all
terms in the equation is (x+ 3)(x−1).
Step 2: Multiply every term by the common denominator Multiplying every
term by the common denominator, we get:
(x+ 3)(x−1) (4
x+ 3 −2
x−1)= (x+ 3)(x−1) (5
(x+ 3)(x−1))
4(x−1) −2(x+ 3) = 5
22
Step 3: Simplify the equation and solve for xSolving the simplified equa-
tion:
4x−4−2x−6 = 5
2x−10 = 5
2x= 15
x=15
2
Step 4: Check for extraneous solutions Check if x=15
2leads to any de-
nominator being equal to zero. Checking x=15
2in the original equation, we
find that x=15
2does not make any denominator zero.
Therefore, the solution to the equation is x=15
2.
Question 34
Question
Solve the rational inequality: 5
x−3≥3.
Solution
Step 1: Begin by setting up the inequality:
5
x−3≥3
Step 2: Multiply both sides of the inequality by (x−3) to remove the fraction:
5≥3(x−3)
Step 3: Distribute on the right side of the inequality:
5≥3x−9
Step 4: Add 9 to both sides of the inequality:
14 ≥3x
Step 5: Divide by 3 to solve for x:
14
3≥x
Step 6: Thus, the solution to the rational inequality is x≤14
3.
23
Question 35
Question
Simplify the rational function:
f(x) = 3x2−5x−2
2x2+ 7x−4
Solution
Step 1: Factor the numerator and the denominator. Step 2: Simplify by can-
celing out common factors.
Step 1: Factor the numerator and denominator of the rational function:
f(x) = 3x2−5x−2
2x2+ 7x−4
Numerator:
3x2−5x−2 = (3x+ 1)(x−2)
Denominator:
2x2+ 7x−4 = (2x−1)(x+ 4)
Therefore, the rational function can be expressed as:
f(x) = (3x+ 1)(x−2)
(2x−1)(x+ 4)
Step 2: Simplify the rational function by canceling out common factors:
f(x) = (3x+ 1)(x−2)
(2x−1)(x+ 4) =3x+ 1
2x−1·x−2
x+ 4
Thus, the simplified form of the rational function f(x)is:
f(x) = 3x+ 1
2x−1·x−2
x+ 4
24
