MATH 121 - COLLEGE ALGEBRA -
Rational Functions
Question Bank - Set 10
Liberty University
Question 1
Question
Find the domain of the rational function:
f(x) = 7x2−2x+ 3
x2−5x+ 6
Solution
Step 1: To find the domain of the rational function, we need to determine all
the values of xfor which the function is defined. The function is defined for all
real numbers except those that make the denominator 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: Set each factor to zero to find the roots:
x−2 = 0 or x−3 = 0
Step 5: Solve for x:
x= 2 or x= 3
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 2 and x= 3.
Domain ={x∈R|x= 2,3}
Question 2
Question
Find the vertical asymptotes, horizontal asymptotes, and the x- and y-intercepts
of the rational function:
f(x) = 2x3−5x2−3x
x2−9
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes of the function, we set the denominator equal
to zero and solve for x:
x2−9 = 0
(x+ 3)(x−3) = 0
So, the vertical asymptotes are x=−3and x= 3.
Step 2: Horizontal Asymptotes
To find the horizontal asymptote, we look at the degrees of the numerator and
denominator:
Since the degree of the numerator (3) is greater than the degree of the
denominator (2), there is no horizontal asymptote.
Step 3: X-intercepts
To find the x-intercepts, we set the numerator equal to zero and solve for x:
2x3−5x2−3x= 0
Factor out an x:
x(2x2−5x−3) = 0
x(2x+ 1)(x−3) = 0
So, the x-intercepts are x= 0,x=−1
2, and x= 3.
Step 4: Y-intercepts
To find the y-intercept, we plug in x= 0 into the function:
f(0) = 2(0)3−5(0)2−3(0)
(0)2−9
f(0) = 0
So, the y-intercept is at the point (0,0).
2
Question 3
Question
Find the domain of the rational function:
f(x) = 4x
x2−9.
Solution
Step 1: The domain of a rational function is all real numbers except those that
make the denominator equal to zero.
Step 2: Set the denominator x2−9equal to zero and solve for x:
x2−9 = 0.
Step 3: Factor the quadratic equation:
(x−3)(x+ 3) = 0.
Step 4: Set each factor equal to zero:
x−3 = 0 and x+ 3 = 0.
Step 5: Solve for xto find the critical points:
x= 3 and x=−3.
Step 6: The domain of the rational function f(x) = 4x
x2−9is all real numbers
except x= 3 and x=−3.
Step 7: Therefore, the domain of f(x)is (−∞,−3) ∪(−3,3) ∪(3,∞).
Question 4
Question
Let f(x) = 2x2−5x−3
x2−9. Find the domain of the function f(x).
Solution
Step 1: To find the domain of a rational function, we need to identify the values
of xwhere the function is undefined. This occurs when the denominator is equal
to zero. In this case, the denominator x2−9is equal to zero when x=−3or
x= 3.
Step 2: Therefore, the domain of the function f(x)is all real numbers except
x=−3and x= 3. We can express this as:
{x∈R|x=−3, x = 3}
So, the domain of the function f(x)is all real numbers except −3and 3.
3
Question 5
Question
Find the domain of the rational function: f(x) = 1
x2−5x+ 6.
Solution
Step 1: Determine the values of xthat would make the denominator equal to
zero, as those values would make the function undefined.
x2−5x+ 6 = 0
Step 2: Factor the quadratic equation.
x2−5x+ 6 = (x−2)(x−3)
Step 3: Set each factor to zero and solve for x.
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 4: The function is undefined at x= 2 and x= 3, so the domain of
the function f(x)is all real numbers except x= 2 and x= 3, which can be
expressed as:
(−∞,2) ∪(2,3) ∪(3,∞)
Question 6
Question
Simplify the rational function:
2x2−x−3
x2−4
Solution
Step 1: Factor the numerator and denominator:
The numerator 2x2−x−3can be factored as (2x+ 1)(x−3).
The denominator x2−4 = (x+ 2)(x−2).
Step 2: Rewrite the rational function:
2x2−x−3
x2−4=(2x+ 1)(x−3)
(x+ 2)(x−2)
Step 3: Simplify the rational function:
4
2x2−x−3
x2−4=(2x+ 1)(x−3)
(x+ 2)(x−2) =2x+ 1
x+ 2
Therefore, the simplified form of the rational function is 2x+1
x+2 .
Question 7
Question
Simplify the rational function:
7x2−2x−3
2x2+ 5x−3
Solution
Step 1: Factor both the numerator and denominator of the rational function.
Step 2: Simplify the rational function by canceling out any common factors in
the numerator and denominator.
Step 1: Factor the numerator and denominator: The numerator, 7x2−2x−
3, can be factored as (7x+ 3)(x−1).
The denominator, 2x2+ 5x−3, can be factored as (2x−1)(x+ 3).
Therefore, the rational function becomes:
(7x+ 3)(x−1)
(2x−1)(x+ 3)
Step 2: Simplify the rational function: Since there are no common factors
that can be canceled out in the numerator and denominator, the simplified form
of the rational function is:
7x+ 3
2x−1·x−1
x+ 3
Therefore, the simplified form of the rational function is (7x+3)(x−1)
(2x−1)(x+3) .
Question 8
Question
Simplify the rational function:
5x3−16x2−5x+ 16
x2−4x−5
5
Solution
Step 1: Factor the numerator and denominator:
5x3−16x2−5x+ 16
x2−4x−5=(5x3−20x2) + (4x2−5x+ 16)
(x−5)(x+ 1)
Step 2: Combine like terms in the numerator:
=5x2(x−4) + 1(4x2−5x+ 16)
(x−5)(x+ 1)
Step 3: Factor out common factors:
=5x2(x−4) + 1(4x−16)
(x−5)(x+ 1) =5x2(x−4) + 4(x−4)
(x−5)(x+ 1)
Step 4: Factor out the common factor of (x−4):
=(5x2+ 4)(x−4)
(x−5)(x+ 1)
Therefore, the simplified form of the rational function is:
(5x2+ 4)(x−4)
(x−5)(x+ 1)
Question 9
Question
Let f(x) = 2x2−3x−2
x2−5x+6 be a rational function. Find the x-intercepts, y-intercepts,
and vertical asymptotes, if any.
Solution
Step 1: Find the x-intercepts by setting f(x) = 0 and solving for x.
0 = 2x2−3x−2
x2−5x+ 6
0 = 2x2−3x−2
We can factor the quadratic equation to find the x-intercepts.
0 = (2x+ 1)(x−2)
Setting each factor to zero gives us x-intercepts at x=−1
2and x= 2.
6
Step 2: Find the y-intercept by setting x= 0.
y=2(0)2−3(0) −2
(0)2−5(0) + 6
y=−2
6
The y-intercept is at (0,−1
3).
Step 3: Find the vertical asymptotes by setting the denominator of f(x)
equal to zero and solving for x.
x2−5x+ 6 = 0
x= 2 or x= 3
Therefore, there are vertical asymptotes at x= 2 and x= 3 for the rational
function f(x).
In summary, the x-intercepts are at x=−1
2and x= 2, the y-intercept is at
(0,−1
3), and there are vertical asymptotes at x= 2 and x= 3 for the rational
function f(x).
Question 10
Question
Given the rational function f(x) = 2x2−7x+3
x2−4x+3 , find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph of f(x).
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes of f(x), we need to determine the values
of xwhere the denominator of the rational function is equal to zero, but the
numerator is not zero.
First, factor the denominator x2−4x+ 3:
x2−4x+ 3 = (x−1)(x−3).
Therefore, the vertical asymptotes occur at x= 1 and x= 3 since these
values make the denominator zero but not the numerator.
Step 2: Horizontal Asymptotes
To find the horizontal asymptotes of f(x), we need to compare the degrees
of the numerators and denominators.
Since the degree of the numerator is 2 and the degree of the denomina-
tor is also 2, we can find the horizontal asymptotes by comparing the leading
coefficients:
lim
x→∞
2x2−7x+ 3
x2−4x+ 3 = lim
x→∞
2
1= 2.
7
Therefore, the horizontal asymptote is at y= 2.
Step 3: Hole in the Graph
To find any holes in the graph of f(x), we need to check for common factors
between the numerator and denominator that can be canceled out.
Factor the numerator 2x2−7x+ 3:
2x2−7x+ 3 = (2x−1)(x−3).
We can see that there is a common factor of (x−3) in both the numerator
and the denominator, which can be canceled out. Therefore, there is a hole in
the graph at x= 3.
Thus, the vertical asymptotes are at x= 1 and x= 3, the horizontal asymp-
tote is at y= 2, and there is a hole in the graph at x= 3.
Question 11
Question
Simplify the following rational expression:
3x2−6x+ 3
x2−1
Solution
Step 1: Factor the numerator and denominator:
3x2−6x+ 3
x2−1=3(x2−2x+ 1)
(x−1)(x+ 1)
Step 2: Simplify the expression further by factoring the quadratic term:
3(x2−2x+ 1)
(x−1)(x+ 1) =3(x−1)2
(x−1)(x+ 1)
Step 3: Cancel out the common factor of (x−1) from the numerator and
denominator: 3(x−1)2
(x−1)(x+ 1) =3(x−1)
x+ 1
Therefore, the simplified form of the rational expression is 3(x−1)
x+1 .
Question 12
Question
The rational function
f(x) = 3x2−5x−2
x2−4x−5
8
can be rewritten as
f(x) = A
x+ 1 +B
x−5
where Aand Bare constants. Find the values of Aand B.
Solution
Step 1: First, we rewrite the given function in partial fraction form:
f(x) = 3x2−5x−2
x2−4x−5=A
x+ 1 +B
x−5
Step 2: Multiply both sides by the denominator (x2−4x−5) to clear the
fractions:
3x2−5x−2 = A(x−5) + B(x+ 1)
Step 3: Expand the right side and group like terms:
3x2−5x−2 = Ax −5A+Bx +B
Step 4: Combine like terms:
3x2−5x−2 = (A+B)x−5A+B
Step 5: Equate the coefficients of like terms on both sides:
3 = A+B
−5 = −5A+B
Step 6: Solve the system of equations by substitution or elimination to find
the values of Aand B. From the first equation, we get B= 3 −A. Substituting
this into the second equation:
−5 = −5A+ (3 −A)
Step 7: Solve for A:
−5 = −5A+ 3 −A
−5 = −6A+ 3
−8 = −6A
A=4
3
Step 8: Substitute A=4
3back into B= 3 −Ato find B:
B= 3 −4
3=5
3
Therefore, the values of Aand Bare A=4
3and B=5
3.
9
Question 13
Question
Let f(x) = 2x3−9x2+9x
x2−4x. Find the following: a) Domain of fb) x-intercepts
(if any) c) y-intercept (if any) d) Asymptotes (vertical, horizontal, and slant)
e) Intervals where f(x)is increasing or decreasing f) Intervals where f(x)is
concave up or down g) Critical points h) Inflection points (if any)
Solution
a) To find the domain of f(x), we need to consider the values that xcan take
without causing division by zero. Thus, the domain of fis all real numbers
except the ones that make the denominator zero. Therefore, the domain of fis
(−∞,0) ∪(0,4) ∪(4,∞).
b) To find the x-intercepts, we set f(x) = 0 and solve for x:
2x3−9x2+ 9x
x2−4x= 0
Notice that we can factor out an xfrom the numerator:
x(2x2−9x+ 9)
x(x−4) = 0
This simplifies to 2x2−9x+9 = 0, which does not have real solutions. Therefore,
there are no x-intercepts.
c) To find the y-intercept, we evaluate f(0):
f(0) = 2(0)3−9(0)2+ 9(0)
(0)2−4(0) = 0
So, the y-intercept is at the point (0,0).
d) To find the asymptotes: Vertical asymptotes occur where the denominator
is zero, so there is a vertical asymptote at x= 0.
Horizontal asymptotes are found by comparing the degrees of the numerator
and denominator. Since the degree of the numerator is equal to the degree of the
denominator, we divide the leading coefficients to find the horizontal asymptote:
lim
x→∞
2x3
x2= lim
x→∞ 2x=∞
Therefore, there is no horizontal asymptote.
Since the degree of the numerator is one higher than the degree of the de-
nominator, there is a slant asymptote. To find it, we perform polynomial long
division: 2x2x3−9x2+ 9x
2x3−8x2
−9x2+36x
−27x
10
Thus, the slant asymptote is given by y= 2x−27.
e) To determine where f(x)is increasing or decreasing, we look for the
critical points. Let f′(x)be the derivative of f(x).
f′(x) = d
dx (2x3−9x2+ 9x
x2−4x)
We can simplify this expression and find f′(x).
f) The concavity of f(x)can be determined by analyzing the sign of the
second derivative f′′ (x).
g) Critical points are the points where the derivative is zero or undefined.
h) To find the inflection points, we look for the points where the concavity
changes. Inflection points occur when f′′ (x) = 0 or is undefined.
Question 14
Question
Let f(x) = 4x2−9
x2−3x−4. Determine the domain of the function f(x).
Solution
Step 1: To find the domain of f(x), we need to identify any values of xthat
would make the denominator equal to zero, as division by zero is undefined.
Step 2: Set the denominator x2−3x−4equal to zero and solve for x:
x2−3x−4 = 0
Step 3: We can factor the quadratic equation as follows:
(x−4)(x+ 1) = 0
Step 4: Set each factor equal to zero to find the values of x:
{x−4 = 0 =⇒x= 4
x+ 1 = 0 =⇒x=−1
Step 5: Therefore, the domain of f(x)is all real numbers except x= 4
and x=−1. In interval notation, the domain can be expressed as (−∞,−1) ∪
(−1,4) ∪(4,∞).
Question 15
Question
Given the rational function f(x) = x2−5x+6
x2−4x−5, determine: (a) the domain of f(x),
(b) the x-intercept(s) of the graph of f(x), (c) the y-intercept of the graph of
f(x), and (d) the vertical asymptote(s) of the graph of f(x).
11
Solution
(a) To find the domain of f(x), we need to identify where the denominator
x2−4x−5is not equal to zero. So, we must solve the equation x2−4x−5 = 0
to find any values of xthat make the denominator equal to zero.
x2−4x−5 = 0
(x−5)(x+ 1) = 0
So, x= 5 and x=−1are the values that make the denominator zero. Therefore,
the domain of f(x)is all real numbers except x= 5 and x=−1.
(b) To find the x-intercepts, we need to set f(x) = 0 and solve for x.
x2−5x+ 6
x2−4x−5= 0
x2−5x+ 6 = 0
(x−2)(x−3) = 0
So, x= 2 and x= 3 are the x-intercepts of the graph of f(x).
(c) To find the y-intercept, we need to evaluate f(0).
f(0) = 02−5(0) + 6
02−4(0) −5
=6
−5
=−6
5
Therefore, the y-intercept is at the point (0,−6
5).
(d) To find the vertical asymptotes, we need to identify the values of xfor
which the rational function is undefined. These are the values that make the
denominator zero. In this case, the vertical asymptotes are at x= 5 and x=−1
(the values that we found in part (a)).
Question 16
Question
Given the rational function f(x) = 2x2−5x−3
x2−4x−5, find the vertical asymptotes (if
any) of the function.
Solution
Step 1: We start by factoring the numerator and denominator of the rational
function. Step 2: Factoring the numerator, we get:
2x2−5x−3 = (2x+ 1)(x−3)
12
Step 3: Factoring the denominator, we get:
x2−4x−5 = (x−5)(x+ 1)
Step 4: Now, the rational function can be expressed as:
f(x) = (2x+ 1)(x−3)
(x−5)(x+ 1)
Step 5: To find the vertical asymptotes of the function, we need to identify the
values of xthat would make the denominator equal to zero. Step 6: Setting the
denominator equal to zero and solving for x, we get:
x−5 = 0 =⇒x= 5
x+ 1 = 0 =⇒x=−1
Step 7: Therefore, the vertical asymptotes of the function f(x)are x= 5 and
x=−1.
Question 17
Question
Let f(x) = 2x2−5x−3
x2−x−6and g(x) = x2−4x−5
x2+x−6. Find the domain of f(x) +
g(x).
Solution
Step 1: Determine the domain of f(x)and g(x).
For f(x): The function f(x)is a rational function, and the domain of a
rational function excludes any xvalues that would make the denominator equal
to zero. Thus, we need to find the values of xthat would make the denominator
x2−x−6equal to zero. Solve x2−x−6 = 0:(x−3)(x+ 2) = 0 x= 3,x=−2
Therefore, the domain of f(x)is all real numbers except x= 3 and x=−2.
For g(x): Similar to f(x), the domain of g(x)is all real numbers except
the xvalues that would make the denominator x2+x−6equal to zero. Solve
x2+x−6 = 0:(x+ 3)(x−2) = 0 x=−3,x= 2
Therefore, the domain of g(x)is all real numbers except x=−3and x= 2.
Step 2: Find the domain of f(x) + g(x).
The sum of two functions is defined at any xvalue where both functions are
defined. The domain of f(x)is all real numbers except x= 3 and x=−2, and
the domain of g(x)is all real numbers except x=−3and x= 2.
Thus, the domain of f(x) + g(x)is all real numbers except x= 3,x=−2,
x=−3, and x= 2.
13
Question 18
Question
Solve the rational equation: 4
x+3 −2
x−1=1
x2+2x−3.
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation.
4(x−1)
(x+ 3)(x−1) −2(x+ 3)
(x+ 3)(x−1) =1
x2+ 2x−3
Step 2: Combine the fractions on the left side of the equation.
4(x−1) −2(x+ 3)
(x+ 3)(x−1) =1
x2+ 2x−3
Step 3: Simplify the numerator on the left side of the equation.
4x−4−2x−6
(x+ 3)(x−1) =1
x2+ 2x−3
2x−10
(x+ 3)(x−1) =1
x2+ 2x−3
Step 4: Factor the denominators.
2x−10
(x+ 3)(x−1) =1
(x+ 3)(x−1)
Step 5: Cross multiply to eliminate the denominators.
2x−10 = 1
Step 6: Solve for x.
2x= 11
x=11
2
Step 7: Check for extraneous solutions. Checking the original equation, we
see that x=11
2does not cause any denominator to be equal to zero, so it is a
valid solution.
Therefore, the solution to the rational equation is x=11
2.
Question 19
Question
Simplify the rational function:
2x3−6x2+ 4x
x2−4
14
Solution
Step 1: Factor out the numerator and denominator if possible.
The numerator can be factored as 2x(x−2)(x−1) and the denominator can
be factored as (x−2)(x+ 2).
Step 2: Cancel out common factors.
2x(x−2)(x−1)
(x−2)(x+ 2) =2x(x−1)
x+ 2
Therefore, the simplified form of the rational function is 2x(x−1)
x+2 .
Question 20
Question
Simplify the following rational expression:
3x2−5x−2
x2−4x−5
Solution
Step 1: Factor the numerator and denominator: The numerator can be factored
as 3x2−5x−2 = (3x+1)(x−2) The denominator can be factored as x2−4x−5 =
(x−5)(x+ 1)
Step 2: Rewrite the rational expression using the factored forms:
(3x+ 1)(x−2)
(x−5)(x+ 1)
Step 3: Simplify by canceling out common factors:
(3x+ 1)(x−2)
(x−5)(x+ 1) =3x+ 1
x−5
Therefore, the simplified form of 3x2−5x−2
x2−4x−5is 3x+1
x−5.
Question 21
Question
Simplify the following rational expression:
4x3−16x2
8x2−32x
15
Solution
Step 1: Factor out a 4x2from the numerator and a 8xfrom the denominator
to simplify the expression:
4x2(x−4)
8x(x−4)
Step 2: Simplify the expression further by canceling out common factors in
the numerator and denominator:
4x2(x−4)
8x(x−4)
Step 3: Simplify the remaining expression:
4x2
8x=x
2
Therefore, the simplified form of the rational expression 4x3−16x2
8x2−32xis x
2.
Question 22
Question
Find the domain of the rational function:
f(x) = 3x+ 1
x2−4
Solution
Step 1: The domain of a rational function is all real numbers except for any
values of xthat would make the denominator equal to zero. Therefore, we need
to find the values of xthat would make the denominator x2−4equal to zero.
Step 2: We solve the equation x2−4 = 0 to find the values that would make
the denominator zero:
x2−4 = (x+ 2)(x−2) = 0
Step 3: Setting each factor to zero, we find that x=−2and x= 2 are the
values that make the denominator zero.
Step 4: Therefore, the domain of the rational function is all real numbers
except x=−2and x= 2.
Step 5: In interval notation, the domain can be expressed as (−∞,−2) ∪
(−2,2) ∪(2,∞).
16
Question 23
Question
Solve the rational inequality:
2
x−5≤3
x+ 2
Solution
Step 1: Begin by finding a common denominator for both fractions. In this
case, the common denominator is (x−5)(x+ 2). Step 2: Rewrite the inequality
with the common denominator:
2(x+ 2)
(x−5)(x+ 2) ≤3(x−5)
(x−5)(x+ 2)
Step 3: Simplify the inequality:
2x+ 4
(x−5)(x+ 2) ≤3x−15
(x−5)(x+ 2)
Step 4: Eliminate the denominators by multiplying both sides of the inequality
by (x−5)(x+ 2), noting that we must consider the restriction x=−2,5which
would make the original denominators equal to zero:
2x+ 4 ≤3x−15
Step 5: Simplify the inequality:
4≤x−15
Step 6: Add 15 to both sides of the inequality:
19 ≤x
Step 7: Thus, the solution to the rational inequality is x≥19, with the restric-
tions x=−2,5.
Question 24
Question
Solve the following rational inequality:
(x−3)(x+ 2) >0
17
Solution
Step 1: Find the critical points by setting the expression equal to zero and
solving for x.
(x−3)(x+ 2) = 0
x−3 = 0 or x+ 2 = 0
x= 3 or x=−2
Step 2: Create a sign chart using the critical points. We have critical points
at x=−2and x= 3.
Step 3: Test the intervals created by the critical points in the original in-
equality.
• Test x=−3:((−3) −3)((−3) + 2) = (−6)(−1) = 6 >0
• Test x= 0:(0 −3)(0 + 2) = (−3)(2) = −6<0
• Test x= 4:(4 −3)(4 + 2) = (1)(6) = 6 >0
Step 4: Determine the solution interval. The solution is x < −2or x > 3.
Question 25
Question
Simplify the following rational expression:
3x2−6x−15
x2−3x−18
Solution
Step 1: Factor both the numerator and denominator.
3x2−6x−15 = 3(x2−2x−5) = 3(x−5)(x+ 1)
x2−3x−18 = (x−6)(x+ 3)
Step 2: Rewrite the expression with factored terms.
3(x−5)(x+ 1)
(x−6)(x+ 3)
Step 3: Cancel out common factors in the numerator and denominator.
3(x−5)(x+ 1)
(x−6)(x+ 3) =3(x+ 1)
x+ 3
Therefore, the simplified form of the rational expression is 3(x+1)
x+3 .
18
Question 26
Question
Simplify the following rational expression:
3x2−5x−2
x2−4
Solution
To simplify the rational expression, we need to factor both the numerator and
the denominator.
Step 1: Factor the numerator
3x2−5x−2 = (3x+ 1)(x−2)
Step 2: Factor the denominator
x2−4 = (x+ 2)(x−2)
Now, we can rewrite the rational expression using the factored forms:
3x2−5x−2
x2−4=(3x+ 1)(x−2)
(x+ 2)(x−2)
Step 3: Simplify the expression Since the (x−2) terms in the numerator
and denominator cancel out, we are left with:
(3x+ 1)
(x+ 2)
Therefore, the simplified form of the given rational expression is 3x+1
x+2 .
Question 27
Question
Given the rational function f(x) = 2x2+5x−3
x2−4, determine the vertical asymptotes,
horizontal asymptote, x-intercepts, and y-intercepts (if any).
Solution
Step 1: Find the vertical asymptotes
Vertical asymptotes occur where the denominator of the rational function is
equal to zero. Set the denominator x2−4equal to zero and solve.
x2−4 = 0
19
x2= 4
x=±2
So, the vertical asymptotes are x= 2 and x=−2.
Step 2: Find the horizontal asymptote
To find the horizontal asymptote, we compare the degrees of the numerator and
denominator. Since both have the same degree (2), we divide the coefficients of
the highest degree terms.
lim
x→∞
2x2+ 5x−3
x2−4= lim
x→∞
2
1= 2
So, the horizontal asymptote is y= 2.
Step 3: Find the x-intercepts
To find the x-intercepts, set f(x) = 0 and solve for x.
2x2+ 5x−3
x2−4= 0
2x2+ 5x−3 = 0
This quadratic equation can be factored as (2x−1)(x+ 3) = 0. So the x-
intercepts are x=1
2and x=−3.
Step 4: Find the y-intercept
To find the y-intercept, set x= 0 in the function f(x).
f(0) = 2(0)2+ 5(0) −3
(0)2−4=−3
−4=3
4
Therefore, the y-intercept is at (0,3
4).
Question 28
Question
Given the rational function f(x) = 3x2−2x−5
2x2+4x−6, determine the equations of the
vertical asymptotes, horizontal asymptotes, and oblique asymptotes (if any).
Solution
Step 1: To find the vertical asymptotes, we need to locate the values of xthat
make the denominator zero, as these would result in division by zero. Setting
the denominator 2x2+ 4x−6equal to zero and solving for x, we get:
2x2+ 4x−6 = 0
x2+ 2x−3 = 0
20
(x+ 3)(x−1) = 0
So, x=−3and x= 1 are the equations of the vertical asymptotes.
Step 2: To find the horizontal asymptote, we’ll compare the degrees of the
numerator and denominator. The degree of the numerator is 2, and the degree
of the denominator is also 2. So, to find the horizontal asymptote, we divide the
leading coefficients of the numerator and denominator. The horizontal asymp-
tote is the line y=3
2.
Step 3: To find the oblique asymptote (if any), we’ll perform long division
on the rational function.
3
2x−7
2
2x2+ 4x−6)3x2−2x−5
−(3x2+6x−9)
0−8x4
Therefore, the oblique asymptote is y=3
2x−7
2.
Question 29
Question
Solve the following rational inequality and express the solution set in interval
notation: x−2
x+ 3 ≤1
Solution
Step 1: Begin by finding the critical points by setting the numerator equal to
the denominator:
x−2 = x+ 3
Step 2: Solve for xto find the critical point:
x−x= 3 + 2 =⇒0 = 5
Since the equation 0 = 5 is never true, there are no critical points.
Step 3: Determine the sign of the inequality between the critical points,
x=−∞ and x= +∞:x−2
x+ 3 ≤1
Step 4: Perform the sign analysis by picking test points within each interval.
Test x=−4:−4−2
−4+3 =−6
−1= 6 >1. Hence, the inequality holds in this
interval.
Step 5: Write the final solution using interval notation:
The solution set for x−2
x+3 ≤1is (−∞,−3).
21
Question 30
Question
Let f(x) = 3x2+x−4
x−2. Find the domain of the function f(x).
Solution
Step 1: The domain of a rational function is all real numbers except for the
values that make the denominator equal to zero. In this case, the denominator
is x−2, so we must find the value of xthat makes x−2 = 0. Step 2: Setting
x−2=0, we find that x= 2. Therefore, xcannot be equal to 2 in order for
the function to be defined. Step 3: Therefore, the domain of f(x)is all real
numbers except x= 2, or in interval notation: (−∞,2) ∪(2,∞).
Question 31
Question
Consider the rational function f(x) = 2x2−5x−3
x2−3x−4. Determine the following: a)
Find the domain of the function f(x). b) Find the vertical asymptotes, if any.
c) Find the horizontal asymptotes, if any. d) Find the x-intercepts, if any. e)
Find the y-intercept, if any.
Solution
a) To find the domain of the function f(x), we need to find the values of xfor
which the denominator is not equal to zero. Therefore, we solve the equation
x2−3x−4= 0:
x2−3x−4= 0
(x−4)(x+ 1) = 0
This quadratic is not equal to zero when x= 4 and x=−1. So, the domain of
f(x)is all real numbers except x= 4 and x=−1.
b) To find the vertical asymptotes, we set the denominator equal to zero and
solve for x:
x2−3x−4 = 0
(x−4)(x+ 1) = 0
Vertical asymptotes occur at x= 4 and x=−1.
c) To find the horizontal asymptotes, we compare the degrees of the numer-
ator and denominator. Since both are of degree 2, the horizontal asymptote is
the ratio of the leading coefficients:
y=2
1= 2
22
So, the horizontal asymptote is y= 2.
d) To find the x-intercepts, we set the numerator equal to zero and solve for
x:
2x2−5x−3 = 0
This quadratic equation can be factored as (2x+1)(x−3) = 0, so the x-intercepts
are at x=−1/2and x= 3.
e) To find the y-intercept, we set x= 0 in the function f(x):
f(0) = 2(0)2−5(0) −3
(0)2−3(0) −4=−3
−4=3
4
Therefore, the y-intercept is at (0,3
4).
Question 32
Question
Simplify the following rational expression:
3x2−2x−5
x2−5x+ 6 ∇ · x2−4
x2−9
Solution
Step 1: Factor the numerators and denominators of the rational expression:
3x2−2x−5
x2−5x+ 6 ∇ · x2−4
x2−9
=(3x+ 1)(x−5)
(x−3)(x−2) ∇ · (x+ 2)(x−2)
(x+ 3)(x−3)
Step 2: Rewrite the division as multiplication by flipping the second fraction:
=(3x+ 1)(x−5)
(x−3)(x−2) ×(x+ 3)(x−3)
(x+ 2)(x−2)
Step 3: Cancel out common factors in the numerator and denominator:
=(3x+ 1)(x−5)
(x−3)(x−2) ×(x+ 3)(x−3)
(x+ 2)(x−2)
Step 4: Multiply the remaining factors in the numerator and denominator:
=3x+ 1
x+ 2
Therefore, the simplified form of the rational expression is 3x+1
x+2 .
23
Question 33
Question
Simplify the rational expression: 4x2−3x−4
2x2−5x−3.
Solution
Step 1: Factor the numerator and denominator of the rational expression.
To factor 4x2−3x−4, we need to find two numbers that multiply to 4(−4) = −16
and add up to −3. The numbers are −4and 1. So, we can write:
4x2−3x−4 = (4x+ 1)(x−4)
Similarly, to factor 2x2−5x−3, we need to find two numbers that multiply to
2(−3) = −6and add up to −5. The numbers are −6and 1. So, we can write:
2x2−5x−3 = (2x+ 1)(x−3)
Step 2: Rewrite the rational expression with the factored forms of the nu-
merator and denominator.
4x2−3x−4
2x2−5x−3=(4x+ 1)(x−4)
(2x+ 1)(x−3)
Step 3: Simplify the expression by canceling out common factors between
the numerator and the denominator.
(4x+ 1)(x−4)
(2x+ 1)(x−3) =4x+ 1
2x+ 1 ·x−4
x−3
Thus, the simplified form of the rational expression is 4x+1
2x+1 ·x−4
x−3.
Question 34
Question
Simplify the rational expression:
4x2−9
x2−5x+ 6 −2x−3
x2−2x−8
Solution
Step 1: Factor the denominators of both rational expressions.
• The denominator x2−5x+ 6 factors as (x−2)(x−3).
• The denominator x2−2x−8factors as (x−4)(x+ 2).
24
Step 2: Write the rational expressions with the factored denominators.
4x2−9
(x−2)(x−3) −2x−3
(x−4)(x+ 2)
Step 3: Determine the least common denominator (LCD) of the two rational
expressions. The LCD is (x−2)(x−3)(x−4)(x+ 2).
Step 4: Rewrite the fractions with the common denominator.
(4x2−9)(x−4)(x+ 2)
(x−2)(x−3)(x−4)(x+ 2) −(2x−3)(x−2)(x−3)
(x−2)(x−3)(x−4)(x+ 2)
Step 5: Simplify the numerators.
• Simplify the first numerator:
(4x2−9)(x−4)(x+ 2) = (2x+ 3)(2x−3)(x−4)(x+ 2)
= (2x−3)(x−2)(x+ 3)(x−4)
• Simplify the second numerator:
(2x−3)(x−2)(x−3) = (2x−3)(x−3)(x−2)
= (2x−3)(x−4)(x+ 2)
Step 6: Substitute the simplified numerators back into the expression.
(2x−3)(x−2)(x+ 3)(x−4)
(x−2)(x−3)(x−4)(x+ 2) −(2x−3)(x−4)(x+ 2)
(x−2)(x−3)(x−4)(x+ 2)
Step 7: Combine the fractions.
(2x−3)(x−2)(x+ 3)(x−4) −(2x−3)(x−4)(x+ 2)
(x−2)(x−3)(x−4)(x+ 2)
Step 8: Factor out the common factor (2x−3)(x−4) from the numerator.
(2x−3)(x−4)(x+ 3 −x+ 2)
(x−2)(x−3)(x−4)(x+ 2) =(2x−3)(x−4)(5)
(x−2)(x−3)(x−4)(x+ 2)
=5(2x−3)
(x−2)(x−3)(x+ 2)
Hence, the simplified form of the given rational expression is 5(2x−3)
(x−2)(x−3)(x+ 2) .
Question 35
Question
Let f(x) = 5x2−7x−2
x2−4x−5be a rational function. Find the vertical asymptotes
of f(x).
25
Question 2
Question
Find the vertical asymptotes, horizontal asymptotes, and the x- and y-intercepts
of the rational function:
f(x) = 2x3−5x2−3x
x2−9
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes of the function, we set the denominator equal
to zero and solve for x:
x2−9 = 0
(x+ 3)(x−3) = 0
So, the vertical asymptotes are x=−3and x= 3.
Step 2: Horizontal Asymptotes
To find the horizontal asymptote, we look at the degrees of the numerator and
denominator:
Since the degree of the numerator (3) is greater than the degree of the
denominator (2), there is no horizontal asymptote.
Step 3: X-intercepts
To find the x-intercepts, we set the numerator equal to zero and solve for x:
2x3−5x2−3x= 0
Factor out an x:
x(2x2−5x−3) = 0
x(2x+ 1)(x−3) = 0
So, the x-intercepts are x= 0,x=−1
2, and x= 3.
Step 4: Y-intercepts
To find the y-intercept, we plug in x= 0 into the function:
f(0) = 2(0)3−5(0)2−3(0)
(0)2−9
f(0) = 0
So, the y-intercept is at the point (0,0).
2
Question 3
Question
Find the domain of the rational function:
f(x) = 4x
x2−9.
Solution
Step 1: The domain of a rational function is all real numbers except those that
make the denominator equal to zero.
Step 2: Set the denominator x2−9equal to zero and solve for x:
x2−9 = 0.
Step 3: Factor the quadratic equation:
(x−3)(x+ 3) = 0.
Step 4: Set each factor equal to zero:
x−3 = 0 and x+ 3 = 0.
Step 5: Solve for xto find the critical points:
x= 3 and x=−3.
Step 6: The domain of the rational function f(x) = 4x
x2−9is all real numbers
except x= 3 and x=−3.
Step 7: Therefore, the domain of f(x)is (−∞,−3) ∪(−3,3) ∪(3,∞).
Question 4
Question
Let f(x) = 2x2−5x−3
x2−9. Find the domain of the function f(x).
Solution
Step 1: To find the domain of a rational function, we need to identify the values
of xwhere the function is undefined. This occurs when the denominator is equal
to zero. In this case, the denominator x2−9is equal to zero when x=−3or
x= 3.
Step 2: Therefore, the domain of the function f(x)is all real numbers except
x=−3and x= 3. We can express this as:
{x∈R|x=−3, x = 3}
So, the domain of the function f(x)is all real numbers except −3and 3.
3
Question 5
Question
Find the domain of the rational function: f(x) = 1
x2−5x+ 6.
Solution
Step 1: Determine the values of xthat would make the denominator equal to
zero, as those values would make the function undefined.
x2−5x+ 6 = 0
Step 2: Factor the quadratic equation.
x2−5x+ 6 = (x−2)(x−3)
Step 3: Set each factor to zero and solve for x.
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 4: The function is undefined at x= 2 and x= 3, so the domain of
the function f(x)is all real numbers except x= 2 and x= 3, which can be
expressed as:
(−∞,2) ∪(2,3) ∪(3,∞)
Question 6
Question
Simplify the rational function:
2x2−x−3
x2−4
Solution
Step 1: Factor the numerator and denominator:
The numerator 2x2−x−3can be factored as (2x+ 1)(x−3).
The denominator x2−4 = (x+ 2)(x−2).
Step 2: Rewrite the rational function:
2x2−x−3
x2−4=(2x+ 1)(x−3)
(x+ 2)(x−2)
Step 3: Simplify the rational function:
4
2x2−x−3
x2−4=(2x+ 1)(x−3)
(x+ 2)(x−2) =2x+ 1
x+ 2
Therefore, the simplified form of the rational function is 2x+1
x+2 .
Question 7
Question
Simplify the rational function:
7x2−2x−3
2x2+ 5x−3
Solution
Step 1: Factor both the numerator and denominator of the rational function.
Step 2: Simplify the rational function by canceling out any common factors in
the numerator and denominator.
Step 1: Factor the numerator and denominator: The numerator, 7x2−2x−
3, can be factored as (7x+ 3)(x−1).
The denominator, 2x2+ 5x−3, can be factored as (2x−1)(x+ 3).
Therefore, the rational function becomes:
(7x+ 3)(x−1)
(2x−1)(x+ 3)
Step 2: Simplify the rational function: Since there are no common factors
that can be canceled out in the numerator and denominator, the simplified form
of the rational function is:
7x+ 3
2x−1·x−1
x+ 3
Therefore, the simplified form of the rational function is (7x+3)(x−1)
(2x−1)(x+3) .
Question 8
Question
Simplify the rational function:
5x3−16x2−5x+ 16
x2−4x−5
5
Solution
Step 1: Factor the numerator and denominator:
5x3−16x2−5x+ 16
x2−4x−5=(5x3−20x2) + (4x2−5x+ 16)
(x−5)(x+ 1)
Step 2: Combine like terms in the numerator:
=5x2(x−4) + 1(4x2−5x+ 16)
(x−5)(x+ 1)
Step 3: Factor out common factors:
=5x2(x−4) + 1(4x−16)
(x−5)(x+ 1) =5x2(x−4) + 4(x−4)
(x−5)(x+ 1)
Step 4: Factor out the common factor of (x−4):
=(5x2+ 4)(x−4)
(x−5)(x+ 1)
Therefore, the simplified form of the rational function is:
(5x2+ 4)(x−4)
(x−5)(x+ 1)
Question 9
Question
Let f(x) = 2x2−3x−2
x2−5x+6 be a rational function. Find the x-intercepts, y-intercepts,
and vertical asymptotes, if any.
Solution
Step 1: Find the x-intercepts by setting f(x) = 0 and solving for x.
0 = 2x2−3x−2
x2−5x+ 6
0 = 2x2−3x−2
We can factor the quadratic equation to find the x-intercepts.
0 = (2x+ 1)(x−2)
Setting each factor to zero gives us x-intercepts at x=−1
2and x= 2.
6
Step 2: Find the y-intercept by setting x= 0.
y=2(0)2−3(0) −2
(0)2−5(0) + 6
y=−2
6
The y-intercept is at (0,−1
3).
Step 3: Find the vertical asymptotes by setting the denominator of f(x)
equal to zero and solving for x.
x2−5x+ 6 = 0
x= 2 or x= 3
Therefore, there are vertical asymptotes at x= 2 and x= 3 for the rational
function f(x).
In summary, the x-intercepts are at x=−1
2and x= 2, the y-intercept is at
(0,−1
3), and there are vertical asymptotes at x= 2 and x= 3 for the rational
function f(x).
Question 10
Question
Given the rational function f(x) = 2x2−7x+3
x2−4x+3 , find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph of f(x).
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes of f(x), we need to determine the values
of xwhere the denominator of the rational function is equal to zero, but the
numerator is not zero.
First, factor the denominator x2−4x+ 3:
x2−4x+ 3 = (x−1)(x−3).
Therefore, the vertical asymptotes occur at x= 1 and x= 3 since these
values make the denominator zero but not the numerator.
Step 2: Horizontal Asymptotes
To find the horizontal asymptotes of f(x), we need to compare the degrees
of the numerators and denominators.
Since the degree of the numerator is 2 and the degree of the denomina-
tor is also 2, we can find the horizontal asymptotes by comparing the leading
coefficients:
lim
x→∞
2x2−7x+ 3
x2−4x+ 3 = lim
x→∞
2
1= 2.
7
Therefore, the horizontal asymptote is at y= 2.
Step 3: Hole in the Graph
To find any holes in the graph of f(x), we need to check for common factors
between the numerator and denominator that can be canceled out.
Factor the numerator 2x2−7x+ 3:
2x2−7x+ 3 = (2x−1)(x−3).
We can see that there is a common factor of (x−3) in both the numerator
and the denominator, which can be canceled out. Therefore, there is a hole in
the graph at x= 3.
Thus, the vertical asymptotes are at x= 1 and x= 3, the horizontal asymp-
tote is at y= 2, and there is a hole in the graph at x= 3.
Question 11
Question
Simplify the following rational expression:
3x2−6x+ 3
x2−1
Solution
Step 1: Factor the numerator and denominator:
3x2−6x+ 3
x2−1=3(x2−2x+ 1)
(x−1)(x+ 1)
Step 2: Simplify the expression further by factoring the quadratic term:
3(x2−2x+ 1)
(x−1)(x+ 1) =3(x−1)2
(x−1)(x+ 1)
Step 3: Cancel out the common factor of (x−1) from the numerator and
denominator: 3(x−1)2
(x−1)(x+ 1) =3(x−1)
x+ 1
Therefore, the simplified form of the rational expression is 3(x−1)
x+1 .
Question 12
Question
The rational function
f(x) = 3x2−5x−2
x2−4x−5
8
can be rewritten as
f(x) = A
x+ 1 +B
x−5
where Aand Bare constants. Find the values of Aand B.
Solution
Step 1: First, we rewrite the given function in partial fraction form:
f(x) = 3x2−5x−2
x2−4x−5=A
x+ 1 +B
x−5
Step 2: Multiply both sides by the denominator (x2−4x−5) to clear the
fractions:
3x2−5x−2 = A(x−5) + B(x+ 1)
Step 3: Expand the right side and group like terms:
3x2−5x−2 = Ax −5A+Bx +B
Step 4: Combine like terms:
3x2−5x−2 = (A+B)x−5A+B
Step 5: Equate the coefficients of like terms on both sides:
3 = A+B
−5 = −5A+B
Step 6: Solve the system of equations by substitution or elimination to find
the values of Aand B. From the first equation, we get B= 3 −A. Substituting
this into the second equation:
−5 = −5A+ (3 −A)
Step 7: Solve for A:
−5 = −5A+ 3 −A
−5 = −6A+ 3
−8 = −6A
A=4
3
Step 8: Substitute A=4
3back into B= 3 −Ato find B:
B= 3 −4
3=5
3
Therefore, the values of Aand Bare A=4
3and B=5
3.
9
Question 13
Question
Let f(x) = 2x3−9x2+9x
x2−4x. Find the following: a) Domain of fb) x-intercepts
(if any) c) y-intercept (if any) d) Asymptotes (vertical, horizontal, and slant)
e) Intervals where f(x)is increasing or decreasing f) Intervals where f(x)is
concave up or down g) Critical points h) Inflection points (if any)
Solution
a) To find the domain of f(x), we need to consider the values that xcan take
without causing division by zero. Thus, the domain of fis all real numbers
except the ones that make the denominator zero. Therefore, the domain of fis
(−∞,0) ∪(0,4) ∪(4,∞).
b) To find the x-intercepts, we set f(x) = 0 and solve for x:
2x3−9x2+ 9x
x2−4x= 0
Notice that we can factor out an xfrom the numerator:
x(2x2−9x+ 9)
x(x−4) = 0
This simplifies to 2x2−9x+9 = 0, which does not have real solutions. Therefore,
there are no x-intercepts.
c) To find the y-intercept, we evaluate f(0):
f(0) = 2(0)3−9(0)2+ 9(0)
(0)2−4(0) = 0
So, the y-intercept is at the point (0,0).
d) To find the asymptotes: Vertical asymptotes occur where the denominator
is zero, so there is a vertical asymptote at x= 0.
Horizontal asymptotes are found by comparing the degrees of the numerator
and denominator. Since the degree of the numerator is equal to the degree of the
denominator, we divide the leading coefficients to find the horizontal asymptote:
lim
x→∞
2x3
x2= lim
x→∞ 2x=∞
Therefore, there is no horizontal asymptote.
Since the degree of the numerator is one higher than the degree of the de-
nominator, there is a slant asymptote. To find it, we perform polynomial long
division: 2x2x3−9x2+ 9x
2x3−8x2
−9x2+36x
−27x
10
Thus, the slant asymptote is given by y= 2x−27.
e) To determine where f(x)is increasing or decreasing, we look for the
critical points. Let f′(x)be the derivative of f(x).
f′(x) = d
dx (2x3−9x2+ 9x
x2−4x)
We can simplify this expression and find f′(x).
f) The concavity of f(x)can be determined by analyzing the sign of the
second derivative f′′ (x).
g) Critical points are the points where the derivative is zero or undefined.
h) To find the inflection points, we look for the points where the concavity
changes. Inflection points occur when f′′ (x) = 0 or is undefined.
Question 14
Question
Let f(x) = 4x2−9
x2−3x−4. Determine the domain of the function f(x).
Solution
Step 1: To find the domain of f(x), we need to identify any values of xthat
would make the denominator equal to zero, as division by zero is undefined.
Step 2: Set the denominator x2−3x−4equal to zero and solve for x:
x2−3x−4 = 0
Step 3: We can factor the quadratic equation as follows:
(x−4)(x+ 1) = 0
Step 4: Set each factor equal to zero to find the values of x:
{x−4 = 0 =⇒x= 4
x+ 1 = 0 =⇒x=−1
Step 5: Therefore, the domain of f(x)is all real numbers except x= 4
and x=−1. In interval notation, the domain can be expressed as (−∞,−1) ∪
(−1,4) ∪(4,∞).
Question 15
Question
Given the rational function f(x) = x2−5x+6
x2−4x−5, determine: (a) the domain of f(x),
(b) the x-intercept(s) of the graph of f(x), (c) the y-intercept of the graph of
f(x), and (d) the vertical asymptote(s) of the graph of f(x).
11
Solution
(a) To find the domain of f(x), we need to identify where the denominator
x2−4x−5is not equal to zero. So, we must solve the equation x2−4x−5 = 0
to find any values of xthat make the denominator equal to zero.
x2−4x−5 = 0
(x−5)(x+ 1) = 0
So, x= 5 and x=−1are the values that make the denominator zero. Therefore,
the domain of f(x)is all real numbers except x= 5 and x=−1.
(b) To find the x-intercepts, we need to set f(x) = 0 and solve for x.
x2−5x+ 6
x2−4x−5= 0
x2−5x+ 6 = 0
(x−2)(x−3) = 0
So, x= 2 and x= 3 are the x-intercepts of the graph of f(x).
(c) To find the y-intercept, we need to evaluate f(0).
f(0) = 02−5(0) + 6
02−4(0) −5
=6
−5
=−6
5
Therefore, the y-intercept is at the point (0,−6
5).
(d) To find the vertical asymptotes, we need to identify the values of xfor
which the rational function is undefined. These are the values that make the
denominator zero. In this case, the vertical asymptotes are at x= 5 and x=−1
(the values that we found in part (a)).
Question 16
Question
Given the rational function f(x) = 2x2−5x−3
x2−4x−5, find the vertical asymptotes (if
any) of the function.
Solution
Step 1: We start by factoring the numerator and denominator of the rational
function. Step 2: Factoring the numerator, we get:
2x2−5x−3 = (2x+ 1)(x−3)
12
Step 3: Factoring the denominator, we get:
x2−4x−5 = (x−5)(x+ 1)
Step 4: Now, the rational function can be expressed as:
f(x) = (2x+ 1)(x−3)
(x−5)(x+ 1)
Step 5: To find the vertical asymptotes of the function, we need to identify the
values of xthat would make the denominator equal to zero. Step 6: Setting the
denominator equal to zero and solving for x, we get:
x−5 = 0 =⇒x= 5
x+ 1 = 0 =⇒x=−1
Step 7: Therefore, the vertical asymptotes of the function f(x)are x= 5 and
x=−1.
Question 17
Question
Let f(x) = 2x2−5x−3
x2−x−6and g(x) = x2−4x−5
x2+x−6. Find the domain of f(x) +
g(x).
Solution
Step 1: Determine the domain of f(x)and g(x).
For f(x): The function f(x)is a rational function, and the domain of a
rational function excludes any xvalues that would make the denominator equal
to zero. Thus, we need to find the values of xthat would make the denominator
x2−x−6equal to zero. Solve x2−x−6 = 0:(x−3)(x+ 2) = 0 x= 3,x=−2
Therefore, the domain of f(x)is all real numbers except x= 3 and x=−2.
For g(x): Similar to f(x), the domain of g(x)is all real numbers except
the xvalues that would make the denominator x2+x−6equal to zero. Solve
x2+x−6 = 0:(x+ 3)(x−2) = 0 x=−3,x= 2
Therefore, the domain of g(x)is all real numbers except x=−3and x= 2.
Step 2: Find the domain of f(x) + g(x).
The sum of two functions is defined at any xvalue where both functions are
defined. The domain of f(x)is all real numbers except x= 3 and x=−2, and
the domain of g(x)is all real numbers except x=−3and x= 2.
Thus, the domain of f(x) + g(x)is all real numbers except x= 3,x=−2,
x=−3, and x= 2.
13
Question 18
Question
Solve the rational equation: 4
x+3 −2
x−1=1
x2+2x−3.
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation.
4(x−1)
(x+ 3)(x−1) −2(x+ 3)
(x+ 3)(x−1) =1
x2+ 2x−3
Step 2: Combine the fractions on the left side of the equation.
4(x−1) −2(x+ 3)
(x+ 3)(x−1) =1
x2+ 2x−3
Step 3: Simplify the numerator on the left side of the equation.
4x−4−2x−6
(x+ 3)(x−1) =1
x2+ 2x−3
2x−10
(x+ 3)(x−1) =1
x2+ 2x−3
Step 4: Factor the denominators.
2x−10
(x+ 3)(x−1) =1
(x+ 3)(x−1)
Step 5: Cross multiply to eliminate the denominators.
2x−10 = 1
Step 6: Solve for x.
2x= 11
x=11
2
Step 7: Check for extraneous solutions. Checking the original equation, we
see that x=11
2does not cause any denominator to be equal to zero, so it is a
valid solution.
Therefore, the solution to the rational equation is x=11
2.
Question 19
Question
Simplify the rational function:
2x3−6x2+ 4x
x2−4
14
Solution
Step 1: Factor out the numerator and denominator if possible.
The numerator can be factored as 2x(x−2)(x−1) and the denominator can
be factored as (x−2)(x+ 2).
Step 2: Cancel out common factors.
2x(x−2)(x−1)
(x−2)(x+ 2) =2x(x−1)
x+ 2
Therefore, the simplified form of the rational function is 2x(x−1)
x+2 .
Question 20
Question
Simplify the following rational expression:
3x2−5x−2
x2−4x−5
Solution
Step 1: Factor the numerator and denominator: The numerator can be factored
as 3x2−5x−2 = (3x+1)(x−2) The denominator can be factored as x2−4x−5 =
(x−5)(x+ 1)
Step 2: Rewrite the rational expression using the factored forms:
(3x+ 1)(x−2)
(x−5)(x+ 1)
Step 3: Simplify by canceling out common factors:
(3x+ 1)(x−2)
(x−5)(x+ 1) =3x+ 1
x−5
Therefore, the simplified form of 3x2−5x−2
x2−4x−5is 3x+1
x−5.
Question 21
Question
Simplify the following rational expression:
4x3−16x2
8x2−32x
15
Solution
Step 1: Factor out a 4x2from the numerator and a 8xfrom the denominator
to simplify the expression:
4x2(x−4)
8x(x−4)
Step 2: Simplify the expression further by canceling out common factors in
the numerator and denominator:
4x2(x−4)
8x(x−4)
Step 3: Simplify the remaining expression:
4x2
8x=x
2
Therefore, the simplified form of the rational expression 4x3−16x2
8x2−32xis x
2.
Question 22
Question
Find the domain of the rational function:
f(x) = 3x+ 1
x2−4
Solution
Step 1: The domain of a rational function is all real numbers except for any
values of xthat would make the denominator equal to zero. Therefore, we need
to find the values of xthat would make the denominator x2−4equal to zero.
Step 2: We solve the equation x2−4 = 0 to find the values that would make
the denominator zero:
x2−4 = (x+ 2)(x−2) = 0
Step 3: Setting each factor to zero, we find that x=−2and x= 2 are the
values that make the denominator zero.
Step 4: Therefore, the domain of the rational function is all real numbers
except x=−2and x= 2.
Step 5: In interval notation, the domain can be expressed as (−∞,−2) ∪
(−2,2) ∪(2,∞).
16
Question 23
Question
Solve the rational inequality:
2
x−5≤3
x+ 2
Solution
Step 1: Begin by finding a common denominator for both fractions. In this
case, the common denominator is (x−5)(x+ 2). Step 2: Rewrite the inequality
with the common denominator:
2(x+ 2)
(x−5)(x+ 2) ≤3(x−5)
(x−5)(x+ 2)
Step 3: Simplify the inequality:
2x+ 4
(x−5)(x+ 2) ≤3x−15
(x−5)(x+ 2)
Step 4: Eliminate the denominators by multiplying both sides of the inequality
by (x−5)(x+ 2), noting that we must consider the restriction x=−2,5which
would make the original denominators equal to zero:
2x+ 4 ≤3x−15
Step 5: Simplify the inequality:
4≤x−15
Step 6: Add 15 to both sides of the inequality:
19 ≤x
Step 7: Thus, the solution to the rational inequality is x≥19, with the restric-
tions x=−2,5.
Question 24
Question
Solve the following rational inequality:
(x−3)(x+ 2) >0
17
Solution
Step 1: Find the critical points by setting the expression equal to zero and
solving for x.
(x−3)(x+ 2) = 0
x−3 = 0 or x+ 2 = 0
x= 3 or x=−2
Step 2: Create a sign chart using the critical points. We have critical points
at x=−2and x= 3.
Step 3: Test the intervals created by the critical points in the original in-
equality.
• Test x=−3:((−3) −3)((−3) + 2) = (−6)(−1) = 6 >0
• Test x= 0:(0 −3)(0 + 2) = (−3)(2) = −6<0
• Test x= 4:(4 −3)(4 + 2) = (1)(6) = 6 >0
Step 4: Determine the solution interval. The solution is x < −2or x > 3.
Question 25
Question
Simplify the following rational expression:
3x2−6x−15
x2−3x−18
Solution
Step 1: Factor both the numerator and denominator.
3x2−6x−15 = 3(x2−2x−5) = 3(x−5)(x+ 1)
x2−3x−18 = (x−6)(x+ 3)
Step 2: Rewrite the expression with factored terms.
3(x−5)(x+ 1)
(x−6)(x+ 3)
Step 3: Cancel out common factors in the numerator and denominator.
3(x−5)(x+ 1)
(x−6)(x+ 3) =3(x+ 1)
x+ 3
Therefore, the simplified form of the rational expression is 3(x+1)
x+3 .
18
Question 26
Question
Simplify the following rational expression:
3x2−5x−2
x2−4
Solution
To simplify the rational expression, we need to factor both the numerator and
the denominator.
Step 1: Factor the numerator
3x2−5x−2 = (3x+ 1)(x−2)
Step 2: Factor the denominator
x2−4 = (x+ 2)(x−2)
Now, we can rewrite the rational expression using the factored forms:
3x2−5x−2
x2−4=(3x+ 1)(x−2)
(x+ 2)(x−2)
Step 3: Simplify the expression Since the (x−2) terms in the numerator
and denominator cancel out, we are left with:
(3x+ 1)
(x+ 2)
Therefore, the simplified form of the given rational expression is 3x+1
x+2 .
Question 27
Question
Given the rational function f(x) = 2x2+5x−3
x2−4, determine the vertical asymptotes,
horizontal asymptote, x-intercepts, and y-intercepts (if any).
Solution
Step 1: Find the vertical asymptotes
Vertical asymptotes occur where the denominator of the rational function is
equal to zero. Set the denominator x2−4equal to zero and solve.
x2−4 = 0
19
x2= 4
x=±2
So, the vertical asymptotes are x= 2 and x=−2.
Step 2: Find the horizontal asymptote
To find the horizontal asymptote, we compare the degrees of the numerator and
denominator. Since both have the same degree (2), we divide the coefficients of
the highest degree terms.
lim
x→∞
2x2+ 5x−3
x2−4= lim
x→∞
2
1= 2
So, the horizontal asymptote is y= 2.
Step 3: Find the x-intercepts
To find the x-intercepts, set f(x) = 0 and solve for x.
2x2+ 5x−3
x2−4= 0
2x2+ 5x−3 = 0
This quadratic equation can be factored as (2x−1)(x+ 3) = 0. So the x-
intercepts are x=1
2and x=−3.
Step 4: Find the y-intercept
To find the y-intercept, set x= 0 in the function f(x).
f(0) = 2(0)2+ 5(0) −3
(0)2−4=−3
−4=3
4
Therefore, the y-intercept is at (0,3
4).
Question 28
Question
Given the rational function f(x) = 3x2−2x−5
2x2+4x−6, determine the equations of the
vertical asymptotes, horizontal asymptotes, and oblique asymptotes (if any).
Solution
Step 1: To find the vertical asymptotes, we need to locate the values of xthat
make the denominator zero, as these would result in division by zero. Setting
the denominator 2x2+ 4x−6equal to zero and solving for x, we get:
2x2+ 4x−6 = 0
x2+ 2x−3 = 0
20
(x+ 3)(x−1) = 0
So, x=−3and x= 1 are the equations of the vertical asymptotes.
Step 2: To find the horizontal asymptote, we’ll compare the degrees of the
numerator and denominator. The degree of the numerator is 2, and the degree
of the denominator is also 2. So, to find the horizontal asymptote, we divide the
leading coefficients of the numerator and denominator. The horizontal asymp-
tote is the line y=3
2.
Step 3: To find the oblique asymptote (if any), we’ll perform long division
on the rational function.
3
2x−7
2
2x2+ 4x−6)3x2−2x−5
−(3x2+6x−9)
0−8x4
Therefore, the oblique asymptote is y=3
2x−7
2.
Question 29
Question
Solve the following rational inequality and express the solution set in interval
notation: x−2
x+ 3 ≤1
Solution
Step 1: Begin by finding the critical points by setting the numerator equal to
the denominator:
x−2 = x+ 3
Step 2: Solve for xto find the critical point:
x−x= 3 + 2 =⇒0 = 5
Since the equation 0 = 5 is never true, there are no critical points.
Step 3: Determine the sign of the inequality between the critical points,
x=−∞ and x= +∞:x−2
x+ 3 ≤1
Step 4: Perform the sign analysis by picking test points within each interval.
Test x=−4:−4−2
−4+3 =−6
−1= 6 >1. Hence, the inequality holds in this
interval.
Step 5: Write the final solution using interval notation:
The solution set for x−2
x+3 ≤1is (−∞,−3).
21
Question 30
Question
Let f(x) = 3x2+x−4
x−2. Find the domain of the function f(x).
Solution
Step 1: The domain of a rational function is all real numbers except for the
values that make the denominator equal to zero. In this case, the denominator
is x−2, so we must find the value of xthat makes x−2 = 0. Step 2: Setting
x−2=0, we find that x= 2. Therefore, xcannot be equal to 2 in order for
the function to be defined. Step 3: Therefore, the domain of f(x)is all real
numbers except x= 2, or in interval notation: (−∞,2) ∪(2,∞).
Question 31
Question
Consider the rational function f(x) = 2x2−5x−3
x2−3x−4. Determine the following: a)
Find the domain of the function f(x). b) Find the vertical asymptotes, if any.
c) Find the horizontal asymptotes, if any. d) Find the x-intercepts, if any. e)
Find the y-intercept, if any.
Solution
a) To find the domain of the function f(x), we need to find the values of xfor
which the denominator is not equal to zero. Therefore, we solve the equation
x2−3x−4= 0:
x2−3x−4= 0
(x−4)(x+ 1) = 0
This quadratic is not equal to zero when x= 4 and x=−1. So, the domain of
f(x)is all real numbers except x= 4 and x=−1.
b) To find the vertical asymptotes, we set the denominator equal to zero and
solve for x:
x2−3x−4 = 0
(x−4)(x+ 1) = 0
Vertical asymptotes occur at x= 4 and x=−1.
c) To find the horizontal asymptotes, we compare the degrees of the numer-
ator and denominator. Since both are of degree 2, the horizontal asymptote is
the ratio of the leading coefficients:
y=2
1= 2
22
So, the horizontal asymptote is y= 2.
d) To find the x-intercepts, we set the numerator equal to zero and solve for
x:
2x2−5x−3 = 0
This quadratic equation can be factored as (2x+1)(x−3) = 0, so the x-intercepts
are at x=−1/2and x= 3.
e) To find the y-intercept, we set x= 0 in the function f(x):
f(0) = 2(0)2−5(0) −3
(0)2−3(0) −4=−3
−4=3
4
Therefore, the y-intercept is at (0,3
4).
Question 32
Question
Simplify the following rational expression:
3x2−2x−5
x2−5x+ 6 ∇ · x2−4
x2−9
Solution
Step 1: Factor the numerators and denominators of the rational expression:
3x2−2x−5
x2−5x+ 6 ∇ · x2−4
x2−9
=(3x+ 1)(x−5)
(x−3)(x−2) ∇ · (x+ 2)(x−2)
(x+ 3)(x−3)
Step 2: Rewrite the division as multiplication by flipping the second fraction:
=(3x+ 1)(x−5)
(x−3)(x−2) ×(x+ 3)(x−3)
(x+ 2)(x−2)
Step 3: Cancel out common factors in the numerator and denominator:
=(3x+ 1)(x−5)
(x−3)(x−2) ×(x+ 3)(x−3)
(x+ 2)(x−2)
Step 4: Multiply the remaining factors in the numerator and denominator:
=3x+ 1
x+ 2
Therefore, the simplified form of the rational expression is 3x+1
x+2 .
23
Question 33
Question
Simplify the rational expression: 4x2−3x−4
2x2−5x−3.
Solution
Step 1: Factor the numerator and denominator of the rational expression.
To factor 4x2−3x−4, we need to find two numbers that multiply to 4(−4) = −16
and add up to −3. The numbers are −4and 1. So, we can write:
4x2−3x−4 = (4x+ 1)(x−4)
Similarly, to factor 2x2−5x−3, we need to find two numbers that multiply to
2(−3) = −6and add up to −5. The numbers are −6and 1. So, we can write:
2x2−5x−3 = (2x+ 1)(x−3)
Step 2: Rewrite the rational expression with the factored forms of the nu-
merator and denominator.
4x2−3x−4
2x2−5x−3=(4x+ 1)(x−4)
(2x+ 1)(x−3)
Step 3: Simplify the expression by canceling out common factors between
the numerator and the denominator.
(4x+ 1)(x−4)
(2x+ 1)(x−3) =4x+ 1
2x+ 1 ·x−4
x−3
Thus, the simplified form of the rational expression is 4x+1
2x+1 ·x−4
x−3.
Question 34
Question
Simplify the rational expression:
4x2−9
x2−5x+ 6 −2x−3
x2−2x−8
Solution
Step 1: Factor the denominators of both rational expressions.
• The denominator x2−5x+ 6 factors as (x−2)(x−3).
• The denominator x2−2x−8factors as (x−4)(x+ 2).
24
Step 2: Write the rational expressions with the factored denominators.
4x2−9
(x−2)(x−3) −2x−3
(x−4)(x+ 2)
Step 3: Determine the least common denominator (LCD) of the two rational
expressions. The LCD is (x−2)(x−3)(x−4)(x+ 2).
Step 4: Rewrite the fractions with the common denominator.
(4x2−9)(x−4)(x+ 2)
(x−2)(x−3)(x−4)(x+ 2) −(2x−3)(x−2)(x−3)
(x−2)(x−3)(x−4)(x+ 2)
Step 5: Simplify the numerators.
• Simplify the first numerator:
(4x2−9)(x−4)(x+ 2) = (2x+ 3)(2x−3)(x−4)(x+ 2)
= (2x−3)(x−2)(x+ 3)(x−4)
• Simplify the second numerator:
(2x−3)(x−2)(x−3) = (2x−3)(x−3)(x−2)
= (2x−3)(x−4)(x+ 2)
Step 6: Substitute the simplified numerators back into the expression.
(2x−3)(x−2)(x+ 3)(x−4)
(x−2)(x−3)(x−4)(x+ 2) −(2x−3)(x−4)(x+ 2)
(x−2)(x−3)(x−4)(x+ 2)
Step 7: Combine the fractions.
(2x−3)(x−2)(x+ 3)(x−4) −(2x−3)(x−4)(x+ 2)
(x−2)(x−3)(x−4)(x+ 2)
Step 8: Factor out the common factor (2x−3)(x−4) from the numerator.
(2x−3)(x−4)(x+ 3 −x+ 2)
(x−2)(x−3)(x−4)(x+ 2) =(2x−3)(x−4)(5)
(x−2)(x−3)(x−4)(x+ 2)
=5(2x−3)
(x−2)(x−3)(x+ 2)
Hence, the simplified form of the given rational expression is 5(2x−3)
(x−2)(x−3)(x+ 2) .
Question 35
Question
Let f(x) = 5x2−7x−2
x2−4x−5be a rational function. Find the vertical asymptotes
of f(x).
25
Solution
Step 1: Determine where the denominator equals zero to find the potential
vertical asymptotes. Set the denominator equal to zero and solve for x.
x2−4x−5 = 0
Step 2: Factor the quadratic equation.
(x−5)(x+ 1) = 0
Step 3: Set each factor equal to zero.
x−5 = 0 or x+ 1 = 0
Step 4: Solve for x.
x= 5 or x=−1
Step 5: Therefore, the vertical asymptotes of f(x)are x= 5 and x=−1.
26
