MATH 121 - COLLEGE ALGEBRA -
Rational Functions
Question Bank - Set 7
Liberty University
Question 2
Question
Let f(x) = 2x3−5x2+3x
x2−4. Find the domain of the function f(x).
Solution
Step 1: Remember that the domain of a rational function is all real numbers
except where the denominator is equal to zero. Step 2: Determine where the
denominator x2−4is equal to zero by setting it equal to zero and solving for
x. Step 3: x2−4 = 0 =⇒x2= 4 =⇒x=±2. Step 4: The values x=±2
make the denominator zero, so they cannot be in the domain of f(x). Step 5:
Therefore, the domain of the function f(x)is all real numbers except x= 2 and
x=−2. Step 6: Hence, the domain of f(x)is (−∞,−2) ∪(−2,2) ∪(2,∞).
Question 3
Question
Let f(x) = 3x2−5x−2
x2−4x−5. Find the domain of the function f(x).
Solution
Step 1: The domain of a function is the set of all real numbers for which the
function is defined. In the case of a rational function like f(x), the function is
defined for all real numbers except where the denominator is zero.
Step 2: To find the domain of f(x), we need to determine where x2−4x−5 =
0.
Step 3: Let’s solve x2−4x−5 = 0 by factoring or using the quadratic
formula.
Step 4: Factoring the quadratic equation x2−4x−5 = 0 gives us (x−5)(x+
1) = 0.
Step 5: Setting each factor to zero gives us x−5 = 0 or x+ 1 = 0.
Step 6: Solving these equations, we get x= 5 and x=−1.
Step 7: Therefore, the domain of f(x)is all real numbers except x= 5 and
x=−1. In interval notation, the domain is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 4
Question
Find the vertical asymptotes of the rational function
f(x) = x2+ 3x−4
x2−1.
Solution
Step 1: Set the denominator equal to 0 and solve for xto find the vertical
asymptotes.
We need to find the values of xfor which the denominator of the rational
function is equal to 0. In this case, the denominator is x2−1, so we set x2−1 = 0
and solve for x.
x2−1 = 0
x2= 1
x=±1
Therefore, the vertical asymptotes are x= 1 and x=−1.
So, the vertical asymptotes of the rational function f(x) = x2+3x−4
x2−1are
x= 1 and x=−1.
Question 5
Question
Find the domain of the rational function:
f(x) = 5
x2−4x−5
2
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xfor which the function is defined. In this case, our function is defined
for all xexcept where the denominator is equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation, we can factor it:
(x−5)(x+ 1) = 0
This gives us two possible values for x:x= 5 and x=−1.
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 5 and x=−1.
Step 5: In interval notation, the domain can be expressed as: (−∞,−1) ∪
(−1,5) ∪(5,∞).
Question 6
Question
Solve the following rational equation for x:2
x−3−3
x+2 =5
x2−x−6.
Solution
Step 1: To get rid of the denominators, we will multiply both sides of the
equation by the least common multiple (LCM) of the denominators. In this
case, the LCM is (x−3)(x+ 2)(x+ 1).
2
x−3·(x−3)(x+2)(x+1)−3
x+ 2·(x−3)(x+2)(x+1) = 5
x2−x−6·(x−3)(x+2)(x+1)
Step 2: Simplify the equation obtained after multiplying.
2(x+ 2)(x+ 1) −3(x−3)(x+ 1) = 5(x−3)(x+ 2)
Step 3: Expand and simplify both sides of the equation.
2(x2+ 3x+ 2) −3(x2−2x−3) = 5(x2−x−6)
2x2+ 6x+ 4 −3x2+ 6x+ 9 = 5x2−5x−30
Step 4: Combine like terms.
−x2+ 12x+ 13 = 5x2−5x−30
Step 5: Move all terms to one side to set the equation to zero.
6x2−17x−43 = 0
3
Step 6: Solve the quadratic equation using the quadratic formula: x=
−b±√b2−4ac
2a.
x=17 ±√(−17)2−4(6)(−43)
2(6)
x=17 ±√289 + 1032
12
x=17 ±√1321
12
x=17 ±37
12
Step 7: Solve for x.x=17+37
12 or x=17−37
12
x=54
12 or x=−20
12
x= 4.5or x=−1.67
Therefore, the solutions to the equation are x= 4.5and x=−1.67.
Question 7
Question
Find the domain of the rational function:
f(x) = 3x
x2−4
Solution
Step 1: Recall that the domain of a function is the set of all real numbers for
which the function is defined. In the case of rational functions, the function is
defined as long as the denominator is not equal to zero.
Step 2: To find the domain of f(x), we need to find the values of xthat
make the denominator, x2−4, equal to zero.
Step 3: The denominator, x2−4, can be factored as (x+ 2)(x−2). To find
the values of xthat make the denominator zero, we set each factor equal to
zero: x+ 2 = 0 and x−2 = 0.
Step 4: Solving x+ 2 = 0, we get x=−2. Solving x−2 = 0, we get x= 2.
Step 5: Therefore, the domain of the function f(x) = 3x
x2−4is all real numbers
except x=−2and x= 2.
Step 6: Combining all the allowed values of x, the domain of the function is
(−∞,−2) ∪(−2,2) ∪(2,∞).
4
Question 8
Question
Find the vertical asymptotes, horizontal asymptotes, and intercepts of the ra-
tional function:
f(x) = 3x2−2x−1
x2+ 2x−3
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to zero and
solve for x.
x2+ 2x−3 = 0
(x+ 3)(x−1) = 0
x=−3or x= 1
So, the vertical asymptotes are x=−3and x= 1.
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and the denominator. Since the degree of the numerator is equal to the
degree of the denominator, divide the leading coefficients to find the horizontal
asymptote.
y=3
1= 3
Therefore, the horizontal asymptote is y= 3.
Step 3: To find the y-intercept, plug in x= 0 into the function.
f(0) = 3(0)2−2(0) −1
(0)2+ 2(0) −3=−1
−3=1
3
Thus, the y-intercept is (0,1
3).
Step 4: To find the x-intercept, set y= 0 and solve for x.
3x2−2x−1
x2+ 2x−3= 0
3x2−2x−1 = 0
(3x+ 1)(x−1) = 0
x=−1
3or x= 1
Therefore, the x-intercepts are (−1
3,0) and (1,0).
Question 9
Question
Let f(x) = 2x3−5x2+3x
x2−3x+2 . Determine the vertical asymptotes of the function f(x).
5
Solution
Step 1: To find the vertical asymptotes of f(x), we need to identify the values
of xfor which the denominator of f(x)equals zero.
Step 2: The denominator of f(x)is x2−3x+ 2. We find the roots of this
quadratic equation by solving x2−3x+ 2 = 0.
Step 3: To solve x2−3x+ 2 = 0, we can factor the quadratic equation or
use the quadratic formula x=−b±√b2−4ac
2a, where the coefficients are a= 1,
b=−3, and c= 2.
Step 4: Using the quadratic formula, we have x=3±√(−3)2−4(1)(2)
2(1) .
Step 5: Simplifying under the square root, we get x=3±√9−8
2.
Step 6: Further simplifying, we have x=3±√1
2.
Step 7: This gives us two possible values for x:x=3+1
2= 2 and x=3−1
2=
1.
Step 8: Therefore, the vertical asymptotes of the function f(x)are x= 2
and x= 1.
Question 10
Question
Simplify the following rational expression:
3x2+ 5x−2
2x2−3x−2
Solution
Step 1: Factor both the numerator and denominator of the rational expression.
Step 2: Factor the numerator: Since 3× −2 = −6and 5 = 3 + 2:
3x2+ 5x−2 = (3x−1)(x+ 2)
Step 3: Factor the denominator: Since 2× −2 = −4and −3 = −4+1:
2x2−3x−2 = (2x+ 1)(x−2)
Step 4: Substitute the factored expressions back into the original rational
expression:
(3x−1)(x+ 2)
(2x+ 1)(x−2)
Therefore, the simplified form of the given rational expression is 3x−1
2x+ 1 .
6
Question 11
Question
Let f(x) = x2−1
x2+ 2x−3and g(x) = x2+x−6
x2+ 2x−3be rational functions. Find
the domain of f(x)·g(x).
Solution
Step 1: Find the product of f(x)and g(x).
f(x)·g(x) = (x2−1
x2+ 2x−3)·(x2+x−6
x2+ 2x−3)
Step 2: Simplify the expression by multiplying the numerators and denomi-
nators.
f(x)·g(x) = (x2−1)(x2+x−6)
(x2+ 2x−3)(x2+ 2x−3)
f(x)·g(x) = x4+x3−6x2−x2+x+ 6
(x2+ 2x−3)2
f(x)·g(x) = x4+x3−7x2+x+ 6
(x2+ 2x−3)2
Step 3: Determine the domain of the function f(x)·g(x)by finding the
values of xfor which the denominator is not equal to zero. The denominator
(x2+ 2x−3)2will be zero when x2+ 2x−3 = 0. Solving x2+ 2x−3 = 0 gives
the solutions: x=−3and x= 1.
Therefore, the domain of f(x)·g(x)is all real numbers except x=−3and
x= 1. In interval notation, the domain is (−∞,−3) ∪(−3,1) ∪(1,∞).
Question 12
Question
Determine the domain of the rational function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: To find the domain of a rational function, we need to identify all real
numbers that make the denominator equal to zero, as division by zero is unde-
fined.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
7
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 or x+ 1 = 0
Step 5: Solve for x:
x= 5 or x=−1
Step 6: Therefore, the domain of f(x)is all real numbers except x= 5 and
x=−1, since these values would make the denominator zero.
Step 7: Thus, the domain of the function f(x) = 2x2−5x−3
x2−4x−5is x∈R\{5,−1}.
Question 13
Question
Simplify the following rational expression:
4x3−8x2+ 4x
2x2−4x
Solution
Step 1: Factor out the common terms in the numerator and the denomina-
tor. Step 2: Cancel out any common factors between the numerator and the
denominator. Step 3: Simplify the resulting expression.
Step 1: Factor out the common terms in the numerator and the denomina-
tor. 4x3−8x2+ 4x
2x2−4x=4x(x2−2x+ 1)
2x(x−2)
Step 2: Cancel out any common factors between the numerator and the
denominator. 4x(x2−2x+ 1)
2x(x−2) =4(x2−2x+ 1)
x−2
Step 3: Simplify the resulting expression.
4(x−1)2
x−2
Therefore, the simplified form of the rational expression is 4(x−1)2
x−2.
8
Question 14
Question
Simplify the following rational expression:
3x2−9x
x2−4∇ · x2−5x+ 6
x2−2x−8
Solution
Step 1: We begin by writing the division of fractions as a multiplication of
fractions by taking the reciprocal of the second fraction:
3x2−9x
x2−4·x2−2x−8
x2−5x+ 6
Step 2: Factor the expressions in both the numerator and denominator of
each fraction: 3x(x−3)
(x+ 2)(x−2) ·(x−4)(x+ 2)
(x−3)(x−2)
Step 3: Cancel out common factors in the numerator and denominator:
3x(x−3)
(x+ 2)(x−2) ·(x−4)(x+ 2)
(x−3)(x−2)
Step 4: Multiply the remaining factors:
3x
x−3
Therefore, the simplified form of the rational expression is 3x
x−3.
Question 15
Question
Simplify the rational function:
f(x) = x2−5x−6
x2−4x−5
Solution
Step 1: Factor the numerator and denominator of the rational function. Step
2: Simplify the rational function by canceling out common factors, if possible.
Step 1: Factor the numerator and denominator of the rational function. The
numerator factors to (x−6)(x+1), and the denominator factors to (x−5)(x+1).
So, f(x) = (x−6)(x+1)
(x−5)(x+1) .
9
Step 2: Simplify the rational function by canceling out common factors, if
possible. We can cancel out the common factor of (x+ 1) in the numerator and
denominator.
So, f(x) = x−6
x−5.
Therefore, the simplified form of the rational function is f(x) = x−6
x−5.
Question 16
Question
Determine all the values of xfor which the rational function x2−4
x2−3x−4is
undefined.
Solution
Step 1: The rational function x2−4
x2−3x−4is undefined when the denominator
is equal to zero, since division by zero is undefined. Therefore, we need to solve
the equation x2−3x−4=0to find the values of xthat make the rational
function undefined.
Step 2: To solve x2−3x−4 = 0, we can factor the quadratic equation by
finding two numbers that multiply to −4and add to −3. These numbers are
−4and 1.
Step 3: Therefore, the factored form of the quadratic equation is (x−4)(x+
1) = 0. Setting each factor to zero gives us x−4 = 0 and x+ 1 = 0.
Step 4: Solving x−4=0gives x= 4, and solving x+ 1 = 0 gives x=−1.
Thus, the values of xfor which the rational function is undefined are x=−1
and x= 4.
Question 17
Question
Find the domain of the function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: To find the domain of the function, we need to identify any 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
10
Step 3: This is a quadratic equation that can be factored:
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero:
x−5 = 0 or x+ 1 = 0
Step 5: Solve for x:
x= 5 or x=−1
Step 6: Therefore, the domain of the function f(x)is all real numbers except
x= 5 and x=−1, since those values make the denominator zero. Step 7: The
domain of the function f(x)is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 18
Question
Let f(x) = 2x2+ 5x−3
x2−4x−5. Find the x-intercepts of the rational function f(x).
Solution
To find the x-intercepts of a function, we set f(x)equal to zero and solve for x.
Step 1: Set f(x)equal to zero:
2x2+ 5x−3
x2−4x−5= 0
Step 2: Factor the numerator and denominator:
(2x−1)(x+ 3)
(x−5)(x+ 1) = 0
Step 3: Use the zero product property:
2x−1 = 0 or x+ 3 = 0 or x−5 = 0 or x+ 1 = 0
Step 4: Solve for xin each equation:
x=1
2, x =−3, x = 5, x =−1
Step 5: The x-intercepts of the rational function f(x)are x=1
2,x=−3,
x= 5, and x=−1.
11
Question 19
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes in the graph
of the rational function given by:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes, we set the denominator equal to zero and
solve for x:
x2−4x−5 = 0
Factoring the quadratic, we have:
(x−5)(x+ 1) = 0
So, x= 5 and x=−1are the vertical asymptotes.
Step 2: Horizontal Asymptotes
To find the horizontal asymptote, we determine what happens as xap-
proaches infinity. Divide the leading coefficient of the numerator by the leading
coefficient of the denominator to find the horizontal asymptote:
y=2
1= 2
So, the horizontal asymptote is y= 2.
Step 3: Holes
To find any holes in the graph, we look for points where the numerator and
denominator can both be factored and canceled out. In this case, we notice both
2x2−5x−3and x2−4x−5can be factored as (2x+1)(x−3) and (x−5)(x+1),
respectively.
Therefore, the function simplifies to:
f(x) = 2x2−5x−3
x2−4x−5=(2x+ 1)(x−3)
(x−5)(x+ 1)
This indicates a hole at x= 5.
In conclusion, the rational function f(x) = 2x2−5x−3
x2−4x−5has vertical asymptotes
at x= 5 and x=−1, a horizontal asymptote at y= 2, and a hole at x= 5.
12
Question 20
Question
Find the vertical and horizontal asymptotes of the rational function: f(x) =
3x2−2x+5
x2+4x−5.
Solution
Step 1: To find the vertical asymptotes of the function, we need to determine
where the denominator is equal to zero (if any). Set x2+ 4x−5=0and solve
for x.
x2+ 4x−5 = 0
(x+ 5)(x−1) = 0
x=−5or x= 1
So, the vertical asymptotes are at x=−5and x= 1.
Step 2: To find the horizontal asymptote, we observe the degrees of the
numerator and denominator. Since the degrees are equal, we look at the leading
coefficients to determine the horizontal asymptote. The horizontal asymptote
is the ratio of the leading coefficients.
lim
x→∞
3x2−2x+ 5
x2+ 4x−5=3
1= 3
Therefore, the horizontal asymptote is y= 3.
In summary, the vertical asymptotes are at x=−5and x= 1, and the
horizontal asymptote is at y= 3 for the rational function f(x) = 3x2−2x+5
x2+4x−5.
Question 21
Question
Find the domain of the rational function:
f(x) = x2−6x+ 8
x2−4x−5
Solution
Step 1: We need to determine the values of xfor which the denominator x2−
4x−5is not equal to zero, since division by zero is undefined in mathematics.
So, we solve the equation x2−4x−5=0to find the values of xthat make
the denominator zero. This is equivalent to finding the roots of the quadratic
equation x2−4x−5 = 0.
13
Step 2: To find the roots of the quadratic equation x2−4x−5 = 0, we can
use the quadratic formula:
x=−b±√b2−4ac
2a
Step 3: Comparing the given equation with the standard form of a quadratic
equation ax2+bx +c= 0, we have a= 1,b=−4, and c=−5. Plugging these
values into the quadratic formula, we get:
x=4±√(−4)2−4∗1∗(−5)
2∗1
x=4±√16 + 20
2
x=4±√36
2
x=4±6
2
Step 4: Therefore, the roots of the quadratic equation are x= 5 and x=−1.
These are the values of xfor which the denominator becomes zero, making the
rational function undefined.
Step 5: Finally, the domain of the rational function f(x) = x2−6x+ 8
x2−4x−5is
all real numbers except x= 5 and x=−1, since these values would make the de-
nominator zero. So, the domain of the function is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 22
Question
Find the domain of the rational function:
f(x) = 5x2−7x+ 2
x2−4x−5
Solution
Step 1: We need to find the values of xfor which the denominator is not equal
to zero, since division by zero is undefined. The denominator of the rational
function is x2−4x−5. We solve the equation x2−4x−5 = 0 to find the values
of xwhere the denominator is zero.
To solve the equation:
x2−4x−5 = 0
we can use the quadratic formula:
x=−(−4) ±√(−4)2−4·1·(−5)
2·1
14
x=4±√16 + 20
2
x=4±√36
2
x=4±6
2
x=4+6
2or x=4−6
2
x= 5 or x=−1
Step 2: The domain of the function f(x)is all real numbers except the values
that make the denominator zero. So, the domain of f(x)is:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 23
Question
Simplify the following rational function:
f(x) = 3x2−5x−2
x2−4x+ 4
Solution
Step 1: Factor the numerator and denominator of the rational function:
f(x) = 3x2−5x−2
(x−2)2
Step 2: Factor the numerator:
f(x) = (3x+ 1)(x−2)
(x−2)2
Step 3: Simplify the rational function:
f(x) = 3x+ 1
x−2
Therefore, the simplified form of the rational function f(x)is 3x+1
x−2.
Question 24
Question
Find the domain of the following rational function:
f(x) = 1
x2−5x+ 6.
15
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that would make the denominator equal to zero. So, we need to find the
values of xthat would make x2−5x+ 6 = 0.
Step 2: To find these values of x, we need to factor the quadratic equation
x2−5x+ 6 = 0. The factored form is (x−2)(x−3) = 0.
Step 3: Setting each factor to zero, we find the roots of the equation: x−2 = 0
or x−3 = 0.
Step 4: Solving these equations gives us x= 2 and x= 3.
Step 5: So, the domain of the function f(x) = 1
x2−5x+ 6 is all real numbers
except x= 2 and x= 3.
Therefore, the domain of the function is x∈R, x = 2,3.
Question 25
Question
Find the domain of the rational function:
f(x) = 5x−1
x2−9.
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. So, we need to find
the values of xthat make the denominator, x2−9, equal to zero.
Step 2: We solve x2−9 = 0 by factoring as the different of squares:
x2−9 = (x+ 3)(x−3) = 0.
Step 3: Setting each factor equal to zero gives us the values of xto exclude
from the domain:
x+ 3 = 0 ⇒x=−3
x−3 = 0 ⇒x= 3
Step 4: Therefore, the domain of the rational function f(x) = 5x−1
x2−9is all
real numbers except x=−3,3, expressed in interval notation as:
(−∞,−3) ∪(−3,3) ∪(3,∞).
Question 26
Question
Find the vertical asymptotes of the rational function f(x) = x2−4
x2−x−6.
16
Solution
Step 1: Determine the values of xthat make the denominator equal to zero, as
these values will give us the vertical asymptotes of the function. Set x2−x−6 = 0
and solve for x.
Step 1: x2−x−6 = 0
(x−3)(x+ 2) = 0
x= 3 or x=−2
Step 2: Therefore, the function f(x)has vertical asymptotes at x= 3 and
x=−2. Step 3: The vertical asymptotes of the function f(x) = x2−4
x2−x−6are x= 3 and x=−2.
Question 27
Question
Let f(x) = 4x2−5x−6
x2−9be a rational function. Find the vertical and horizontal
asymptotes of the graph of f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator of the rational
function equal to zero and solve for x:
x2−9 = 0
(x+ 3)(x−3) = 0
This gives us x=−3and x= 3 as vertical asymptotes.
Step 2: To find the horizontal asymptote, compare the degree of the nu-
merator and the denominator of the rational function. Since the degree of the
numerator is 2 and the degree of the denominator is also 2, we need to compare
the leading coefficients. Divide the leading coefficient of the numerator by the
leading coefficient of the denominator.
4
1= 4
Therefore, the horizontal asymptote is y= 4.
Question 28
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes in the graph
of the function:
f(x) = 3x2−9x−18
x2−4x−21
17
Solution
Step 1: To find the vertical asymptotes of the function, set the denominator
equal to zero and solve for x.
x2−4x−21 = 0
Factoring the quadratic equation gives:
(x−7)(x+ 3) = 0
So, x= 7 and x=−3are the vertical asymptotes of the function.
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and the denominator. Since both have the same degree, the horizontal
asymptote will be the ratio of the leading coefficients.
The horizontal asymptote is:
y=3
1= 3
Step 3: To find any holes in the graph, simplify the function and check for
common factors in the numerator and denominator.
Factorizing the numerator as 3(x−3)(x+ 2) and the denominator as (x−
7)(x+ 3), we can cancel out the common factor (x+ 3).
The simplified function is:
f(x) = 3(x−3)
x−7
There are no holes in the graph of the function.
Therefore, the function has vertical asymptotes at x= 7 and x=−3, a
horizontal asymptote at y= 3, and no holes in the graph.
Question 29
Question
Simplify the following rational expression:
4x3−7x2−5x
2x2−3x−2∇ · 5x2−7x−6
2x2+ 3x−2
Solution
Step 1: Factor both the numerator and denominator of each fraction.
For the first fraction, we have:
4x3−7x2−5x=x(4x2−7x−5) = x(4x+ 3)(x−5)
2x2−3x−2 = (2x−1)(x+ 2)
18
Therefore, the first fraction simplifies to:
x(4x+ 3)(x−5)
(2x−1)(x+ 2)
For the second fraction, we have:
5x2−7x−6 = (5x+ 2)(x−3)
2x2+ 3x−2 = (2x−1)(x+ 2)
Therefore, the second fraction simplifies to:
(5x+ 2)(x−3)
(2x−1)(x+ 2)
Step 2: Rewrite the division as multiplication by the reciprocal. Then sim-
plify by multiplying by the reciprocal of the second fraction.
x(4x+ 3)(x−5)
(2x−1)(x+ 2) ·(2x−1)(x+ 2)
(5x+ 2)(x−3)
Step 3: Cancel out common factors in the numerator and denominator.
Multiply the remaining terms.
x(4x+ 3)(x−5)
(5x+ 2)(x−3)
Therefore, the simplified expression is:
x(4x+ 3)(x−5)
(5x+ 2)(x−3)
Question 30
Question
Simplify the rational function:
2x3+ 6x2−8x
4x2−16
Solution
Step 1: Factor out common terms in the numerator and denominator. Step 2:
Simplify the rational function by canceling out common factors.
Step 1: First, factor out common terms in the numerator and denominator:
2x3+ 6x2−8x
4x2−16 =2x(x2+ 3x−4)
4(x2−4)
19
Now, factor each quadratic expression:
x2+ 3x−4 = (x+ 4)(x−1)
x2−4 = (x+ 2)(x−2)
So, the rational function becomes:
2x(x+ 4)(x−1)
4(x+ 2)(x−2)
Step 2: Next, simplify the rational function by canceling out common fac-
tors: 2x(x+ 4)(x−1)
4(x+ 2)(x−2) =x(x+ 4)(x−1)
2(x+ 2)(x−2)
Therefore, the simplified form of the rational function is:
x(x+ 4)(x−1)
2(x+ 2)(x−2)
Question 31
Question
Find the equations of the asymptotes of the rational function:
f(x) = 2x2+ 3x−2
x2−4
Solution
Step 1: Determine the vertical asymptotes by setting the denominator equal to
zero and solving for x.
x2−4 = 0
(x−2)(x+ 2) = 0
x= 2 or x=−2
So, the vertical asymptotes are x= 2 and x=−2.
Step 2: Determine the horizontal asymptote by comparing the degrees of
the numerator and the denominator. Since the degree of the numerator is equal
to the degree of the denominator, the horizontal asymptote is given by the ratio
of the leading coefficients. Therefore, the horizontal asymptote is y= 2.
Step 3: Determine the slant (or oblique) asymptote, if it exists. To find
the slant asymptote, perform polynomial long division of the numerator by the
denominator: 2x+7 +1
x2−4 2x2+3x−2
−2x2+8x
0−5x−2
20
The slant asymptote is given by the quotient, which is 2x+ 7.
Therefore, the equations of the asymptotes are:
• Vertical asymptotes: x= 2 and x=−2
• Horizontal asymptote: y= 2
• Slant asymptote: y= 2x+ 7
Question 32
Question
Solve the following rational equation for x:
2
x−3−3
x+ 2 =1
x2−x−6
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation. Since the denominators are (x−3),(x+ 2), and (x2−x−6), the
common denominator is (x−3)(x+ 2).
Step 2: Rewrite the equation with the common denominator:
2(x+ 2)
(x−3)(x+ 2) −3(x−3)
(x−3)(x+ 2) =1
x2−x−6
Step 3: Combine the fractions on the left side of the equation:
2(x+ 2) −3(x−3)
(x−3)(x+ 2) =1
x2−x−6
Step 4: Simplify the equation by expanding and combining like terms:
2x+ 4 −3x+ 9
(x−3)(x+ 2) =1
x2−x−6
−x+ 13
(x−3)(x+ 2) =1
x2−x−6
Step 5: Multiply both sides of the equation by x2−x−6to get rid of the
denominators:
(−x+ 13)(x2−x−6) = 1
−x3+ 13x2−6x+ 13x2−169x+ 78 = 1
−x3+ 26x2−175x+ 78 = 1
21
−x3+ 26x2−175x+ 77 = 0
This is a cubic equation and may not have simple integer solutions. The
roots of this cubic equation can be found using numerical methods or a graphing
calculator.
Question 33
Question
Solve the rational equation: 5
x−3−3
x+2 =8
x2−x−6.
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation.
Step 2: Multiply each fraction by the appropriate factor to obtain the com-
mon denominator.
Step 3: Combine the fractions on the left side by adding or subtracting
numerators.
Step 4: Simplify the resulting fraction.
Step 5: Factor the denominator on the right side of the equation.
Step 6: Rewrite the equation with the factored form of the denominator.
Step 7: Multiply both sides of the equation by the factored form of the
denominator to eliminate the fraction.
Step 8: Solve the resulting equation.
Step 9: Check the solutions in the original equation to ensure they are valid.
Question 34
Question
Find the vertical asymptotes of the rational function:
f(x) = x3−2x2−x+ 2
x2−2x−3
Solution
To find the vertical asymptotes of the rational function f(x), we need to deter-
mine where the denominator is equal to zero, but the numerator is not equal to
zero at those points.
Step 1: Find where the denominator is equal to zero. Setting the
denominator equal to zero gives:
x2−2x−3 = 0
22
We can factor the quadratic equation as:
(x−3)(x+ 1) = 0
So, x= 3 and x=−1are the values where the denominator is equal to zero.
Step 2: Determine where the numerator is not equal to zero at
those points. Substitute x= 3 first:
f(3) = 33−2(3)2−3+2
32−2(3) −3=27 −18 −3+2
9−6−3=8
0
Since the numerator is not equal to zero at x= 3, there is a vertical asymp-
tote at x= 3.
Next, substitute x=−1:
f(−1) = (−1)3−2(−1)2−(−1) + 2
(−1)2−2(−1) −3=−1+2+1+2
1+2−3=4
0
Since the numerator is not equal to zero at x=−1, there is a vertical
asymptote at x=−1.
Therefore, the vertical asymptotes of the rational function f(x)are x= 3
and x=−1.
Question 35
Question
Find the domain of the rational function:
f(x) = 3x2−7x−6
x2−5x−14
Solution
Step 1: To find the domain of a rational function, we need to determine the
values that xcannot take in order to avoid division by zero.
Step 2: The function f(x)is defined for all real numbers xexcept for those
that make the denominator equal to zero.
Step 3: Set the denominator equal to zero and solve for x:
x2−5x−14 = 0
(x−7)(x+ 2) = 0
x= 7 or x=−2
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except for x= 7 and x=−2.
Step 5: In interval notation, the domain can be expressed as (−∞,−2) ∪
(−2,7) ∪(7,∞).
23
Step 3: Let’s solve x2−4x−5 = 0 by factoring or using the quadratic
formula.
Step 4: Factoring the quadratic equation x2−4x−5 = 0 gives us (x−5)(x+
1) = 0.
Step 5: Setting each factor to zero gives us x−5 = 0 or x+ 1 = 0.
Step 6: Solving these equations, we get x= 5 and x=−1.
Step 7: Therefore, the domain of f(x)is all real numbers except x= 5 and
x=−1. In interval notation, the domain is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 4
Question
Find the vertical asymptotes of the rational function
f(x) = x2+ 3x−4
x2−1.
Solution
Step 1: Set the denominator equal to 0 and solve for xto find the vertical
asymptotes.
We need to find the values of xfor which the denominator of the rational
function is equal to 0. In this case, the denominator is x2−1, so we set x2−1 = 0
and solve for x.
x2−1 = 0
x2= 1
x=±1
Therefore, the vertical asymptotes are x= 1 and x=−1.
So, the vertical asymptotes of the rational function f(x) = x2+3x−4
x2−1are
x= 1 and x=−1.
Question 5
Question
Find the domain of the rational function:
f(x) = 5
x2−4x−5
2
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xfor which the function is defined. In this case, our function is defined
for all xexcept where the denominator is equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation, we can factor it:
(x−5)(x+ 1) = 0
This gives us two possible values for x:x= 5 and x=−1.
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 5 and x=−1.
Step 5: In interval notation, the domain can be expressed as: (−∞,−1) ∪
(−1,5) ∪(5,∞).
Question 6
Question
Solve the following rational equation for x:2
x−3−3
x+2 =5
x2−x−6.
Solution
Step 1: To get rid of the denominators, we will multiply both sides of the
equation by the least common multiple (LCM) of the denominators. In this
case, the LCM is (x−3)(x+ 2)(x+ 1).
2
x−3·(x−3)(x+2)(x+1)−3
x+ 2·(x−3)(x+2)(x+1) = 5
x2−x−6·(x−3)(x+2)(x+1)
Step 2: Simplify the equation obtained after multiplying.
2(x+ 2)(x+ 1) −3(x−3)(x+ 1) = 5(x−3)(x+ 2)
Step 3: Expand and simplify both sides of the equation.
2(x2+ 3x+ 2) −3(x2−2x−3) = 5(x2−x−6)
2x2+ 6x+ 4 −3x2+ 6x+ 9 = 5x2−5x−30
Step 4: Combine like terms.
−x2+ 12x+ 13 = 5x2−5x−30
Step 5: Move all terms to one side to set the equation to zero.
6x2−17x−43 = 0
3
Step 6: Solve the quadratic equation using the quadratic formula: x=
−b±√b2−4ac
2a.
x=17 ±√(−17)2−4(6)(−43)
2(6)
x=17 ±√289 + 1032
12
x=17 ±√1321
12
x=17 ±37
12
Step 7: Solve for x.x=17+37
12 or x=17−37
12
x=54
12 or x=−20
12
x= 4.5or x=−1.67
Therefore, the solutions to the equation are x= 4.5and x=−1.67.
Question 7
Question
Find the domain of the rational function:
f(x) = 3x
x2−4
Solution
Step 1: Recall that the domain of a function is the set of all real numbers for
which the function is defined. In the case of rational functions, the function is
defined as long as the denominator is not equal to zero.
Step 2: To find the domain of f(x), we need to find the values of xthat
make the denominator, x2−4, equal to zero.
Step 3: The denominator, x2−4, can be factored as (x+ 2)(x−2). To find
the values of xthat make the denominator zero, we set each factor equal to
zero: x+ 2 = 0 and x−2 = 0.
Step 4: Solving x+ 2 = 0, we get x=−2. Solving x−2 = 0, we get x= 2.
Step 5: Therefore, the domain of the function f(x) = 3x
x2−4is all real numbers
except x=−2and x= 2.
Step 6: Combining all the allowed values of x, the domain of the function is
(−∞,−2) ∪(−2,2) ∪(2,∞).
4
Question 8
Question
Find the vertical asymptotes, horizontal asymptotes, and intercepts of the ra-
tional function:
f(x) = 3x2−2x−1
x2+ 2x−3
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to zero and
solve for x.
x2+ 2x−3 = 0
(x+ 3)(x−1) = 0
x=−3or x= 1
So, the vertical asymptotes are x=−3and x= 1.
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and the denominator. Since the degree of the numerator is equal to the
degree of the denominator, divide the leading coefficients to find the horizontal
asymptote.
y=3
1= 3
Therefore, the horizontal asymptote is y= 3.
Step 3: To find the y-intercept, plug in x= 0 into the function.
f(0) = 3(0)2−2(0) −1
(0)2+ 2(0) −3=−1
−3=1
3
Thus, the y-intercept is (0,1
3).
Step 4: To find the x-intercept, set y= 0 and solve for x.
3x2−2x−1
x2+ 2x−3= 0
3x2−2x−1 = 0
(3x+ 1)(x−1) = 0
x=−1
3or x= 1
Therefore, the x-intercepts are (−1
3,0) and (1,0).
Question 9
Question
Let f(x) = 2x3−5x2+3x
x2−3x+2 . Determine the vertical asymptotes of the function f(x).
5
Solution
Step 1: To find the vertical asymptotes of f(x), we need to identify the values
of xfor which the denominator of f(x)equals zero.
Step 2: The denominator of f(x)is x2−3x+ 2. We find the roots of this
quadratic equation by solving x2−3x+ 2 = 0.
Step 3: To solve x2−3x+ 2 = 0, we can factor the quadratic equation or
use the quadratic formula x=−b±√b2−4ac
2a, where the coefficients are a= 1,
b=−3, and c= 2.
Step 4: Using the quadratic formula, we have x=3±√(−3)2−4(1)(2)
2(1) .
Step 5: Simplifying under the square root, we get x=3±√9−8
2.
Step 6: Further simplifying, we have x=3±√1
2.
Step 7: This gives us two possible values for x:x=3+1
2= 2 and x=3−1
2=
1.
Step 8: Therefore, the vertical asymptotes of the function f(x)are x= 2
and x= 1.
Question 10
Question
Simplify the following rational expression:
3x2+ 5x−2
2x2−3x−2
Solution
Step 1: Factor both the numerator and denominator of the rational expression.
Step 2: Factor the numerator: Since 3× −2 = −6and 5 = 3 + 2:
3x2+ 5x−2 = (3x−1)(x+ 2)
Step 3: Factor the denominator: Since 2× −2 = −4and −3 = −4+1:
2x2−3x−2 = (2x+ 1)(x−2)
Step 4: Substitute the factored expressions back into the original rational
expression:
(3x−1)(x+ 2)
(2x+ 1)(x−2)
Therefore, the simplified form of the given rational expression is 3x−1
2x+ 1 .
6
Question 11
Question
Let f(x) = x2−1
x2+ 2x−3and g(x) = x2+x−6
x2+ 2x−3be rational functions. Find
the domain of f(x)·g(x).
Solution
Step 1: Find the product of f(x)and g(x).
f(x)·g(x) = (x2−1
x2+ 2x−3)·(x2+x−6
x2+ 2x−3)
Step 2: Simplify the expression by multiplying the numerators and denomi-
nators.
f(x)·g(x) = (x2−1)(x2+x−6)
(x2+ 2x−3)(x2+ 2x−3)
f(x)·g(x) = x4+x3−6x2−x2+x+ 6
(x2+ 2x−3)2
f(x)·g(x) = x4+x3−7x2+x+ 6
(x2+ 2x−3)2
Step 3: Determine the domain of the function f(x)·g(x)by finding the
values of xfor which the denominator is not equal to zero. The denominator
(x2+ 2x−3)2will be zero when x2+ 2x−3 = 0. Solving x2+ 2x−3 = 0 gives
the solutions: x=−3and x= 1.
Therefore, the domain of f(x)·g(x)is all real numbers except x=−3and
x= 1. In interval notation, the domain is (−∞,−3) ∪(−3,1) ∪(1,∞).
Question 12
Question
Determine the domain of the rational function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: To find the domain of a rational function, we need to identify all real
numbers that make the denominator equal to zero, as division by zero is unde-
fined.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
7
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 or x+ 1 = 0
Step 5: Solve for x:
x= 5 or x=−1
Step 6: Therefore, the domain of f(x)is all real numbers except x= 5 and
x=−1, since these values would make the denominator zero.
Step 7: Thus, the domain of the function f(x) = 2x2−5x−3
x2−4x−5is x∈R\{5,−1}.
Question 13
Question
Simplify the following rational expression:
4x3−8x2+ 4x
2x2−4x
Solution
Step 1: Factor out the common terms in the numerator and the denomina-
tor. Step 2: Cancel out any common factors between the numerator and the
denominator. Step 3: Simplify the resulting expression.
Step 1: Factor out the common terms in the numerator and the denomina-
tor. 4x3−8x2+ 4x
2x2−4x=4x(x2−2x+ 1)
2x(x−2)
Step 2: Cancel out any common factors between the numerator and the
denominator. 4x(x2−2x+ 1)
2x(x−2) =4(x2−2x+ 1)
x−2
Step 3: Simplify the resulting expression.
4(x−1)2
x−2
Therefore, the simplified form of the rational expression is 4(x−1)2
x−2.
8
Question 14
Question
Simplify the following rational expression:
3x2−9x
x2−4∇ · x2−5x+ 6
x2−2x−8
Solution
Step 1: We begin by writing the division of fractions as a multiplication of
fractions by taking the reciprocal of the second fraction:
3x2−9x
x2−4·x2−2x−8
x2−5x+ 6
Step 2: Factor the expressions in both the numerator and denominator of
each fraction: 3x(x−3)
(x+ 2)(x−2) ·(x−4)(x+ 2)
(x−3)(x−2)
Step 3: Cancel out common factors in the numerator and denominator:
3x(x−3)
(x+ 2)(x−2) ·(x−4)(x+ 2)
(x−3)(x−2)
Step 4: Multiply the remaining factors:
3x
x−3
Therefore, the simplified form of the rational expression is 3x
x−3.
Question 15
Question
Simplify the rational function:
f(x) = x2−5x−6
x2−4x−5
Solution
Step 1: Factor the numerator and denominator of the rational function. Step
2: Simplify the rational function by canceling out common factors, if possible.
Step 1: Factor the numerator and denominator of the rational function. The
numerator factors to (x−6)(x+1), and the denominator factors to (x−5)(x+1).
So, f(x) = (x−6)(x+1)
(x−5)(x+1) .
9
Step 2: Simplify the rational function by canceling out common factors, if
possible. We can cancel out the common factor of (x+ 1) in the numerator and
denominator.
So, f(x) = x−6
x−5.
Therefore, the simplified form of the rational function is f(x) = x−6
x−5.
Question 16
Question
Determine all the values of xfor which the rational function x2−4
x2−3x−4is
undefined.
Solution
Step 1: The rational function x2−4
x2−3x−4is undefined when the denominator
is equal to zero, since division by zero is undefined. Therefore, we need to solve
the equation x2−3x−4=0to find the values of xthat make the rational
function undefined.
Step 2: To solve x2−3x−4 = 0, we can factor the quadratic equation by
finding two numbers that multiply to −4and add to −3. These numbers are
−4and 1.
Step 3: Therefore, the factored form of the quadratic equation is (x−4)(x+
1) = 0. Setting each factor to zero gives us x−4 = 0 and x+ 1 = 0.
Step 4: Solving x−4=0gives x= 4, and solving x+ 1 = 0 gives x=−1.
Thus, the values of xfor which the rational function is undefined are x=−1
and x= 4.
Question 17
Question
Find the domain of the function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: To find the domain of the function, we need to identify any 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
10
Step 3: This is a quadratic equation that can be factored:
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero:
x−5 = 0 or x+ 1 = 0
Step 5: Solve for x:
x= 5 or x=−1
Step 6: Therefore, the domain of the function f(x)is all real numbers except
x= 5 and x=−1, since those values make the denominator zero. Step 7: The
domain of the function f(x)is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 18
Question
Let f(x) = 2x2+ 5x−3
x2−4x−5. Find the x-intercepts of the rational function f(x).
Solution
To find the x-intercepts of a function, we set f(x)equal to zero and solve for x.
Step 1: Set f(x)equal to zero:
2x2+ 5x−3
x2−4x−5= 0
Step 2: Factor the numerator and denominator:
(2x−1)(x+ 3)
(x−5)(x+ 1) = 0
Step 3: Use the zero product property:
2x−1 = 0 or x+ 3 = 0 or x−5 = 0 or x+ 1 = 0
Step 4: Solve for xin each equation:
x=1
2, x =−3, x = 5, x =−1
Step 5: The x-intercepts of the rational function f(x)are x=1
2,x=−3,
x= 5, and x=−1.
11
Question 19
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes in the graph
of the rational function given by:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes, we set the denominator equal to zero and
solve for x:
x2−4x−5 = 0
Factoring the quadratic, we have:
(x−5)(x+ 1) = 0
So, x= 5 and x=−1are the vertical asymptotes.
Step 2: Horizontal Asymptotes
To find the horizontal asymptote, we determine what happens as xap-
proaches infinity. Divide the leading coefficient of the numerator by the leading
coefficient of the denominator to find the horizontal asymptote:
y=2
1= 2
So, the horizontal asymptote is y= 2.
Step 3: Holes
To find any holes in the graph, we look for points where the numerator and
denominator can both be factored and canceled out. In this case, we notice both
2x2−5x−3and x2−4x−5can be factored as (2x+1)(x−3) and (x−5)(x+1),
respectively.
Therefore, the function simplifies to:
f(x) = 2x2−5x−3
x2−4x−5=(2x+ 1)(x−3)
(x−5)(x+ 1)
This indicates a hole at x= 5.
In conclusion, the rational function f(x) = 2x2−5x−3
x2−4x−5has vertical asymptotes
at x= 5 and x=−1, a horizontal asymptote at y= 2, and a hole at x= 5.
12
Question 20
Question
Find the vertical and horizontal asymptotes of the rational function: f(x) =
3x2−2x+5
x2+4x−5.
Solution
Step 1: To find the vertical asymptotes of the function, we need to determine
where the denominator is equal to zero (if any). Set x2+ 4x−5=0and solve
for x.
x2+ 4x−5 = 0
(x+ 5)(x−1) = 0
x=−5or x= 1
So, the vertical asymptotes are at x=−5and x= 1.
Step 2: To find the horizontal asymptote, we observe the degrees of the
numerator and denominator. Since the degrees are equal, we look at the leading
coefficients to determine the horizontal asymptote. The horizontal asymptote
is the ratio of the leading coefficients.
lim
x→∞
3x2−2x+ 5
x2+ 4x−5=3
1= 3
Therefore, the horizontal asymptote is y= 3.
In summary, the vertical asymptotes are at x=−5and x= 1, and the
horizontal asymptote is at y= 3 for the rational function f(x) = 3x2−2x+5
x2+4x−5.
Question 21
Question
Find the domain of the rational function:
f(x) = x2−6x+ 8
x2−4x−5
Solution
Step 1: We need to determine the values of xfor which the denominator x2−
4x−5is not equal to zero, since division by zero is undefined in mathematics.
So, we solve the equation x2−4x−5=0to find the values of xthat make
the denominator zero. This is equivalent to finding the roots of the quadratic
equation x2−4x−5 = 0.
13
Step 2: To find the roots of the quadratic equation x2−4x−5 = 0, we can
use the quadratic formula:
x=−b±√b2−4ac
2a
Step 3: Comparing the given equation with the standard form of a quadratic
equation ax2+bx +c= 0, we have a= 1,b=−4, and c=−5. Plugging these
values into the quadratic formula, we get:
x=4±√(−4)2−4∗1∗(−5)
2∗1
x=4±√16 + 20
2
x=4±√36
2
x=4±6
2
Step 4: Therefore, the roots of the quadratic equation are x= 5 and x=−1.
These are the values of xfor which the denominator becomes zero, making the
rational function undefined.
Step 5: Finally, the domain of the rational function f(x) = x2−6x+ 8
x2−4x−5is
all real numbers except x= 5 and x=−1, since these values would make the de-
nominator zero. So, the domain of the function is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 22
Question
Find the domain of the rational function:
f(x) = 5x2−7x+ 2
x2−4x−5
Solution
Step 1: We need to find the values of xfor which the denominator is not equal
to zero, since division by zero is undefined. The denominator of the rational
function is x2−4x−5. We solve the equation x2−4x−5 = 0 to find the values
of xwhere the denominator is zero.
To solve the equation:
x2−4x−5 = 0
we can use the quadratic formula:
x=−(−4) ±√(−4)2−4·1·(−5)
2·1
14
x=4±√16 + 20
2
x=4±√36
2
x=4±6
2
x=4+6
2or x=4−6
2
x= 5 or x=−1
Step 2: The domain of the function f(x)is all real numbers except the values
that make the denominator zero. So, the domain of f(x)is:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 23
Question
Simplify the following rational function:
f(x) = 3x2−5x−2
x2−4x+ 4
Solution
Step 1: Factor the numerator and denominator of the rational function:
f(x) = 3x2−5x−2
(x−2)2
Step 2: Factor the numerator:
f(x) = (3x+ 1)(x−2)
(x−2)2
Step 3: Simplify the rational function:
f(x) = 3x+ 1
x−2
Therefore, the simplified form of the rational function f(x)is 3x+1
x−2.
Question 24
Question
Find the domain of the following rational function:
f(x) = 1
x2−5x+ 6.
15
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that would make the denominator equal to zero. So, we need to find the
values of xthat would make x2−5x+ 6 = 0.
Step 2: To find these values of x, we need to factor the quadratic equation
x2−5x+ 6 = 0. The factored form is (x−2)(x−3) = 0.
Step 3: Setting each factor to zero, we find the roots of the equation: x−2 = 0
or x−3 = 0.
Step 4: Solving these equations gives us x= 2 and x= 3.
Step 5: So, the domain of the function f(x) = 1
x2−5x+ 6 is all real numbers
except x= 2 and x= 3.
Therefore, the domain of the function is x∈R, x = 2,3.
Question 25
Question
Find the domain of the rational function:
f(x) = 5x−1
x2−9.
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. So, we need to find
the values of xthat make the denominator, x2−9, equal to zero.
Step 2: We solve x2−9 = 0 by factoring as the different of squares:
x2−9 = (x+ 3)(x−3) = 0.
Step 3: Setting each factor equal to zero gives us the values of xto exclude
from the domain:
x+ 3 = 0 ⇒x=−3
x−3 = 0 ⇒x= 3
Step 4: Therefore, the domain of the rational function f(x) = 5x−1
x2−9is all
real numbers except x=−3,3, expressed in interval notation as:
(−∞,−3) ∪(−3,3) ∪(3,∞).
Question 26
Question
Find the vertical asymptotes of the rational function f(x) = x2−4
x2−x−6.
16
Solution
Step 1: Determine the values of xthat make the denominator equal to zero, as
these values will give us the vertical asymptotes of the function. Set x2−x−6 = 0
and solve for x.
Step 1: x2−x−6 = 0
(x−3)(x+ 2) = 0
x= 3 or x=−2
Step 2: Therefore, the function f(x)has vertical asymptotes at x= 3 and
x=−2. Step 3: The vertical asymptotes of the function f(x) = x2−4
x2−x−6are x= 3 and x=−2.
Question 27
Question
Let f(x) = 4x2−5x−6
x2−9be a rational function. Find the vertical and horizontal
asymptotes of the graph of f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator of the rational
function equal to zero and solve for x:
x2−9 = 0
(x+ 3)(x−3) = 0
This gives us x=−3and x= 3 as vertical asymptotes.
Step 2: To find the horizontal asymptote, compare the degree of the nu-
merator and the denominator of the rational function. Since the degree of the
numerator is 2 and the degree of the denominator is also 2, we need to compare
the leading coefficients. Divide the leading coefficient of the numerator by the
leading coefficient of the denominator.
4
1= 4
Therefore, the horizontal asymptote is y= 4.
Question 28
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes in the graph
of the function:
f(x) = 3x2−9x−18
x2−4x−21
17
Solution
Step 1: To find the vertical asymptotes of the function, set the denominator
equal to zero and solve for x.
x2−4x−21 = 0
Factoring the quadratic equation gives:
(x−7)(x+ 3) = 0
So, x= 7 and x=−3are the vertical asymptotes of the function.
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and the denominator. Since both have the same degree, the horizontal
asymptote will be the ratio of the leading coefficients.
The horizontal asymptote is:
y=3
1= 3
Step 3: To find any holes in the graph, simplify the function and check for
common factors in the numerator and denominator.
Factorizing the numerator as 3(x−3)(x+ 2) and the denominator as (x−
7)(x+ 3), we can cancel out the common factor (x+ 3).
The simplified function is:
f(x) = 3(x−3)
x−7
There are no holes in the graph of the function.
Therefore, the function has vertical asymptotes at x= 7 and x=−3, a
horizontal asymptote at y= 3, and no holes in the graph.
Question 29
Question
Simplify the following rational expression:
4x3−7x2−5x
2x2−3x−2∇ · 5x2−7x−6
2x2+ 3x−2
Solution
Step 1: Factor both the numerator and denominator of each fraction.
For the first fraction, we have:
4x3−7x2−5x=x(4x2−7x−5) = x(4x+ 3)(x−5)
2x2−3x−2 = (2x−1)(x+ 2)
18
Therefore, the first fraction simplifies to:
x(4x+ 3)(x−5)
(2x−1)(x+ 2)
For the second fraction, we have:
5x2−7x−6 = (5x+ 2)(x−3)
2x2+ 3x−2 = (2x−1)(x+ 2)
Therefore, the second fraction simplifies to:
(5x+ 2)(x−3)
(2x−1)(x+ 2)
Step 2: Rewrite the division as multiplication by the reciprocal. Then sim-
plify by multiplying by the reciprocal of the second fraction.
x(4x+ 3)(x−5)
(2x−1)(x+ 2) ·(2x−1)(x+ 2)
(5x+ 2)(x−3)
Step 3: Cancel out common factors in the numerator and denominator.
Multiply the remaining terms.
x(4x+ 3)(x−5)
(5x+ 2)(x−3)
Therefore, the simplified expression is:
x(4x+ 3)(x−5)
(5x+ 2)(x−3)
Question 30
Question
Simplify the rational function:
2x3+ 6x2−8x
4x2−16
Solution
Step 1: Factor out common terms in the numerator and denominator. Step 2:
Simplify the rational function by canceling out common factors.
Step 1: First, factor out common terms in the numerator and denominator:
2x3+ 6x2−8x
4x2−16 =2x(x2+ 3x−4)
4(x2−4)
19
Now, factor each quadratic expression:
x2+ 3x−4 = (x+ 4)(x−1)
x2−4 = (x+ 2)(x−2)
So, the rational function becomes:
2x(x+ 4)(x−1)
4(x+ 2)(x−2)
Step 2: Next, simplify the rational function by canceling out common fac-
tors: 2x(x+ 4)(x−1)
4(x+ 2)(x−2) =x(x+ 4)(x−1)
2(x+ 2)(x−2)
Therefore, the simplified form of the rational function is:
x(x+ 4)(x−1)
2(x+ 2)(x−2)
Question 31
Question
Find the equations of the asymptotes of the rational function:
f(x) = 2x2+ 3x−2
x2−4
Solution
Step 1: Determine the vertical asymptotes by setting the denominator equal to
zero and solving for x.
x2−4 = 0
(x−2)(x+ 2) = 0
x= 2 or x=−2
So, the vertical asymptotes are x= 2 and x=−2.
Step 2: Determine the horizontal asymptote by comparing the degrees of
the numerator and the denominator. Since the degree of the numerator is equal
to the degree of the denominator, the horizontal asymptote is given by the ratio
of the leading coefficients. Therefore, the horizontal asymptote is y= 2.
Step 3: Determine the slant (or oblique) asymptote, if it exists. To find
the slant asymptote, perform polynomial long division of the numerator by the
denominator: 2x+7 +1
x2−4 2x2+3x−2
−2x2+8x
0−5x−2
20
The slant asymptote is given by the quotient, which is 2x+ 7.
Therefore, the equations of the asymptotes are:
• Vertical asymptotes: x= 2 and x=−2
• Horizontal asymptote: y= 2
• Slant asymptote: y= 2x+ 7
Question 32
Question
Solve the following rational equation for x:
2
x−3−3
x+ 2 =1
x2−x−6
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation. Since the denominators are (x−3),(x+ 2), and (x2−x−6), the
common denominator is (x−3)(x+ 2).
Step 2: Rewrite the equation with the common denominator:
2(x+ 2)
(x−3)(x+ 2) −3(x−3)
(x−3)(x+ 2) =1
x2−x−6
Step 3: Combine the fractions on the left side of the equation:
2(x+ 2) −3(x−3)
(x−3)(x+ 2) =1
x2−x−6
Step 4: Simplify the equation by expanding and combining like terms:
2x+ 4 −3x+ 9
(x−3)(x+ 2) =1
x2−x−6
−x+ 13
(x−3)(x+ 2) =1
x2−x−6
Step 5: Multiply both sides of the equation by x2−x−6to get rid of the
denominators:
(−x+ 13)(x2−x−6) = 1
−x3+ 13x2−6x+ 13x2−169x+ 78 = 1
−x3+ 26x2−175x+ 78 = 1
21
−x3+ 26x2−175x+ 77 = 0
This is a cubic equation and may not have simple integer solutions. The
roots of this cubic equation can be found using numerical methods or a graphing
calculator.
Question 33
Question
Solve the rational equation: 5
x−3−3
x+2 =8
x2−x−6.
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation.
Step 2: Multiply each fraction by the appropriate factor to obtain the com-
mon denominator.
Step 3: Combine the fractions on the left side by adding or subtracting
numerators.
Step 4: Simplify the resulting fraction.
Step 5: Factor the denominator on the right side of the equation.
Step 6: Rewrite the equation with the factored form of the denominator.
Step 7: Multiply both sides of the equation by the factored form of the
denominator to eliminate the fraction.
Step 8: Solve the resulting equation.
Step 9: Check the solutions in the original equation to ensure they are valid.
Question 34
Question
Find the vertical asymptotes of the rational function:
f(x) = x3−2x2−x+ 2
x2−2x−3
Solution
To find the vertical asymptotes of the rational function f(x), we need to deter-
mine where the denominator is equal to zero, but the numerator is not equal to
zero at those points.
Step 1: Find where the denominator is equal to zero. Setting the
denominator equal to zero gives:
x2−2x−3 = 0
22
We can factor the quadratic equation as:
(x−3)(x+ 1) = 0
So, x= 3 and x=−1are the values where the denominator is equal to zero.
Step 2: Determine where the numerator is not equal to zero at
those points. Substitute x= 3 first:
f(3) = 33−2(3)2−3+2
32−2(3) −3=27 −18 −3+2
9−6−3=8
0
Since the numerator is not equal to zero at x= 3, there is a vertical asymp-
tote at x= 3.
Next, substitute x=−1:
f(−1) = (−1)3−2(−1)2−(−1) + 2
(−1)2−2(−1) −3=−1+2+1+2
1+2−3=4
0
Since the numerator is not equal to zero at x=−1, there is a vertical
asymptote at x=−1.
Therefore, the vertical asymptotes of the rational function f(x)are x= 3
and x=−1.
Question 35
Question
Find the domain of the rational function:
f(x) = 3x2−7x−6
x2−5x−14
Solution
Step 1: To find the domain of a rational function, we need to determine the
values that xcannot take in order to avoid division by zero.
Step 2: The function f(x)is defined for all real numbers xexcept for those
that make the denominator equal to zero.
Step 3: Set the denominator equal to zero and solve for x:
x2−5x−14 = 0
(x−7)(x+ 2) = 0
x= 7 or x=−2
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except for x= 7 and x=−2.
Step 5: In interval notation, the domain can be expressed as (−∞,−2) ∪
(−2,7) ∪(7,∞).
23
Step 3: Let’s solve x2−4x−5 = 0 by factoring or using the quadratic
formula.
Step 4: Factoring the quadratic equation x2−4x−5 = 0 gives us (x−5)(x+
1) = 0.
Step 5: Setting each factor to zero gives us x−5 = 0 or x+ 1 = 0.
Step 6: Solving these equations, we get x= 5 and x=−1.
Step 7: Therefore, the domain of f(x)is all real numbers except x= 5 and
x=−1. In interval notation, the domain is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 4
Question
Find the vertical asymptotes of the rational function
f(x) = x2+ 3x−4
x2−1.
Solution
Step 1: Set the denominator equal to 0 and solve for xto find the vertical
asymptotes.
We need to find the values of xfor which the denominator of the rational
function is equal to 0. In this case, the denominator is x2−1, so we set x2−1 = 0
and solve for x.
x2−1 = 0
x2= 1
x=±1
Therefore, the vertical asymptotes are x= 1 and x=−1.
So, the vertical asymptotes of the rational function f(x) = x2+3x−4
x2−1are
x= 1 and x=−1.
Question 5
Question
Find the domain of the rational function:
f(x) = 5
x2−4x−5
2
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xfor which the function is defined. In this case, our function is defined
for all xexcept where the denominator is equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation, we can factor it:
(x−5)(x+ 1) = 0
This gives us two possible values for x:x= 5 and x=−1.
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 5 and x=−1.
Step 5: In interval notation, the domain can be expressed as: (−∞,−1) ∪
(−1,5) ∪(5,∞).
Question 6
Question
Solve the following rational equation for x:2
x−3−3
x+2 =5
x2−x−6.
Solution
Step 1: To get rid of the denominators, we will multiply both sides of the
equation by the least common multiple (LCM) of the denominators. In this
case, the LCM is (x−3)(x+ 2)(x+ 1).
2
x−3·(x−3)(x+2)(x+1)−3
x+ 2·(x−3)(x+2)(x+1) = 5
x2−x−6·(x−3)(x+2)(x+1)
Step 2: Simplify the equation obtained after multiplying.
2(x+ 2)(x+ 1) −3(x−3)(x+ 1) = 5(x−3)(x+ 2)
Step 3: Expand and simplify both sides of the equation.
2(x2+ 3x+ 2) −3(x2−2x−3) = 5(x2−x−6)
2x2+ 6x+ 4 −3x2+ 6x+ 9 = 5x2−5x−30
Step 4: Combine like terms.
−x2+ 12x+ 13 = 5x2−5x−30
Step 5: Move all terms to one side to set the equation to zero.
6x2−17x−43 = 0
3
Step 6: Solve the quadratic equation using the quadratic formula: x=
−b±√b2−4ac
2a.
x=17 ±√(−17)2−4(6)(−43)
2(6)
x=17 ±√289 + 1032
12
x=17 ±√1321
12
x=17 ±37
12
Step 7: Solve for x.x=17+37
12 or x=17−37
12
x=54
12 or x=−20
12
x= 4.5or x=−1.67
Therefore, the solutions to the equation are x= 4.5and x=−1.67.
Question 7
Question
Find the domain of the rational function:
f(x) = 3x
x2−4
Solution
Step 1: Recall that the domain of a function is the set of all real numbers for
which the function is defined. In the case of rational functions, the function is
defined as long as the denominator is not equal to zero.
Step 2: To find the domain of f(x), we need to find the values of xthat
make the denominator, x2−4, equal to zero.
Step 3: The denominator, x2−4, can be factored as (x+ 2)(x−2). To find
the values of xthat make the denominator zero, we set each factor equal to
zero: x+ 2 = 0 and x−2 = 0.
Step 4: Solving x+ 2 = 0, we get x=−2. Solving x−2 = 0, we get x= 2.
Step 5: Therefore, the domain of the function f(x) = 3x
x2−4is all real numbers
except x=−2and x= 2.
Step 6: Combining all the allowed values of x, the domain of the function is
(−∞,−2) ∪(−2,2) ∪(2,∞).
4
Question 8
Question
Find the vertical asymptotes, horizontal asymptotes, and intercepts of the ra-
tional function:
f(x) = 3x2−2x−1
x2+ 2x−3
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to zero and
solve for x.
x2+ 2x−3 = 0
(x+ 3)(x−1) = 0
x=−3or x= 1
So, the vertical asymptotes are x=−3and x= 1.
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and the denominator. Since the degree of the numerator is equal to the
degree of the denominator, divide the leading coefficients to find the horizontal
asymptote.
y=3
1= 3
Therefore, the horizontal asymptote is y= 3.
Step 3: To find the y-intercept, plug in x= 0 into the function.
f(0) = 3(0)2−2(0) −1
(0)2+ 2(0) −3=−1
−3=1
3
Thus, the y-intercept is (0,1
3).
Step 4: To find the x-intercept, set y= 0 and solve for x.
3x2−2x−1
x2+ 2x−3= 0
3x2−2x−1 = 0
(3x+ 1)(x−1) = 0
x=−1
3or x= 1
Therefore, the x-intercepts are (−1
3,0) and (1,0).
Question 9
Question
Let f(x) = 2x3−5x2+3x
x2−3x+2 . Determine the vertical asymptotes of the function f(x).
5
Solution
Step 1: To find the vertical asymptotes of f(x), we need to identify the values
of xfor which the denominator of f(x)equals zero.
Step 2: The denominator of f(x)is x2−3x+ 2. We find the roots of this
quadratic equation by solving x2−3x+ 2 = 0.
Step 3: To solve x2−3x+ 2 = 0, we can factor the quadratic equation or
use the quadratic formula x=−b±√b2−4ac
2a, where the coefficients are a= 1,
b=−3, and c= 2.
Step 4: Using the quadratic formula, we have x=3±√(−3)2−4(1)(2)
2(1) .
Step 5: Simplifying under the square root, we get x=3±√9−8
2.
Step 6: Further simplifying, we have x=3±√1
2.
Step 7: This gives us two possible values for x:x=3+1
2= 2 and x=3−1
2=
1.
Step 8: Therefore, the vertical asymptotes of the function f(x)are x= 2
and x= 1.
Question 10
Question
Simplify the following rational expression:
3x2+ 5x−2
2x2−3x−2
Solution
Step 1: Factor both the numerator and denominator of the rational expression.
Step 2: Factor the numerator: Since 3× −2 = −6and 5 = 3 + 2:
3x2+ 5x−2 = (3x−1)(x+ 2)
Step 3: Factor the denominator: Since 2× −2 = −4and −3 = −4+1:
2x2−3x−2 = (2x+ 1)(x−2)
Step 4: Substitute the factored expressions back into the original rational
expression:
(3x−1)(x+ 2)
(2x+ 1)(x−2)
Therefore, the simplified form of the given rational expression is 3x−1
2x+ 1 .
6
Question 11
Question
Let f(x) = x2−1
x2+ 2x−3and g(x) = x2+x−6
x2+ 2x−3be rational functions. Find
the domain of f(x)·g(x).
Solution
Step 1: Find the product of f(x)and g(x).
f(x)·g(x) = (x2−1
x2+ 2x−3)·(x2+x−6
x2+ 2x−3)
Step 2: Simplify the expression by multiplying the numerators and denomi-
nators.
f(x)·g(x) = (x2−1)(x2+x−6)
(x2+ 2x−3)(x2+ 2x−3)
f(x)·g(x) = x4+x3−6x2−x2+x+ 6
(x2+ 2x−3)2
f(x)·g(x) = x4+x3−7x2+x+ 6
(x2+ 2x−3)2
Step 3: Determine the domain of the function f(x)·g(x)by finding the
values of xfor which the denominator is not equal to zero. The denominator
(x2+ 2x−3)2will be zero when x2+ 2x−3 = 0. Solving x2+ 2x−3 = 0 gives
the solutions: x=−3and x= 1.
Therefore, the domain of f(x)·g(x)is all real numbers except x=−3and
x= 1. In interval notation, the domain is (−∞,−3) ∪(−3,1) ∪(1,∞).
Question 12
Question
Determine the domain of the rational function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: To find the domain of a rational function, we need to identify all real
numbers that make the denominator equal to zero, as division by zero is unde-
fined.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
7
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 or x+ 1 = 0
Step 5: Solve for x:
x= 5 or x=−1
Step 6: Therefore, the domain of f(x)is all real numbers except x= 5 and
x=−1, since these values would make the denominator zero.
Step 7: Thus, the domain of the function f(x) = 2x2−5x−3
x2−4x−5is x∈R\{5,−1}.
Question 13
Question
Simplify the following rational expression:
4x3−8x2+ 4x
2x2−4x
Solution
Step 1: Factor out the common terms in the numerator and the denomina-
tor. Step 2: Cancel out any common factors between the numerator and the
denominator. Step 3: Simplify the resulting expression.
Step 1: Factor out the common terms in the numerator and the denomina-
tor. 4x3−8x2+ 4x
2x2−4x=4x(x2−2x+ 1)
2x(x−2)
Step 2: Cancel out any common factors between the numerator and the
denominator. 4x(x2−2x+ 1)
2x(x−2) =4(x2−2x+ 1)
x−2
Step 3: Simplify the resulting expression.
4(x−1)2
x−2
Therefore, the simplified form of the rational expression is 4(x−1)2
x−2.
8
Question 14
Question
Simplify the following rational expression:
3x2−9x
x2−4∇ · x2−5x+ 6
x2−2x−8
Solution
Step 1: We begin by writing the division of fractions as a multiplication of
fractions by taking the reciprocal of the second fraction:
3x2−9x
x2−4·x2−2x−8
x2−5x+ 6
Step 2: Factor the expressions in both the numerator and denominator of
each fraction: 3x(x−3)
(x+ 2)(x−2) ·(x−4)(x+ 2)
(x−3)(x−2)
Step 3: Cancel out common factors in the numerator and denominator:
3x(x−3)
(x+ 2)(x−2) ·(x−4)(x+ 2)
(x−3)(x−2)
Step 4: Multiply the remaining factors:
3x
x−3
Therefore, the simplified form of the rational expression is 3x
x−3.
Question 15
Question
Simplify the rational function:
f(x) = x2−5x−6
x2−4x−5
Solution
Step 1: Factor the numerator and denominator of the rational function. Step
2: Simplify the rational function by canceling out common factors, if possible.
Step 1: Factor the numerator and denominator of the rational function. The
numerator factors to (x−6)(x+1), and the denominator factors to (x−5)(x+1).
So, f(x) = (x−6)(x+1)
(x−5)(x+1) .
9
Step 2: Simplify the rational function by canceling out common factors, if
possible. We can cancel out the common factor of (x+ 1) in the numerator and
denominator.
So, f(x) = x−6
x−5.
Therefore, the simplified form of the rational function is f(x) = x−6
x−5.
Question 16
Question
Determine all the values of xfor which the rational function x2−4
x2−3x−4is
undefined.
Solution
Step 1: The rational function x2−4
x2−3x−4is undefined when the denominator
is equal to zero, since division by zero is undefined. Therefore, we need to solve
the equation x2−3x−4=0to find the values of xthat make the rational
function undefined.
Step 2: To solve x2−3x−4 = 0, we can factor the quadratic equation by
finding two numbers that multiply to −4and add to −3. These numbers are
−4and 1.
Step 3: Therefore, the factored form of the quadratic equation is (x−4)(x+
1) = 0. Setting each factor to zero gives us x−4 = 0 and x+ 1 = 0.
Step 4: Solving x−4=0gives x= 4, and solving x+ 1 = 0 gives x=−1.
Thus, the values of xfor which the rational function is undefined are x=−1
and x= 4.
Question 17
Question
Find the domain of the function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: To find the domain of the function, we need to identify any 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
10
Step 3: This is a quadratic equation that can be factored:
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero:
x−5 = 0 or x+ 1 = 0
Step 5: Solve for x:
x= 5 or x=−1
Step 6: Therefore, the domain of the function f(x)is all real numbers except
x= 5 and x=−1, since those values make the denominator zero. Step 7: The
domain of the function f(x)is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 18
Question
Let f(x) = 2x2+ 5x−3
x2−4x−5. Find the x-intercepts of the rational function f(x).
Solution
To find the x-intercepts of a function, we set f(x)equal to zero and solve for x.
Step 1: Set f(x)equal to zero:
2x2+ 5x−3
x2−4x−5= 0
Step 2: Factor the numerator and denominator:
(2x−1)(x+ 3)
(x−5)(x+ 1) = 0
Step 3: Use the zero product property:
2x−1 = 0 or x+ 3 = 0 or x−5 = 0 or x+ 1 = 0
Step 4: Solve for xin each equation:
x=1
2, x =−3, x = 5, x =−1
Step 5: The x-intercepts of the rational function f(x)are x=1
2,x=−3,
x= 5, and x=−1.
11
Question 19
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes in the graph
of the rational function given by:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes, we set the denominator equal to zero and
solve for x:
x2−4x−5 = 0
Factoring the quadratic, we have:
(x−5)(x+ 1) = 0
So, x= 5 and x=−1are the vertical asymptotes.
Step 2: Horizontal Asymptotes
To find the horizontal asymptote, we determine what happens as xap-
proaches infinity. Divide the leading coefficient of the numerator by the leading
coefficient of the denominator to find the horizontal asymptote:
y=2
1= 2
So, the horizontal asymptote is y= 2.
Step 3: Holes
To find any holes in the graph, we look for points where the numerator and
denominator can both be factored and canceled out. In this case, we notice both
2x2−5x−3and x2−4x−5can be factored as (2x+1)(x−3) and (x−5)(x+1),
respectively.
Therefore, the function simplifies to:
f(x) = 2x2−5x−3
x2−4x−5=(2x+ 1)(x−3)
(x−5)(x+ 1)
This indicates a hole at x= 5.
In conclusion, the rational function f(x) = 2x2−5x−3
x2−4x−5has vertical asymptotes
at x= 5 and x=−1, a horizontal asymptote at y= 2, and a hole at x= 5.
12
Question 20
Question
Find the vertical and horizontal asymptotes of the rational function: f(x) =
3x2−2x+5
x2+4x−5.
Solution
Step 1: To find the vertical asymptotes of the function, we need to determine
where the denominator is equal to zero (if any). Set x2+ 4x−5=0and solve
for x.
x2+ 4x−5 = 0
(x+ 5)(x−1) = 0
x=−5or x= 1
So, the vertical asymptotes are at x=−5and x= 1.
Step 2: To find the horizontal asymptote, we observe the degrees of the
numerator and denominator. Since the degrees are equal, we look at the leading
coefficients to determine the horizontal asymptote. The horizontal asymptote
is the ratio of the leading coefficients.
lim
x→∞
3x2−2x+ 5
x2+ 4x−5=3
1= 3
Therefore, the horizontal asymptote is y= 3.
In summary, the vertical asymptotes are at x=−5and x= 1, and the
horizontal asymptote is at y= 3 for the rational function f(x) = 3x2−2x+5
x2+4x−5.
Question 21
Question
Find the domain of the rational function:
f(x) = x2−6x+ 8
x2−4x−5
Solution
Step 1: We need to determine the values of xfor which the denominator x2−
4x−5is not equal to zero, since division by zero is undefined in mathematics.
So, we solve the equation x2−4x−5=0to find the values of xthat make
the denominator zero. This is equivalent to finding the roots of the quadratic
equation x2−4x−5 = 0.
13
Step 2: To find the roots of the quadratic equation x2−4x−5 = 0, we can
use the quadratic formula:
x=−b±√b2−4ac
2a
Step 3: Comparing the given equation with the standard form of a quadratic
equation ax2+bx +c= 0, we have a= 1,b=−4, and c=−5. Plugging these
values into the quadratic formula, we get:
x=4±√(−4)2−4∗1∗(−5)
2∗1
x=4±√16 + 20
2
x=4±√36
2
x=4±6
2
Step 4: Therefore, the roots of the quadratic equation are x= 5 and x=−1.
These are the values of xfor which the denominator becomes zero, making the
rational function undefined.
Step 5: Finally, the domain of the rational function f(x) = x2−6x+ 8
x2−4x−5is
all real numbers except x= 5 and x=−1, since these values would make the de-
nominator zero. So, the domain of the function is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 22
Question
Find the domain of the rational function:
f(x) = 5x2−7x+ 2
x2−4x−5
Solution
Step 1: We need to find the values of xfor which the denominator is not equal
to zero, since division by zero is undefined. The denominator of the rational
function is x2−4x−5. We solve the equation x2−4x−5 = 0 to find the values
of xwhere the denominator is zero.
To solve the equation:
x2−4x−5 = 0
we can use the quadratic formula:
x=−(−4) ±√(−4)2−4·1·(−5)
2·1
14
x=4±√16 + 20
2
x=4±√36
2
x=4±6
2
x=4+6
2or x=4−6
2
x= 5 or x=−1
Step 2: The domain of the function f(x)is all real numbers except the values
that make the denominator zero. So, the domain of f(x)is:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 23
Question
Simplify the following rational function:
f(x) = 3x2−5x−2
x2−4x+ 4
Solution
Step 1: Factor the numerator and denominator of the rational function:
f(x) = 3x2−5x−2
(x−2)2
Step 2: Factor the numerator:
f(x) = (3x+ 1)(x−2)
(x−2)2
Step 3: Simplify the rational function:
f(x) = 3x+ 1
x−2
Therefore, the simplified form of the rational function f(x)is 3x+1
x−2.
Question 24
Question
Find the domain of the following rational function:
f(x) = 1
x2−5x+ 6.
15
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that would make the denominator equal to zero. So, we need to find the
values of xthat would make x2−5x+ 6 = 0.
Step 2: To find these values of x, we need to factor the quadratic equation
x2−5x+ 6 = 0. The factored form is (x−2)(x−3) = 0.
Step 3: Setting each factor to zero, we find the roots of the equation: x−2 = 0
or x−3 = 0.
Step 4: Solving these equations gives us x= 2 and x= 3.
Step 5: So, the domain of the function f(x) = 1
x2−5x+ 6 is all real numbers
except x= 2 and x= 3.
Therefore, the domain of the function is x∈R, x = 2,3.
Question 25
Question
Find the domain of the rational function:
f(x) = 5x−1
x2−9.
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. So, we need to find
the values of xthat make the denominator, x2−9, equal to zero.
Step 2: We solve x2−9 = 0 by factoring as the different of squares:
x2−9 = (x+ 3)(x−3) = 0.
Step 3: Setting each factor equal to zero gives us the values of xto exclude
from the domain:
x+ 3 = 0 ⇒x=−3
x−3 = 0 ⇒x= 3
Step 4: Therefore, the domain of the rational function f(x) = 5x−1
x2−9is all
real numbers except x=−3,3, expressed in interval notation as:
(−∞,−3) ∪(−3,3) ∪(3,∞).
Question 26
Question
Find the vertical asymptotes of the rational function f(x) = x2−4
x2−x−6.
16
Solution
Step 1: Determine the values of xthat make the denominator equal to zero, as
these values will give us the vertical asymptotes of the function. Set x2−x−6 = 0
and solve for x.
Step 1: x2−x−6 = 0
(x−3)(x+ 2) = 0
x= 3 or x=−2
Step 2: Therefore, the function f(x)has vertical asymptotes at x= 3 and
x=−2. Step 3: The vertical asymptotes of the function f(x) = x2−4
x2−x−6are x= 3 and x=−2.
Question 27
Question
Let f(x) = 4x2−5x−6
x2−9be a rational function. Find the vertical and horizontal
asymptotes of the graph of f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator of the rational
function equal to zero and solve for x:
x2−9 = 0
(x+ 3)(x−3) = 0
This gives us x=−3and x= 3 as vertical asymptotes.
Step 2: To find the horizontal asymptote, compare the degree of the nu-
merator and the denominator of the rational function. Since the degree of the
numerator is 2 and the degree of the denominator is also 2, we need to compare
the leading coefficients. Divide the leading coefficient of the numerator by the
leading coefficient of the denominator.
4
1= 4
Therefore, the horizontal asymptote is y= 4.
Question 28
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes in the graph
of the function:
f(x) = 3x2−9x−18
x2−4x−21
17
Solution
Step 1: To find the vertical asymptotes of the function, set the denominator
equal to zero and solve for x.
x2−4x−21 = 0
Factoring the quadratic equation gives:
(x−7)(x+ 3) = 0
So, x= 7 and x=−3are the vertical asymptotes of the function.
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and the denominator. Since both have the same degree, the horizontal
asymptote will be the ratio of the leading coefficients.
The horizontal asymptote is:
y=3
1= 3
Step 3: To find any holes in the graph, simplify the function and check for
common factors in the numerator and denominator.
Factorizing the numerator as 3(x−3)(x+ 2) and the denominator as (x−
7)(x+ 3), we can cancel out the common factor (x+ 3).
The simplified function is:
f(x) = 3(x−3)
x−7
There are no holes in the graph of the function.
Therefore, the function has vertical asymptotes at x= 7 and x=−3, a
horizontal asymptote at y= 3, and no holes in the graph.
Question 29
Question
Simplify the following rational expression:
4x3−7x2−5x
2x2−3x−2∇ · 5x2−7x−6
2x2+ 3x−2
Solution
Step 1: Factor both the numerator and denominator of each fraction.
For the first fraction, we have:
4x3−7x2−5x=x(4x2−7x−5) = x(4x+ 3)(x−5)
2x2−3x−2 = (2x−1)(x+ 2)
18
Therefore, the first fraction simplifies to:
x(4x+ 3)(x−5)
(2x−1)(x+ 2)
For the second fraction, we have:
5x2−7x−6 = (5x+ 2)(x−3)
2x2+ 3x−2 = (2x−1)(x+ 2)
Therefore, the second fraction simplifies to:
(5x+ 2)(x−3)
(2x−1)(x+ 2)
Step 2: Rewrite the division as multiplication by the reciprocal. Then sim-
plify by multiplying by the reciprocal of the second fraction.
x(4x+ 3)(x−5)
(2x−1)(x+ 2) ·(2x−1)(x+ 2)
(5x+ 2)(x−3)
Step 3: Cancel out common factors in the numerator and denominator.
Multiply the remaining terms.
x(4x+ 3)(x−5)
(5x+ 2)(x−3)
Therefore, the simplified expression is:
x(4x+ 3)(x−5)
(5x+ 2)(x−3)
Question 30
Question
Simplify the rational function:
2x3+ 6x2−8x
4x2−16
Solution
Step 1: Factor out common terms in the numerator and denominator. Step 2:
Simplify the rational function by canceling out common factors.
Step 1: First, factor out common terms in the numerator and denominator:
2x3+ 6x2−8x
4x2−16 =2x(x2+ 3x−4)
4(x2−4)
19
Now, factor each quadratic expression:
x2+ 3x−4 = (x+ 4)(x−1)
x2−4 = (x+ 2)(x−2)
So, the rational function becomes:
2x(x+ 4)(x−1)
4(x+ 2)(x−2)
Step 2: Next, simplify the rational function by canceling out common fac-
tors: 2x(x+ 4)(x−1)
4(x+ 2)(x−2) =x(x+ 4)(x−1)
2(x+ 2)(x−2)
Therefore, the simplified form of the rational function is:
x(x+ 4)(x−1)
2(x+ 2)(x−2)
Question 31
Question
Find the equations of the asymptotes of the rational function:
f(x) = 2x2+ 3x−2
x2−4
Solution
Step 1: Determine the vertical asymptotes by setting the denominator equal to
zero and solving for x.
x2−4 = 0
(x−2)(x+ 2) = 0
x= 2 or x=−2
So, the vertical asymptotes are x= 2 and x=−2.
Step 2: Determine the horizontal asymptote by comparing the degrees of
the numerator and the denominator. Since the degree of the numerator is equal
to the degree of the denominator, the horizontal asymptote is given by the ratio
of the leading coefficients. Therefore, the horizontal asymptote is y= 2.
Step 3: Determine the slant (or oblique) asymptote, if it exists. To find
the slant asymptote, perform polynomial long division of the numerator by the
denominator: 2x+7 +1
x2−4 2x2+3x−2
−2x2+8x
0−5x−2
20
The slant asymptote is given by the quotient, which is 2x+ 7.
Therefore, the equations of the asymptotes are:
• Vertical asymptotes: x= 2 and x=−2
• Horizontal asymptote: y= 2
• Slant asymptote: y= 2x+ 7
Question 32
Question
Solve the following rational equation for x:
2
x−3−3
x+ 2 =1
x2−x−6
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation. Since the denominators are (x−3),(x+ 2), and (x2−x−6), the
common denominator is (x−3)(x+ 2).
Step 2: Rewrite the equation with the common denominator:
2(x+ 2)
(x−3)(x+ 2) −3(x−3)
(x−3)(x+ 2) =1
x2−x−6
Step 3: Combine the fractions on the left side of the equation:
2(x+ 2) −3(x−3)
(x−3)(x+ 2) =1
x2−x−6
Step 4: Simplify the equation by expanding and combining like terms:
2x+ 4 −3x+ 9
(x−3)(x+ 2) =1
x2−x−6
−x+ 13
(x−3)(x+ 2) =1
x2−x−6
Step 5: Multiply both sides of the equation by x2−x−6to get rid of the
denominators:
(−x+ 13)(x2−x−6) = 1
−x3+ 13x2−6x+ 13x2−169x+ 78 = 1
−x3+ 26x2−175x+ 78 = 1
21
−x3+ 26x2−175x+ 77 = 0
This is a cubic equation and may not have simple integer solutions. The
roots of this cubic equation can be found using numerical methods or a graphing
calculator.
Question 33
Question
Solve the rational equation: 5
x−3−3
x+2 =8
x2−x−6.
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation.
Step 2: Multiply each fraction by the appropriate factor to obtain the com-
mon denominator.
Step 3: Combine the fractions on the left side by adding or subtracting
numerators.
Step 4: Simplify the resulting fraction.
Step 5: Factor the denominator on the right side of the equation.
Step 6: Rewrite the equation with the factored form of the denominator.
Step 7: Multiply both sides of the equation by the factored form of the
denominator to eliminate the fraction.
Step 8: Solve the resulting equation.
Step 9: Check the solutions in the original equation to ensure they are valid.
Question 34
Question
Find the vertical asymptotes of the rational function:
f(x) = x3−2x2−x+ 2
x2−2x−3
Solution
To find the vertical asymptotes of the rational function f(x), we need to deter-
mine where the denominator is equal to zero, but the numerator is not equal to
zero at those points.
Step 1: Find where the denominator is equal to zero. Setting the
denominator equal to zero gives:
x2−2x−3 = 0
22
We can factor the quadratic equation as:
(x−3)(x+ 1) = 0
So, x= 3 and x=−1are the values where the denominator is equal to zero.
Step 2: Determine where the numerator is not equal to zero at
those points. Substitute x= 3 first:
f(3) = 33−2(3)2−3+2
32−2(3) −3=27 −18 −3+2
9−6−3=8
0
Since the numerator is not equal to zero at x= 3, there is a vertical asymp-
tote at x= 3.
Next, substitute x=−1:
f(−1) = (−1)3−2(−1)2−(−1) + 2
(−1)2−2(−1) −3=−1+2+1+2
1+2−3=4
0
Since the numerator is not equal to zero at x=−1, there is a vertical
asymptote at x=−1.
Therefore, the vertical asymptotes of the rational function f(x)are x= 3
and x=−1.
Question 35
Question
Find the domain of the rational function:
f(x) = 3x2−7x−6
x2−5x−14
Solution
Step 1: To find the domain of a rational function, we need to determine the
values that xcannot take in order to avoid division by zero.
Step 2: The function f(x)is defined for all real numbers xexcept for those
that make the denominator equal to zero.
Step 3: Set the denominator equal to zero and solve for x:
x2−5x−14 = 0
(x−7)(x+ 2) = 0
x= 7 or x=−2
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except for x= 7 and x=−2.
Step 5: In interval notation, the domain can be expressed as (−∞,−2) ∪
(−2,7) ∪(7,∞).
23
