MATH 121 - COLLEGE ALGEBRA -
Rational Functions
Question Bank - Set 9
Liberty University
Question 1
Question
Let f(x) = 2x3
−5x2
−3x+2
x2−3x+2 . Find the domain of f(x).
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that would make the denominator zero. So, we need to find the values of
xthat would result in a zero denominator.
Step 2: We set the denominator x2−3x+ 2 equal to zero and solve for x.
x2−3x+ 2 = 0
Step 3: To solve the quadratic equation, we can factor or use the quadratic
formula. Let’s factor this equation.
(x−1)(x−2) = 0
Step 4: Setting each factor to zero gives us two possible values for x:
x−1 = 0 =⇒x= 1
x−2 = 0 =⇒x= 2
Step 5: Therefore, the values x= 1 and x= 2 would make the denominator
zero. Hence, the domain of f(x)is all real numbers except x= 1 and x= 2,
which can be expressed as (−∞,1) ∪(1,2) ∪(2,∞).
Question 2
Question
Given the rational function f(x) = 2x2
−x−3
x2−5x+6 , find the following: 1. Domain of
the function f(x). 2. x-intercepts, if any. 3. y-intercept, if any. 4. Vertical
asymptotes, if any. 5. Horizontal asymptotes, if any.
Solution
1. Domain of the function f(x):
The domain of a rational function includes all real numbers except for the
values that make the denominator equal to zero. Set the denominator equal to
zero and solve for x:
x2−5x+ 6 = 0
Factoring, we get:
(x−2)(x−3) = 0
So, x= 2 or x= 3. Therefore, the domain of f(x)is all real numbers except
x= 2 and x= 3, or in interval notation: (−∞,2) ∪(2,3) ∪(3,∞).
2. x-intercepts:
To find the x-intercepts, set f(x) = 0 and solve for x:
2x2−x−3
x2−5x+ 6 = 0
This occurs when the numerator is zero. So, 2x2−x−3 = 0. Factoring, we
get:
(2x+ 1)(x−3) = 0
So, x=−1
2or x= 3. Therefore, the x-intercepts are (−1
2,0) and (3,0).
3. y-intercept:
To find the y-intercept, set x= 0 in the function:
f(0) = 2(0)2−0−3
(0)2−5(0) + 6 =−3
6=−1
2
Therefore, the y-intercept is (0,−1
2).
4. Vertical asymptotes:
Vertical asymptotes occur at the values of xthat make the denominator of
the function equal to zero. We found earlier that the denominator is zero at
x= 2 and x= 3. So, the vertical asymptotes are x= 2 and x= 3.
5. Horizontal asymptotes:
2
To find the horizontal asymptotes, we compare the degrees of the numerator
and denominator of the function. Since the degrees are the same, we look at
the leading coefficients:
The leading coefficient of the numerator is 2, and the leading coefficient of
the denominator is 1. Therefore, there is a horizontal asymptote at y= 2/1 = 2.
In summary: - Domain: (−∞,2) ∪(2,3) ∪(3,∞)-x-intercepts: (−1
2,0) and
(3,0) -y-intercept: (0,−1
2)- Vertical asymptotes: x= 2 and x= 3 - Horizontal
asymptote: y= 2
Question 3
Question
Find the domain of the rational function:
f(x) = 3x2−6x
x2−4
Solution
To find the domain of the rational function, we need to determine which values
of xmake the denominator equal to zero, since division by zero is undefined.
Step 1: Set the denominator equal to zero and solve for x:
x2−4 = 0
Step 2: Factor the quadratic equation:
(x−2)(x+ 2) = 0
Step 3: Use the zero product property to find the values of x:
x−2 = 0 OR x+ 2 = 0
Step 4: Solve for xin each equation:
x= 2 OR x=−2
Thus, the domain of the rational function f(x)is all real numbers except
x=−2and x= 2, since these values would make the denominator equal to
zero. Therefore, the domain is (−∞,−2) ∪(−2,2) ∪(2,∞).
Question 4
Question
Find the domain of the rational function:
f(x) = 5x2−4x−3
x2−9
3
Solution
Step 1: To find the domain of the rational function, we need to identify the
values of xthat would make the denominator equal to zero. Since division by
zero is undefined, these values must be excluded from the domain.
Step 2: Set the denominator equal to zero and solve for x:
x2−9 = 0
Step 3: Factor the quadratic equation:
(x+ 3)(x−3) = 0
Step 4: Use the zero-product property to find the solutions:
x+ 3 = 0 or x−3 = 0
Step 5: Solve for x:
For x+ 3 = 0:
x=−3
For x−3 = 0:
x= 3
Step 6: The domain of the rational function f(x)is all real numbers except
x=−3and x= 3, since these values make the denominator zero.
Step 7: Therefore, the domain of the rational function f(x) = 5x2
−4x−3
x2−9is
(−∞,−3) ∪(−3,3) ∪(3,∞).
Question 5
Question
Determine 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 the values of xfor
which the function is defined. In a rational function, the function is undefined
when the denominator is equal to zero. Therefore, we need to find the values of
xthat make the denominator x2−4x−5equal to zero.
Step 2: Let’s solve the equation x2−4x−5 = 0 to find the values that make
the denominator zero. This can be factored as (x−5)(x+ 1) = 0. Setting each
factor to zero gives us x−5 = 0 and x+ 1 = 0. Thus, x= 5 and x=−1.
4
Step 3: Therefore, the domain of the function f(x)is all real numbers except
x= 5 and x=−1. The domain is given by (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 6
Question
Simplify the following rational expression:
3x2−2x−8
x2−4
Solution
Step 1: Factor the numerator and denominator.
The numerator factors as 3x2−2x−8 = (3x+ 4)(x−2).
The denominator factors as x2−4 = (x+ 2)(x−2).
Step 2: Rewrite the expression with factored numerator and denominator.
3x2−2x−8
x2−4=(3x+ 4)(x−2)
(x+ 2)(x−2)
Step 3: Simplify by canceling out the common factor (x−2) in the numerator
and denominator.
(3x+ 4)(x−2)
(x+ 2)(x−2) =3x+ 4
x+ 2
Therefore, the simplified form of the rational expression is 3x+4
x+2 .
Question 7
Question
Find the domain of the rational function:
f(x) = 2x2−5x−3
x2−4x−7
Solution
Step 1: We start by identifying the values of xthat would make the denominator
equal to zero since division by zero is undefined. Set the denominator equal to
zero and solve for x:
x2−4x−7 = 0
5
Step 2: To solve this quadratic equation, we can use the quadratic formula:
x=−(−4) ±√(−4)2−4(1)(−7)
2(1)
Step 3: Simplifying under the square root:
x=4±√16 + 28
2=4±√44
2=4±2√11
2= 2 ±√11
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except 2 + √11 and 2−√11.
Domain = (−∞,2−√11) ∪(2 −√11,2 + √11) ∪(2 + √11,∞)
Question 8
Question
Let f(x) = 2x2
−x−3
x2−4x−5. Find the domain of the function f(x).
Solution
Step 1: The domain of a rational function consists of all real numbers except
the values that make the denominator equal to zero. Thus, we need to find the
values of xthat make x2−4x−5equal to zero.
Step 2: To find these values, we can factor the denominator. The factored
form of x2−4x−5is (x−5)(x+ 1), so the denominator becomes (x−5)(x+ 1).
Step 3: Setting the denominator equal to zero, we get (x−5)(x+ 1) = 0.
This equation is true when either x−5 = 0 or x+ 1 = 0.
Step 4: Solving x−5=0gives x= 5, and solving x+ 1 = 0 gives x=−1.
Therefore, x= 5 and x=−1are the values that make the denominator zero.
Step 5: Hence, the domain of the function f(x)is all real numbers except x=
5and x=−1. In interval notation, the domain is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 9
Question
Find the vertical asymptotes of the rational function:
f(x) = 2x2−x−3
x2−4x−32
6
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero
and solving for x.
x2−4x−32 = 0
Step 2: Factor the quadratic equation.
(x−8)(x+ 4) = 0
Step 3: Set each factor equal to zero and solve for x.
x−8 = 0 =⇒x= 8
x+ 4 = 0 =⇒x=−4
Step 4: Therefore, the vertical asymptotes of the rational function f(x) =
2x2
−x−3
x2−4x−32 are x= 8 and x=−4.
Question 10
Question
Find the vertical asymptotes of the rational function:
f(x) = 3x2−5x−2
x2−9
Solution
Step 1: To find the vertical asymptotes of the rational function, we need to
determine the values of xthat make the denominator equal to zero.
Step 2: Set the denominator equal 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 to zero:
x+ 3 = 0 or x−3 = 0
Step 5: Solve for xin each equation:
x=−3or x= 3
Step 6: Therefore, the vertical asymptotes of the rational function f(x) =
3x2
−5x−2
x2−9are x=−3and x= 3.
7
Question 11
Question
Simplify the following rational function:
f(x) = 3x3−15x2+ 24x
x2−4x−2x2−6x+ 4
x2−4
Solution
Step 1: Factor out common terms in each fraction
f(x) = 3x(x2−5x+ 8)
x(x−4) −2(x2−3x+ 2)
x2−4
=3x(x−4)(x−2)
x(x−4) −2(x−2)(x−1)
(x+ 2)(x−2)
Step 2: Simplify the expression by canceling out common factors
f(x) = 3(x−2)
1−2(x−1)
x+ 2
= 3(x−2) −2(x−1)
x+ 2
= 3x−6−2x−2
x+ 2
= 3x−6−2x−2
x+ 2
Therefore, the simplified form of the given rational function is f(x)=3x−
6−2x−2
x+ 2 .
Question 12
Question
Find the domain of the following rational function:
f(x) = 3x2−7x+ 2
x2−4x−5
Solution
Step 1: We need to find the values of xfor which the denominator is not equal to
zero. This is because division by zero is undefined in mathematics. Therefore,
we set the denominator equal to zero and solve for x:
x2−4x−5 = 0
8
Step 2: We can factor the quadratic equation to find its roots:
(x−5)(x+ 1) = 0
x= 5 or x=−1
Step 3: The values x= 5 and x=−1make the denominator equal to zero.
Therefore, the domain of the function f(x)is all real numbers except x= 5 and
x=−1.
Step 4: Writing the domain in interval notation, we have:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 13
Question
Given the rational function f(x) = x2
−3x−4
x2−2x−8, determine the following:
1. The domain of the function.
2. The x-intercepts of the function (if they exist).
3. The y-intercept of the function.
4. Any vertical asymptotes of the function.
5. Any horizontal asymptotes of the function.
Solution
1. Step 1: Find the domain of the function by determining where the de-
nominator is not equal to zero.
For our function f(x) = x2
−3x−4
x2−2x−8, the denominator x2−2x−8is equal to
zero when x=−2or x= 4. Therefore, the domain of the function is all
real numbers except x=−2and x= 4.
2. Step 2: Find the x-intercepts of the function by setting f(x) = 0 and
solving for x.
Setting f(x) = 0, we have:
x2−3x−4
x2−2x−8= 0
This fraction is zero when the numerator is zero, so we solve x2−3x−4 = 0
to find the x-intercepts. Factoring or using the quadratic formula gives us
x=−1and x= 4.
9
3. Step 3: Find the y-intercept of the function by setting x= 0 and evalu-
ating the function.
Setting x= 0 in the function f(x) = x2
−3x−4
x2−2x−8gives us:
f(0) = 02−3(0) −4
02−2(0) −8=−4
−8=1
2
So, the y-intercept is at the point (0, 0.5).
4. Step 4: Find any vertical asymptotes of the function by setting the de-
nominator equal to zero.
The vertical asymptotes occur where the denominator is equal to zero, at
x=−2and x= 4.
5. Step 5: Find any horizontal asymptotes of the function by comparing the
degrees of the numerator and denominator.
Since the degree of the numerator is equal to the degree of the denomina-
tor, we compare the leading coefficients. The horizontal asymptote is the
ratio of the leading coefficients, which gives no horizontal asymptote for
this function.
Question 14
Question
Find the domain of the rational function: f(x) = x2−4
5x2−10x−15.
Solution
Step 1: Set the denominator equal to zero and solve for xto find the values that
make the function undefined.
5x2−10x−15 = 0
Step 2: Factor the quadratic expression to solve for x.
5x2−10x−15 = 5(x2−2x−3) = 5(x−3)(x+ 1)
Step 3: Set each factor equal to zero and solve for x.
x−3 = 0 or x+ 1 = 0
Step 4: Solve for xin each equation.
x= 3 or x=−1
Step 5: So, the values x= 3 and x=−1make the denominator zero and
should be excluded from the domain of the function. Therefore, the domain
of the function is all real numbers except x= 3 and x=−1, or in interval
notation: (−∞,−1) ∪(−1,3) ∪(3,∞).
10
Question 15
Question
Let f(x) = x2
−4
x−2. Determine the domain of f(x).
Solution
Step 1: The domain of a rational function is all real numbers except for the
values that make the denominator zero. Therefore, we need to find the values
of xthat make the denominator x−2equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
(x−2) = 0
Step 3: Solve for x:
x= 2
Step 4: Since xcannot equal 2 in order to avoid division by zero, the domain
of f(x)is all real numbers except for x= 2. Therefore, the domain of f(x)is
given by:
(−∞,2) ∪(2,∞)
Question 16
Question
Find the domain of the rational function: f(x) = 2x2−5x−3
x2−9.
Solution
Step 1: Determine the values that make the denominator zero (since division by
zero is undefined). The denominator can be factored as (x+ 3)(x−3). Setting
the denominator equal to zero gives:
(x+ 3)(x−3) = 0
This implies x=−3or x= 3.
Step 2: The domain of the function is all real numbers except the values that
make the denominator zero. Therefore, the domain of f(x)is all real numbers
except x=−3and x= 3. This can be expressed in interval notation as:
(−∞,−3) ∪(−3,3) ∪(3,∞)
11
Question 17
Question
Simplify the following rational expression:
2x2+ 4x−6
4x2−9
Solution
Step 1: Factor both the numerator and the denominator.
Step 2: Factor the numerator 2x2+ 4x−6:
We look for two numbers that multiply to 2∗(−6) = −12 and add up to 4.
These numbers are 6 and -2. So, we can rewrite the numerator as:
2x2+4x−6 = 2x2+6x−2x−6 = 2x(x+3)−2(x+3) = (2x−2)(x+3) = 2(x−1)(x+3)
Therefore, the numerator is factored as 2(x−1)(x+ 3).
Step 3: Factor the denominator 4x2−9:
Using the difference of squares formula, we have:
4x2−9 = (2x)2−32= (2x+ 3)(2x−3)
Therefore, the denominator is factored as (2x+ 3)(2x−3).
Step 4: Rewrite the original expression using the factored forms of the nu-
merator and denominator:
2(x−1)(x+ 3)
(2x+ 3)(2x−3)
Step 5: Cancel out common factors from the numerator and denominator:
2(x−1)(x+ 3)
(2x+ 3)(2x−3)
Step 6: Simplify the expression:
2(x+ 3)
2x−3
So the simplified form of the given rational expression is 2(x+3)
2x−3.
Question 18
Question
Solve the following rational inequality:
1
x−2≥3
x+ 1
12
Solution
Step 1: Begin by finding a common denominator. In this case, the common
denominator is (x−2)(x+ 1).
Step 2: Rewrite the inequality using the common denominator:
(x+ 1)
(x−2)(x+ 1) ≥3(x−2)
(x−2)(x+ 1)
Step 3: Simplify the inequality:
x+ 1
x2−x−2≥3x−6
x2−x−2
Step 4: Now that the denominators are equal, we can drop them from both
sides of the inequality:
x+ 1 ≥3x−6
Step 5: Subtract xfrom both sides of the inequality:
1≥2x−6
Step 6: Add 6 to both sides of the inequality:
7≥2x
Step 7: Divide by 2:
7
2≥x
Step 8: Therefore, the solution to the inequality is x≤7
2.
Question 19
Question
Find the domain of the rational function: f(x) = x2
−9
x2−x−6.
Solution
Step 1: First, we need to find the values of xthat make the denominator equal
to zero, because division by zero is undefined.
Step 2: Factor the denominator x2−x−6to identify the values of xthat
make it zero.
x2−x−6 = (x−3)(x+ 2)
Step 3: Set the factors equal to zero and solve for x:
x−3 = 0 ⇒x= 3
x+ 2 = 0 ⇒x=−2
13
Step 4: Now, we know that the function is undefined for x= 3 and x=−2,
so the domain of fis all real numbers except x= 3 and x=−2.
Step 5: Therefore, the domain of the function f(x) = x2
−9
x2−x−6is {x∈R|x=
3,−2}.
Question 20
Question
Simplify the rational function: 4x2−9
x2−4x−12.
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify the rational
function by cancelling out the common factors, if possible.
Step 1:
Factor the numerator: 4x2−9can be factored as (2x+ 3)(2x−3).
Factor the denominator: x2−4x−12 can be factored as (x−6)(x+ 2).
Step 2:
Now, the rational function becomes:
4x2−9
x2−4x−12 =(2x+ 3)(2x−3)
(x−6)(x+ 2) .
Therefore, the simplified form of the rational function is 2x+ 3
x−6.
Question 21
Question
Let f(x) = x2
−3x−4
x2−5x+6 be a rational function. Find the equation of the horizontal
asymptote of f(x).
Solution
Step 1: To find the horizontal asymptote of a rational function, we compare the
degrees of the numerator and denominator polynomials.
Step 2: Since the degree of the numerator (x2−3x−4) is equal to the degree
of the denominator (x2−5x+ 6), we first need to divide the coefficients of the
leading terms to determine the horizontal asymptote.
Step 3: Divide the coefficient of the leading term of the numerator by the
coefficient of the leading term of the denominator. This gives us 1
1= 1.
Step 4: Therefore, the equation of the horizontal asymptote is y= 1.
14
Question 22
Question
Let f(x) = 2x2
−5x−3
x2−3x−4be a rational function. Determine the x-intercepts, y-
intercepts, vertical asymptotes, horizontal asymptotes, and the domain of f(x).
Solution
Step 1: To find the x-intercepts, set f(x) = 0 and solve for x.
2x2−5x−3
x2−3x−4= 0
2x2−5x−3 = 0
This is a quadratic equation that can be factored:
(2x+ 1)(x−3) = 0
So, x=−1
2or x= 3. Therefore, the x-intercepts are at (−1
2,0) and (3,0).
Step 2: To find the y-intercept, evaluate f(0).
f(0) = 2(0)2−5(0) −3
(0)2−3(0) −4=−3
−4=3
4
The y-intercept is at (0,3
4).
Step 3: To find the vertical asymptotes, set the denominator equal to zero
and solve for x.
x2−3x−4 = 0
This quadratic equation can be factored:
(x−4)(x+ 1) = 0
So, x= 4 or x=−1. Therefore, the vertical asymptotes are at x= 4 and
x=−1.
Step 4: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator. Since they are the same, divide the leading coefficient
of the numerator by the leading coefficient of the denominator. The horizontal
asymptote is y=2
1= 2.
Step 5: The domain of f(x)is all real numbers except where the denominator
is equal to zero. So, the domain is {x∈R:x=−1,4}or (−∞,−1) ∪(−1,4) ∪
(4,∞).
Question 23
Question
Solve for x:x−3
x+2 + 2 = 4x+7
x+2 .
15
Solution
Step 1: Start by subtracting x−3
x+2 from both sides of the equation:
x−3
x+ 2 + 2 −x−3
x+ 2 =4x+ 7
x+ 2 −x−3
x+ 2
Step 2: Simplify the left side of the equation:
2 = 4x+ 7
x+ 2 −x−3
x+ 2
Step 3: Combine the fractions on the right side of the equation:
2 = (4x+ 7) −(x−3)
x+ 2
Step 4: Simplify the expression in the numerator on the right side of the
equation:
2 = 4x+ 7 −x+ 3
x+ 2
2 = 3x+ 10
x+ 2
Step 5: Multiply both sides of the equation by x+2 to clear the denominator:
2(x+ 2) = 3x+ 10
Step 6: Expand and simplify the left side of the equation:
2x+ 4 = 3x+ 10
Step 7: Subtract 2xfrom both sides of the equation:
4 = x+ 10
Step 8: Subtract 10 from both sides of the equation to solve for x:
x=−6
Therefore, the solution to the equation x−3
x+2 + 2 = 4x+7
x+2 is x=−6.
Question 24
Question
Given the rational function f(x) = 2x3
−x2
−13x+6
x2−4x−5, find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph of f(x).
16
Solution
Step 1: First, let’s factor both the numerator and denominator of the rational
function f(x). The numerator can be factored as: 2x3−x2−13x+6 = (x−2)(x−
1)(2x−3). The denominator can be factored as: x2−4x−5 = (x−5)(x+ 1).
Step 2: Now, rewrite the rational function using the factored forms: f(x) =
(x−2)(x−1)(2x−3)
(x−5)(x+1) .
Step 3: To find the vertical asymptotes, set the denominator equal to zero
and solve for x:x−5 = 0 or x+ 1 = 0.
Therefore, the vertical asymptotes are x= 5 and x=−1.
Step 4: To find the horizontal asymptote, compare the degrees of the nu-
merator and denominator. Since the degree of the numerator is equal to the
degree of the denominator, the horizontal asymptote is the ratio of the leading
coefficients: Horizontal asymptote: y= 2.
Step 5: To find any holes in the graph, check if there are any common
factors in the numerator and denominator: At x= 2, we have a common factor
of (x−2) which cancels out, resulting in a hole in the graph. The hole occurs
at the point (2, f(2)).
Therefore, the vertical asymptotes are x= 5 and x=−1, the horizontal
asymptote is y= 2, and there is a hole at the point (2, f(2)).
Question 25
Question
Simplify the following rational expression:
5x2−3x−2
3x2+ 7x−6∇ · 15x2−4x−4
9x2+ 16x−4
Solution
Step 1: To divide two fractions, we multiply by the reciprocal of the divisor.
Step 2: Find the reciprocal of the second fraction. Step 3: Multiply the first
fraction by the reciprocal of the second fraction. Step 4: Factor all quadratics in
the resulting fraction. Step 5: Cancel out any common factors in the numerator
and denominator. Step 6: Simplify the resulting expression as much as possible.
Step 1: Multiply by the reciprocal of the second fraction:
5x2−3x−2
3x2+ 7x−6×9x2+ 16x−4
15x2−4x−4
Step 2: Find the reciprocal of the second fraction:
5x2−3x−2
3x2+ 7x−6×9x2+ 16x−4
15x2−4x−4=5x2−3x−2
3x2+ 7x−6×9x2+ 16x−4
15x2−4x−4×1
1
17
Step 3: Multiply the first fraction by the reciprocal of the second fraction:
(5x2−3x−2)(9x2+ 16x−4)
(3x2+ 7x−6)(15x2−4x−4)
Step 4: Factor all quadratics in the resulting fraction: The numerator
factors to (5x−2)(x+ 2)(9x−2) and the denominator factors to (3x−2)(x−
3)(15x+ 1).
Step 5: Cancel out any common factors in the numerator and denominator:
Cancelling out common factors gives us:
5(x+ 2)(9x−2)
(3x−2)(x−3)(15x+ 1)
Step 6: Simplify the resulting expression as much as possible: No further
simplification is possible, so the fully simplified expression is:
5(x+ 2)(9x−2)
(3x−2)(x−3)(15x+ 1)
Question 26
Question
Find the domain of the rational function:
f(x) = 5x−2
x2−9
Solution
Step 1: To find the domain of the rational function, we need to identify any
values of xthat would make the denominator equal to zero, since division by
zero is undefined. The denominator x2−9will equal zero when:
x2−9 = 0
x2= 9
x=±3
Step 2: Therefore, the domain of the rational function f(x)is all real num-
bers except for x=−3and x= 3, since these values would make the denomi-
nator zero. So, the domain of f(x)is:
(−∞,−3) ∪(−3,3) ∪(3,∞)
18
Question 27
Question
Simplify the rational function 2x2−6x−16
x2−8x+ 12 .
Solution
Step 1: Factor the numerator and denominator:
2x2−6x−16 = 2(x2−3x−8)
= 2(x−4)(x+ 2)
x2−8x+ 12 = (x−6)(x−2)
Step 2: Rewrite the rational function with factored expressions in the nu-
merator and denominator: 2(x−4)(x+ 2)
(x−6)(x−2)
Step 3: Cancel out common factors in the numerator and denominator:
2(x−4)(x+ 2)
(x−6)(x−2)
Step 4: Simplify the expression:
2(x+ 2)
x−2
Therefore, the simplified form of the rational function 2x2−6x−16
x2−8x+ 12 is
2(x+ 2)
x−2.
Question 28
Question
Solve the rational inequality and express the solution set in interval notation:
3x−5
x+ 2 ≤1
19
Solution
Step 1: Find the critical points by setting the numerator equal to the denomi-
nator and solving for x:
3x−5 = x+ 2
Step 2: Simplify the equation:
2x= 7
Step 3: Solve for x:
x=7
2= 3.5
So, x= 3.5is a critical point for the given rational inequality.
Step 4: Test the intervals determined by the critical point x= 3.5.
For x < 3.5: Substitute x= 3 into the inequality:
3(3) −5
3+2 ≤1
4
5≤1
Since 4
5≤1is true, x < 3.5is part of the solution set.
For x > 3.5: Substitute x= 4 into the inequality:
3(4) −5
4+2 ≤1
7
6≤1
Since 7
6≤1is false, x > 3.5is not part of the solution set.
Step 5: Write the final solution in interval notation:
(−∞,3.5]
Therefore, the solution set in interval notation for the given rational inequal-
ity is (−∞,3.5].
Question 29
Question
Simplify the rational expression: 3x3
−8
x2+x−2.
20
Solution
Step 1: Factor the numerator and the denominator. Step 2: Simplify the ra-
tional expression by canceling out any common factors between the numerator
and the denominator.
Step 1: To factor the numerator and denominator: We have: Numerator:
3x3−8
=3x3−23(Difference of cubes)
=(3x−2)(3x2+ 6x+ 4)
Denominator: x2+x−2
=x2+ 2x−x−2
=x(x+ 2) −1(x+ 2)
=(x−1)(x+ 2)
So, the factored form of the rational expression is (3x−2)(3x2+6x+4)
(x−1)(x+2) .
Step 2: To simplify the rational expression:
(3x−2)(3x2+ 6x+ 4)
(x−1)(x+ 2) =3x−2
x−1
Therefore, the simplified form of the given rational expression is 3x−2
x−1.
Question 30
Question
Let f(x) = 3x3
−7x2+2x−5
2x2+x−3. Find the vertical asymptotes, horizontal asymptotes,
and any holes in the graph of f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to 0 and
solve for x.
2x2+x−3 = 0
Using the quadratic formula, we get:
x=−1±√12−4(2)(−3)
2(2)
x=−1±√1 + 24
4
x=−1±√25
4
x=−1±5
4
So, the vertical asymptotes are x=−1+5
4=4
4= 1 and x=−1−5
4=−6
4=−3
2.
21
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and denominator. Since the degree of the numerator is equal to the
degree of the denominator, the horizontal asymptote is the ratio of the leading
coefficients. The horizontal asymptote is y=3
2.
Step 3: To find any holes in the graph, factor the numerator and denominator
and see if any common factors cancel out. The numerator can be factored as
3x3−7x2+ 2x−5 = (x−1)(3x2−4x+ 5). The denominator can be factored
as 2x2+x−3 = (2x−3)(x+ 1). Since 3x2−4x+ 5 does not have any common
factors with 2x−3or x+ 1, there are no holes in the graph.
Therefore, the vertical asymptotes are x= 1 and x=−3
2, the horizontal
asymptote is y=3
2, and there are no holes in the graph of f(x).
Question 31
Question
Simplify the rational function:
f(x) = 3x2−5x−2
2x2+ 3x−2
Solution
Step 1: Factor the numerator and denominator:
f(x) = (3x+ 1)(x−2)
(2x−1)(x+ 2)
Step 2: Since the function is already in factored form, we can simplify the
expression further:
f(x) = 3x+ 1
2x−1·x−2
x+ 2
Step 3: Now, we can simplify the expression by canceling out common fac-
tors:
f(x) = 3(x+ 1)
2·x−2
x+ 2
Step 4: Simplifying further, we get:
f(x) = 3
2·x−2
x+ 2
Step 5: Therefore, the simplified form of the rational function is:
f(x) = 3(x−2)
2(x+ 2)
22
Question 32
Question
Simplify the rational function:
f(x) = 2x2−3x−2
x2−5x+ 6
Solution
Step 1: Factor the numerator and denominator.
2x2−3x−2 = 2x2−4x+x−2
= 2x(x−2) + 1(x−2)
= (2x+ 1)(x−2)
x2−5x+ 6 = x2−3x−2x+ 6
=x(x−3) −2(x−3)
= (x−2)(x−3)
Step 2: Rewrite the function using the factored form of the numerator and
denominator.
f(x) = (2x+ 1)(x−2)
(x−2)(x−3)
Step 3: Cancel out the common factor in the numerator and denominator.
f(x) = 2x+ 1
x−3
Therefore, the simplified form of the rational function is f(x) = 2x+ 1
x−3.
Question 33
Question
Find the domain of the following rational function:
f(x) = x2−4
x2−x−6
23
Solution
Step 1: We first need to find the values of xthat would make the denominator
x2−x−6equal to zero, as these values would lead to division by zero in
the rational function. So, we solve the equation x2−x−6=0to find the
critical points. Step 2: Factor the quadratic equation x2−x−6 = 0 into
(x−3)(x+ 2) = 0. Step 3: Set each factor to zero and solve for x:
x−3 = 0 or x+ 2 = 0
x= 3 or x=−2
Step 4: So, the values x= 3 and x=−2make the denominator equal to zero,
which means they are not in the domain of the function. Step 5: Therefore,
the domain of the function f(x) = x2
−4
x2−x−6is all real numbers except x= 3 and
x=−2, which can be expressed as:
(−∞,−2) ∪(−2,3) ∪(3,∞)
Question 34
Question
Find the domain of the rational function:
f(x) = x2−4x−5
x2+ 3x−10.
Solution
Step 1: To find the domain of a rational function, we need to identify the values
of xfor which the function is defined. In this case, the function is defined for all
values of xexcept those that would make the denominator equal to zero, since
division by zero is undefined.
Step 2: To find where the denominator x2+ 3x−10 is equal to zero, we
solve the equation x2+ 3x−10 = 0 for x.
Step 3: This equation can be factored as (x+ 5)(x−2) = 0.
Step 4: Setting each factor to zero, we find the values of xthat make the
denominator zero: x+ 5 = 0 ⇒x=−5and x−2 = 0 ⇒x= 2.
Step 5: Therefore, the domain of the rational function f(x) = x2
−4x−5
x2+3x−10 is
all real numbers except x=−5and x= 2.
Step 6: In interval notation, the domain can be expressed as (−∞,−5) ∪
(−5,2) ∪(2,∞).
24
Question 35
Question
Simplify the rational function:
f(x) = 3x2−2x−1
2x2+ 5x−3
Solution
Step 1: Factor the numerator and denominator of the rational function.
The numerator can be factored as:
3x2−2x−1 = (3x+ 1)(x−1)
The denominator can be factored as:
2x2+ 5x−3 = (2x−1)(x+ 3)
Step 2: Rewrite the rational function using the factored forms of the numer-
ator and denominator.
f(x) = (3x+ 1)(x−1)
(2x−1)(x+ 3)
Step 3: Simplify the rational function by canceling out common factors in
the numerator and denominator.
f(x) = 3x+ 1
2x−1·x−1
x+ 3
Step 4: The simplified form of the rational function is:
f(x) = (3x+ 1)(x−1)
(2x−1)(x+ 3)
25
Question 2
Question
Given the rational function f(x) = 2x2
−x−3
x2−5x+6 , find the following: 1. Domain of
the function f(x). 2. x-intercepts, if any. 3. y-intercept, if any. 4. Vertical
asymptotes, if any. 5. Horizontal asymptotes, if any.
Solution
1. Domain of the function f(x):
The domain of a rational function includes all real numbers except for the
values that make the denominator equal to zero. Set the denominator equal to
zero and solve for x:
x2−5x+ 6 = 0
Factoring, we get:
(x−2)(x−3) = 0
So, x= 2 or x= 3. Therefore, the domain of f(x)is all real numbers except
x= 2 and x= 3, or in interval notation: (−∞,2) ∪(2,3) ∪(3,∞).
2. x-intercepts:
To find the x-intercepts, set f(x) = 0 and solve for x:
2x2−x−3
x2−5x+ 6 = 0
This occurs when the numerator is zero. So, 2x2−x−3 = 0. Factoring, we
get:
(2x+ 1)(x−3) = 0
So, x=−1
2or x= 3. Therefore, the x-intercepts are (−1
2,0) and (3,0).
3. y-intercept:
To find the y-intercept, set x= 0 in the function:
f(0) = 2(0)2−0−3
(0)2−5(0) + 6 =−3
6=−1
2
Therefore, the y-intercept is (0,−1
2).
4. Vertical asymptotes:
Vertical asymptotes occur at the values of xthat make the denominator of
the function equal to zero. We found earlier that the denominator is zero at
x= 2 and x= 3. So, the vertical asymptotes are x= 2 and x= 3.
5. Horizontal asymptotes:
2
To find the horizontal asymptotes, we compare the degrees of the numerator
and denominator of the function. Since the degrees are the same, we look at
the leading coefficients:
The leading coefficient of the numerator is 2, and the leading coefficient of
the denominator is 1. Therefore, there is a horizontal asymptote at y= 2/1 = 2.
In summary: - Domain: (−∞,2) ∪(2,3) ∪(3,∞)-x-intercepts: (−1
2,0) and
(3,0) -y-intercept: (0,−1
2)- Vertical asymptotes: x= 2 and x= 3 - Horizontal
asymptote: y= 2
Question 3
Question
Find the domain of the rational function:
f(x) = 3x2−6x
x2−4
Solution
To find the domain of the rational function, we need to determine which values
of xmake the denominator equal to zero, since division by zero is undefined.
Step 1: Set the denominator equal to zero and solve for x:
x2−4 = 0
Step 2: Factor the quadratic equation:
(x−2)(x+ 2) = 0
Step 3: Use the zero product property to find the values of x:
x−2 = 0 OR x+ 2 = 0
Step 4: Solve for xin each equation:
x= 2 OR x=−2
Thus, the domain of the rational function f(x)is all real numbers except
x=−2and x= 2, since these values would make the denominator equal to
zero. Therefore, the domain is (−∞,−2) ∪(−2,2) ∪(2,∞).
Question 4
Question
Find the domain of the rational function:
f(x) = 5x2−4x−3
x2−9
3
Solution
Step 1: To find the domain of the rational function, we need to identify the
values of xthat would make the denominator equal to zero. Since division by
zero is undefined, these values must be excluded from the domain.
Step 2: Set the denominator equal to zero and solve for x:
x2−9 = 0
Step 3: Factor the quadratic equation:
(x+ 3)(x−3) = 0
Step 4: Use the zero-product property to find the solutions:
x+ 3 = 0 or x−3 = 0
Step 5: Solve for x:
For x+ 3 = 0:
x=−3
For x−3 = 0:
x= 3
Step 6: The domain of the rational function f(x)is all real numbers except
x=−3and x= 3, since these values make the denominator zero.
Step 7: Therefore, the domain of the rational function f(x) = 5x2
−4x−3
x2−9is
(−∞,−3) ∪(−3,3) ∪(3,∞).
Question 5
Question
Determine 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 the values of xfor
which the function is defined. In a rational function, the function is undefined
when the denominator is equal to zero. Therefore, we need to find the values of
xthat make the denominator x2−4x−5equal to zero.
Step 2: Let’s solve the equation x2−4x−5 = 0 to find the values that make
the denominator zero. This can be factored as (x−5)(x+ 1) = 0. Setting each
factor to zero gives us x−5 = 0 and x+ 1 = 0. Thus, x= 5 and x=−1.
4
Step 3: Therefore, the domain of the function f(x)is all real numbers except
x= 5 and x=−1. The domain is given by (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 6
Question
Simplify the following rational expression:
3x2−2x−8
x2−4
Solution
Step 1: Factor the numerator and denominator.
The numerator factors as 3x2−2x−8 = (3x+ 4)(x−2).
The denominator factors as x2−4 = (x+ 2)(x−2).
Step 2: Rewrite the expression with factored numerator and denominator.
3x2−2x−8
x2−4=(3x+ 4)(x−2)
(x+ 2)(x−2)
Step 3: Simplify by canceling out the common factor (x−2) in the numerator
and denominator.
(3x+ 4)(x−2)
(x+ 2)(x−2) =3x+ 4
x+ 2
Therefore, the simplified form of the rational expression is 3x+4
x+2 .
Question 7
Question
Find the domain of the rational function:
f(x) = 2x2−5x−3
x2−4x−7
Solution
Step 1: We start by identifying the values of xthat would make the denominator
equal to zero since division by zero is undefined. Set the denominator equal to
zero and solve for x:
x2−4x−7 = 0
5
Step 2: To solve this quadratic equation, we can use the quadratic formula:
x=−(−4) ±√(−4)2−4(1)(−7)
2(1)
Step 3: Simplifying under the square root:
x=4±√16 + 28
2=4±√44
2=4±2√11
2= 2 ±√11
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except 2 + √11 and 2−√11.
Domain = (−∞,2−√11) ∪(2 −√11,2 + √11) ∪(2 + √11,∞)
Question 8
Question
Let f(x) = 2x2
−x−3
x2−4x−5. Find the domain of the function f(x).
Solution
Step 1: The domain of a rational function consists of all real numbers except
the values that make the denominator equal to zero. Thus, we need to find the
values of xthat make x2−4x−5equal to zero.
Step 2: To find these values, we can factor the denominator. The factored
form of x2−4x−5is (x−5)(x+ 1), so the denominator becomes (x−5)(x+ 1).
Step 3: Setting the denominator equal to zero, we get (x−5)(x+ 1) = 0.
This equation is true when either x−5 = 0 or x+ 1 = 0.
Step 4: Solving x−5=0gives x= 5, and solving x+ 1 = 0 gives x=−1.
Therefore, x= 5 and x=−1are the values that make the denominator zero.
Step 5: Hence, the domain of the function f(x)is all real numbers except x=
5and x=−1. In interval notation, the domain is (−∞,−1) ∪(−1,5) ∪(5,∞).
Question 9
Question
Find the vertical asymptotes of the rational function:
f(x) = 2x2−x−3
x2−4x−32
6
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero
and solving for x.
x2−4x−32 = 0
Step 2: Factor the quadratic equation.
(x−8)(x+ 4) = 0
Step 3: Set each factor equal to zero and solve for x.
x−8 = 0 =⇒x= 8
x+ 4 = 0 =⇒x=−4
Step 4: Therefore, the vertical asymptotes of the rational function f(x) =
2x2
−x−3
x2−4x−32 are x= 8 and x=−4.
Question 10
Question
Find the vertical asymptotes of the rational function:
f(x) = 3x2−5x−2
x2−9
Solution
Step 1: To find the vertical asymptotes of the rational function, we need to
determine the values of xthat make the denominator equal to zero.
Step 2: Set the denominator equal 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 to zero:
x+ 3 = 0 or x−3 = 0
Step 5: Solve for xin each equation:
x=−3or x= 3
Step 6: Therefore, the vertical asymptotes of the rational function f(x) =
3x2
−5x−2
x2−9are x=−3and x= 3.
7
Question 11
Question
Simplify the following rational function:
f(x) = 3x3−15x2+ 24x
x2−4x−2x2−6x+ 4
x2−4
Solution
Step 1: Factor out common terms in each fraction
f(x) = 3x(x2−5x+ 8)
x(x−4) −2(x2−3x+ 2)
x2−4
=3x(x−4)(x−2)
x(x−4) −2(x−2)(x−1)
(x+ 2)(x−2)
Step 2: Simplify the expression by canceling out common factors
f(x) = 3(x−2)
1−2(x−1)
x+ 2
= 3(x−2) −2(x−1)
x+ 2
= 3x−6−2x−2
x+ 2
= 3x−6−2x−2
x+ 2
Therefore, the simplified form of the given rational function is f(x)=3x−
6−2x−2
x+ 2 .
Question 12
Question
Find the domain of the following rational function:
f(x) = 3x2−7x+ 2
x2−4x−5
Solution
Step 1: We need to find the values of xfor which the denominator is not equal to
zero. This is because division by zero is undefined in mathematics. Therefore,
we set the denominator equal to zero and solve for x:
x2−4x−5 = 0
8
Step 2: We can factor the quadratic equation to find its roots:
(x−5)(x+ 1) = 0
x= 5 or x=−1
Step 3: The values x= 5 and x=−1make the denominator equal to zero.
Therefore, the domain of the function f(x)is all real numbers except x= 5 and
x=−1.
Step 4: Writing the domain in interval notation, we have:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 13
Question
Given the rational function f(x) = x2
−3x−4
x2−2x−8, determine the following:
1. The domain of the function.
2. The x-intercepts of the function (if they exist).
3. The y-intercept of the function.
4. Any vertical asymptotes of the function.
5. Any horizontal asymptotes of the function.
Solution
1. Step 1: Find the domain of the function by determining where the de-
nominator is not equal to zero.
For our function f(x) = x2
−3x−4
x2−2x−8, the denominator x2−2x−8is equal to
zero when x=−2or x= 4. Therefore, the domain of the function is all
real numbers except x=−2and x= 4.
2. Step 2: Find the x-intercepts of the function by setting f(x) = 0 and
solving for x.
Setting f(x) = 0, we have:
x2−3x−4
x2−2x−8= 0
This fraction is zero when the numerator is zero, so we solve x2−3x−4 = 0
to find the x-intercepts. Factoring or using the quadratic formula gives us
x=−1and x= 4.
9
3. Step 3: Find the y-intercept of the function by setting x= 0 and evalu-
ating the function.
Setting x= 0 in the function f(x) = x2
−3x−4
x2−2x−8gives us:
f(0) = 02−3(0) −4
02−2(0) −8=−4
−8=1
2
So, the y-intercept is at the point (0, 0.5).
4. Step 4: Find any vertical asymptotes of the function by setting the de-
nominator equal to zero.
The vertical asymptotes occur where the denominator is equal to zero, at
x=−2and x= 4.
5. Step 5: Find any horizontal asymptotes of the function by comparing the
degrees of the numerator and denominator.
Since the degree of the numerator is equal to the degree of the denomina-
tor, we compare the leading coefficients. The horizontal asymptote is the
ratio of the leading coefficients, which gives no horizontal asymptote for
this function.
Question 14
Question
Find the domain of the rational function: f(x) = x2−4
5x2−10x−15.
Solution
Step 1: Set the denominator equal to zero and solve for xto find the values that
make the function undefined.
5x2−10x−15 = 0
Step 2: Factor the quadratic expression to solve for x.
5x2−10x−15 = 5(x2−2x−3) = 5(x−3)(x+ 1)
Step 3: Set each factor equal to zero and solve for x.
x−3 = 0 or x+ 1 = 0
Step 4: Solve for xin each equation.
x= 3 or x=−1
Step 5: So, the values x= 3 and x=−1make the denominator zero and
should be excluded from the domain of the function. Therefore, the domain
of the function is all real numbers except x= 3 and x=−1, or in interval
notation: (−∞,−1) ∪(−1,3) ∪(3,∞).
10
Question 15
Question
Let f(x) = x2
−4
x−2. Determine the domain of f(x).
Solution
Step 1: The domain of a rational function is all real numbers except for the
values that make the denominator zero. Therefore, we need to find the values
of xthat make the denominator x−2equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
(x−2) = 0
Step 3: Solve for x:
x= 2
Step 4: Since xcannot equal 2 in order to avoid division by zero, the domain
of f(x)is all real numbers except for x= 2. Therefore, the domain of f(x)is
given by:
(−∞,2) ∪(2,∞)
Question 16
Question
Find the domain of the rational function: f(x) = 2x2−5x−3
x2−9.
Solution
Step 1: Determine the values that make the denominator zero (since division by
zero is undefined). The denominator can be factored as (x+ 3)(x−3). Setting
the denominator equal to zero gives:
(x+ 3)(x−3) = 0
This implies x=−3or x= 3.
Step 2: The domain of the function is all real numbers except the values that
make the denominator zero. Therefore, the domain of f(x)is all real numbers
except x=−3and x= 3. This can be expressed in interval notation as:
(−∞,−3) ∪(−3,3) ∪(3,∞)
11
Question 17
Question
Simplify the following rational expression:
2x2+ 4x−6
4x2−9
Solution
Step 1: Factor both the numerator and the denominator.
Step 2: Factor the numerator 2x2+ 4x−6:
We look for two numbers that multiply to 2∗(−6) = −12 and add up to 4.
These numbers are 6 and -2. So, we can rewrite the numerator as:
2x2+4x−6 = 2x2+6x−2x−6 = 2x(x+3)−2(x+3) = (2x−2)(x+3) = 2(x−1)(x+3)
Therefore, the numerator is factored as 2(x−1)(x+ 3).
Step 3: Factor the denominator 4x2−9:
Using the difference of squares formula, we have:
4x2−9 = (2x)2−32= (2x+ 3)(2x−3)
Therefore, the denominator is factored as (2x+ 3)(2x−3).
Step 4: Rewrite the original expression using the factored forms of the nu-
merator and denominator:
2(x−1)(x+ 3)
(2x+ 3)(2x−3)
Step 5: Cancel out common factors from the numerator and denominator:
2(x−1)(x+ 3)
(2x+ 3)(2x−3)
Step 6: Simplify the expression:
2(x+ 3)
2x−3
So the simplified form of the given rational expression is 2(x+3)
2x−3.
Question 18
Question
Solve the following rational inequality:
1
x−2≥3
x+ 1
12
Solution
Step 1: Begin by finding a common denominator. In this case, the common
denominator is (x−2)(x+ 1).
Step 2: Rewrite the inequality using the common denominator:
(x+ 1)
(x−2)(x+ 1) ≥3(x−2)
(x−2)(x+ 1)
Step 3: Simplify the inequality:
x+ 1
x2−x−2≥3x−6
x2−x−2
Step 4: Now that the denominators are equal, we can drop them from both
sides of the inequality:
x+ 1 ≥3x−6
Step 5: Subtract xfrom both sides of the inequality:
1≥2x−6
Step 6: Add 6 to both sides of the inequality:
7≥2x
Step 7: Divide by 2:
7
2≥x
Step 8: Therefore, the solution to the inequality is x≤7
2.
Question 19
Question
Find the domain of the rational function: f(x) = x2
−9
x2−x−6.
Solution
Step 1: First, we need to find the values of xthat make the denominator equal
to zero, because division by zero is undefined.
Step 2: Factor the denominator x2−x−6to identify the values of xthat
make it zero.
x2−x−6 = (x−3)(x+ 2)
Step 3: Set the factors equal to zero and solve for x:
x−3 = 0 ⇒x= 3
x+ 2 = 0 ⇒x=−2
13
Step 4: Now, we know that the function is undefined for x= 3 and x=−2,
so the domain of fis all real numbers except x= 3 and x=−2.
Step 5: Therefore, the domain of the function f(x) = x2
−9
x2−x−6is {x∈R|x=
3,−2}.
Question 20
Question
Simplify the rational function: 4x2−9
x2−4x−12.
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify the rational
function by cancelling out the common factors, if possible.
Step 1:
Factor the numerator: 4x2−9can be factored as (2x+ 3)(2x−3).
Factor the denominator: x2−4x−12 can be factored as (x−6)(x+ 2).
Step 2:
Now, the rational function becomes:
4x2−9
x2−4x−12 =(2x+ 3)(2x−3)
(x−6)(x+ 2) .
Therefore, the simplified form of the rational function is 2x+ 3
x−6.
Question 21
Question
Let f(x) = x2
−3x−4
x2−5x+6 be a rational function. Find the equation of the horizontal
asymptote of f(x).
Solution
Step 1: To find the horizontal asymptote of a rational function, we compare the
degrees of the numerator and denominator polynomials.
Step 2: Since the degree of the numerator (x2−3x−4) is equal to the degree
of the denominator (x2−5x+ 6), we first need to divide the coefficients of the
leading terms to determine the horizontal asymptote.
Step 3: Divide the coefficient of the leading term of the numerator by the
coefficient of the leading term of the denominator. This gives us 1
1= 1.
Step 4: Therefore, the equation of the horizontal asymptote is y= 1.
14
Question 22
Question
Let f(x) = 2x2
−5x−3
x2−3x−4be a rational function. Determine the x-intercepts, y-
intercepts, vertical asymptotes, horizontal asymptotes, and the domain of f(x).
Solution
Step 1: To find the x-intercepts, set f(x) = 0 and solve for x.
2x2−5x−3
x2−3x−4= 0
2x2−5x−3 = 0
This is a quadratic equation that can be factored:
(2x+ 1)(x−3) = 0
So, x=−1
2or x= 3. Therefore, the x-intercepts are at (−1
2,0) and (3,0).
Step 2: To find the y-intercept, evaluate f(0).
f(0) = 2(0)2−5(0) −3
(0)2−3(0) −4=−3
−4=3
4
The y-intercept is at (0,3
4).
Step 3: To find the vertical asymptotes, set the denominator equal to zero
and solve for x.
x2−3x−4 = 0
This quadratic equation can be factored:
(x−4)(x+ 1) = 0
So, x= 4 or x=−1. Therefore, the vertical asymptotes are at x= 4 and
x=−1.
Step 4: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator. Since they are the same, divide the leading coefficient
of the numerator by the leading coefficient of the denominator. The horizontal
asymptote is y=2
1= 2.
Step 5: The domain of f(x)is all real numbers except where the denominator
is equal to zero. So, the domain is {x∈R:x=−1,4}or (−∞,−1) ∪(−1,4) ∪
(4,∞).
Question 23
Question
Solve for x:x−3
x+2 + 2 = 4x+7
x+2 .
15
Solution
Step 1: Start by subtracting x−3
x+2 from both sides of the equation:
x−3
x+ 2 + 2 −x−3
x+ 2 =4x+ 7
x+ 2 −x−3
x+ 2
Step 2: Simplify the left side of the equation:
2 = 4x+ 7
x+ 2 −x−3
x+ 2
Step 3: Combine the fractions on the right side of the equation:
2 = (4x+ 7) −(x−3)
x+ 2
Step 4: Simplify the expression in the numerator on the right side of the
equation:
2 = 4x+ 7 −x+ 3
x+ 2
2 = 3x+ 10
x+ 2
Step 5: Multiply both sides of the equation by x+2 to clear the denominator:
2(x+ 2) = 3x+ 10
Step 6: Expand and simplify the left side of the equation:
2x+ 4 = 3x+ 10
Step 7: Subtract 2xfrom both sides of the equation:
4 = x+ 10
Step 8: Subtract 10 from both sides of the equation to solve for x:
x=−6
Therefore, the solution to the equation x−3
x+2 + 2 = 4x+7
x+2 is x=−6.
Question 24
Question
Given the rational function f(x) = 2x3
−x2
−13x+6
x2−4x−5, find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph of f(x).
16
Solution
Step 1: First, let’s factor both the numerator and denominator of the rational
function f(x). The numerator can be factored as: 2x3−x2−13x+6 = (x−2)(x−
1)(2x−3). The denominator can be factored as: x2−4x−5 = (x−5)(x+ 1).
Step 2: Now, rewrite the rational function using the factored forms: f(x) =
(x−2)(x−1)(2x−3)
(x−5)(x+1) .
Step 3: To find the vertical asymptotes, set the denominator equal to zero
and solve for x:x−5 = 0 or x+ 1 = 0.
Therefore, the vertical asymptotes are x= 5 and x=−1.
Step 4: To find the horizontal asymptote, compare the degrees of the nu-
merator and denominator. Since the degree of the numerator is equal to the
degree of the denominator, the horizontal asymptote is the ratio of the leading
coefficients: Horizontal asymptote: y= 2.
Step 5: To find any holes in the graph, check if there are any common
factors in the numerator and denominator: At x= 2, we have a common factor
of (x−2) which cancels out, resulting in a hole in the graph. The hole occurs
at the point (2, f(2)).
Therefore, the vertical asymptotes are x= 5 and x=−1, the horizontal
asymptote is y= 2, and there is a hole at the point (2, f(2)).
Question 25
Question
Simplify the following rational expression:
5x2−3x−2
3x2+ 7x−6∇ · 15x2−4x−4
9x2+ 16x−4
Solution
Step 1: To divide two fractions, we multiply by the reciprocal of the divisor.
Step 2: Find the reciprocal of the second fraction. Step 3: Multiply the first
fraction by the reciprocal of the second fraction. Step 4: Factor all quadratics in
the resulting fraction. Step 5: Cancel out any common factors in the numerator
and denominator. Step 6: Simplify the resulting expression as much as possible.
Step 1: Multiply by the reciprocal of the second fraction:
5x2−3x−2
3x2+ 7x−6×9x2+ 16x−4
15x2−4x−4
Step 2: Find the reciprocal of the second fraction:
5x2−3x−2
3x2+ 7x−6×9x2+ 16x−4
15x2−4x−4=5x2−3x−2
3x2+ 7x−6×9x2+ 16x−4
15x2−4x−4×1
1
17
Step 3: Multiply the first fraction by the reciprocal of the second fraction:
(5x2−3x−2)(9x2+ 16x−4)
(3x2+ 7x−6)(15x2−4x−4)
Step 4: Factor all quadratics in the resulting fraction: The numerator
factors to (5x−2)(x+ 2)(9x−2) and the denominator factors to (3x−2)(x−
3)(15x+ 1).
Step 5: Cancel out any common factors in the numerator and denominator:
Cancelling out common factors gives us:
5(x+ 2)(9x−2)
(3x−2)(x−3)(15x+ 1)
Step 6: Simplify the resulting expression as much as possible: No further
simplification is possible, so the fully simplified expression is:
5(x+ 2)(9x−2)
(3x−2)(x−3)(15x+ 1)
Question 26
Question
Find the domain of the rational function:
f(x) = 5x−2
x2−9
Solution
Step 1: To find the domain of the rational function, we need to identify any
values of xthat would make the denominator equal to zero, since division by
zero is undefined. The denominator x2−9will equal zero when:
x2−9 = 0
x2= 9
x=±3
Step 2: Therefore, the domain of the rational function f(x)is all real num-
bers except for x=−3and x= 3, since these values would make the denomi-
nator zero. So, the domain of f(x)is:
(−∞,−3) ∪(−3,3) ∪(3,∞)
18
Question 27
Question
Simplify the rational function 2x2−6x−16
x2−8x+ 12 .
Solution
Step 1: Factor the numerator and denominator:
2x2−6x−16 = 2(x2−3x−8)
= 2(x−4)(x+ 2)
x2−8x+ 12 = (x−6)(x−2)
Step 2: Rewrite the rational function with factored expressions in the nu-
merator and denominator: 2(x−4)(x+ 2)
(x−6)(x−2)
Step 3: Cancel out common factors in the numerator and denominator:
2(x−4)(x+ 2)
(x−6)(x−2)
Step 4: Simplify the expression:
2(x+ 2)
x−2
Therefore, the simplified form of the rational function 2x2−6x−16
x2−8x+ 12 is
2(x+ 2)
x−2.
Question 28
Question
Solve the rational inequality and express the solution set in interval notation:
3x−5
x+ 2 ≤1
19
Solution
Step 1: Find the critical points by setting the numerator equal to the denomi-
nator and solving for x:
3x−5 = x+ 2
Step 2: Simplify the equation:
2x= 7
Step 3: Solve for x:
x=7
2= 3.5
So, x= 3.5is a critical point for the given rational inequality.
Step 4: Test the intervals determined by the critical point x= 3.5.
For x < 3.5: Substitute x= 3 into the inequality:
3(3) −5
3+2 ≤1
4
5≤1
Since 4
5≤1is true, x < 3.5is part of the solution set.
For x > 3.5: Substitute x= 4 into the inequality:
3(4) −5
4+2 ≤1
7
6≤1
Since 7
6≤1is false, x > 3.5is not part of the solution set.
Step 5: Write the final solution in interval notation:
(−∞,3.5]
Therefore, the solution set in interval notation for the given rational inequal-
ity is (−∞,3.5].
Question 29
Question
Simplify the rational expression: 3x3
−8
x2+x−2.
20
Solution
Step 1: Factor the numerator and the denominator. Step 2: Simplify the ra-
tional expression by canceling out any common factors between the numerator
and the denominator.
Step 1: To factor the numerator and denominator: We have: Numerator:
3x3−8
=3x3−23(Difference of cubes)
=(3x−2)(3x2+ 6x+ 4)
Denominator: x2+x−2
=x2+ 2x−x−2
=x(x+ 2) −1(x+ 2)
=(x−1)(x+ 2)
So, the factored form of the rational expression is (3x−2)(3x2+6x+4)
(x−1)(x+2) .
Step 2: To simplify the rational expression:
(3x−2)(3x2+ 6x+ 4)
(x−1)(x+ 2) =3x−2
x−1
Therefore, the simplified form of the given rational expression is 3x−2
x−1.
Question 30
Question
Let f(x) = 3x3
−7x2+2x−5
2x2+x−3. Find the vertical asymptotes, horizontal asymptotes,
and any holes in the graph of f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to 0 and
solve for x.
2x2+x−3 = 0
Using the quadratic formula, we get:
x=−1±√12−4(2)(−3)
2(2)
x=−1±√1 + 24
4
x=−1±√25
4
x=−1±5
4
So, the vertical asymptotes are x=−1+5
4=4
4= 1 and x=−1−5
4=−6
4=−3
2.
21
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and denominator. Since the degree of the numerator is equal to the
degree of the denominator, the horizontal asymptote is the ratio of the leading
coefficients. The horizontal asymptote is y=3
2.
Step 3: To find any holes in the graph, factor the numerator and denominator
and see if any common factors cancel out. The numerator can be factored as
3x3−7x2+ 2x−5 = (x−1)(3x2−4x+ 5). The denominator can be factored
as 2x2+x−3 = (2x−3)(x+ 1). Since 3x2−4x+ 5 does not have any common
factors with 2x−3or x+ 1, there are no holes in the graph.
Therefore, the vertical asymptotes are x= 1 and x=−3
2, the horizontal
asymptote is y=3
2, and there are no holes in the graph of f(x).
Question 31
Question
Simplify the rational function:
f(x) = 3x2−5x−2
2x2+ 3x−2
Solution
Step 1: Factor the numerator and denominator:
f(x) = (3x+ 1)(x−2)
(2x−1)(x+ 2)
Step 2: Since the function is already in factored form, we can simplify the
expression further:
f(x) = 3x+ 1
2x−1·x−2
x+ 2
Step 3: Now, we can simplify the expression by canceling out common fac-
tors:
f(x) = 3(x+ 1)
2·x−2
x+ 2
Step 4: Simplifying further, we get:
f(x) = 3
2·x−2
x+ 2
Step 5: Therefore, the simplified form of the rational function is:
f(x) = 3(x−2)
2(x+ 2)
22
Question 32
Question
Simplify the rational function:
f(x) = 2x2−3x−2
x2−5x+ 6
Solution
Step 1: Factor the numerator and denominator.
2x2−3x−2 = 2x2−4x+x−2
= 2x(x−2) + 1(x−2)
= (2x+ 1)(x−2)
x2−5x+ 6 = x2−3x−2x+ 6
=x(x−3) −2(x−3)
= (x−2)(x−3)
Step 2: Rewrite the function using the factored form of the numerator and
denominator.
f(x) = (2x+ 1)(x−2)
(x−2)(x−3)
Step 3: Cancel out the common factor in the numerator and denominator.
f(x) = 2x+ 1
x−3
Therefore, the simplified form of the rational function is f(x) = 2x+ 1
x−3.
Question 33
Question
Find the domain of the following rational function:
f(x) = x2−4
x2−x−6
23
Solution
Step 1: We first need to find the values of xthat would make the denominator
x2−x−6equal to zero, as these values would lead to division by zero in
the rational function. So, we solve the equation x2−x−6=0to find the
critical points. Step 2: Factor the quadratic equation x2−x−6 = 0 into
(x−3)(x+ 2) = 0. Step 3: Set each factor to zero and solve for x:
x−3 = 0 or x+ 2 = 0
x= 3 or x=−2
Step 4: So, the values x= 3 and x=−2make the denominator equal to zero,
which means they are not in the domain of the function. Step 5: Therefore,
the domain of the function f(x) = x2
−4
x2−x−6is all real numbers except x= 3 and
x=−2, which can be expressed as:
(−∞,−2) ∪(−2,3) ∪(3,∞)
Question 34
Question
Find the domain of the rational function:
f(x) = x2−4x−5
x2+ 3x−10.
Solution
Step 1: To find the domain of a rational function, we need to identify the values
of xfor which the function is defined. In this case, the function is defined for all
values of xexcept those that would make the denominator equal to zero, since
division by zero is undefined.
Step 2: To find where the denominator x2+ 3x−10 is equal to zero, we
solve the equation x2+ 3x−10 = 0 for x.
Step 3: This equation can be factored as (x+ 5)(x−2) = 0.
Step 4: Setting each factor to zero, we find the values of xthat make the
denominator zero: x+ 5 = 0 ⇒x=−5and x−2 = 0 ⇒x= 2.
Step 5: Therefore, the domain of the rational function f(x) = x2
−4x−5
x2+3x−10 is
all real numbers except x=−5and x= 2.
Step 6: In interval notation, the domain can be expressed as (−∞,−5) ∪
(−5,2) ∪(2,∞).
24
Question 35
Question
Simplify the rational function:
f(x) = 3x2−2x−1
2x2+ 5x−3
Solution
Step 1: Factor the numerator and denominator of the rational function.
The numerator can be factored as:
3x2−2x−1 = (3x+ 1)(x−1)
The denominator can be factored as:
2x2+ 5x−3 = (2x−1)(x+ 3)
Step 2: Rewrite the rational function using the factored forms of the numer-
ator and denominator.
f(x) = (3x+ 1)(x−1)
(2x−1)(x+ 3)
Step 3: Simplify the rational function by canceling out common factors in
the numerator and denominator.
f(x) = 3x+ 1
2x−1·x−1
x+ 3
Step 4: The simplified form of the rational function is:
f(x) = (3x+ 1)(x−1)
(2x−1)(x+ 3)
25
