MATH 121 - COLLEGE ALGEBRA -
Rational Functions
Question Bank - Set 8
Liberty University
Question 1
Question
Solve the rational inequality: x+ 4
x−3≥2
Solution
Step 1: Find the critical points by setting the numerator equal to zero:
x+ 4 = 0
x=−4
Step 2: Find the critical points by setting the denominator equal to zero:
x−3 = 0
x= 3
Step 3: Create a sign chart using the critical points -4 and 3:
x < −4−4< x < 3x > 3
x+ 4 −+ +
x−3− − +
x+4
x−3+−+
Step 4: Analyze the sign chart to solve the inequality: Since we are looking
for values where x+4
x−3≥2, the solution is:
x∈(−∞,−4) ∪(3,∞)
Question 2
Question
Let f(x) = 5x3−4x2+2x−1
3x2−x−2. Find the x-intercepts of the function f(x).
Solution
To find the x-intercepts of a function, we need to solve for the values of xwhere
f(x) = 0.
Step 1: Set f(x) = 0 Set 5x3−4x2+2x−1
3x2−x−2= 0.
Step 2: Find Common Denominator To simplify, we find a common
denominator for the fractions in the numerator and denominator:
0 = 5x3−4x2+ 2x−1
3x2−x−2=(5x3−4x2+ 2x−1)(3x2−x−2)
3x2−x−2
Step 3: Use Numerator Setting the numerator equal to zero:
5x3−4x2+ 2x−1 = 0
Step 4: Factorize Quadratic Factorizing the cubic polynomial is challeng-
ing and may involve the Rational Root Theorem or synthetic division. Once we
identify one root, we can divide by the corresponding factor to find the other
roots.
Step 5: Solve for xAfter factoring or using other methods to find the
roots of the equation, we obtain the x-intercepts of the function f(x).
Question 3
Question
Let f(x) = 3x2−5x+2
2x2+x−3be a rational function. Find the horizontal asymptotes,
vertical asymptotes, and any holes in the graph of f(x).
Solution
Step 1: To find the horizontal asymptotes, we compare the degrees of the nu-
merator and denominator. If the degree of the numerator is less than the degree
of the denominator, the x-axis (i.e., y= 0) is a horizontal asymptote. If the de-
grees are equal, the horizontal asymptote is the ratio of the leading coefficients.
If the degree of the numerator is greater, there is no horizontal asymptote.
Step 2: The degree of the numerator is 2 and the degree of the denominator
is 2. Therefore, to find the horizontal asymptote(s), we compare the leading
coefficients: 3
2. Thus, the horizontal asymptote is y=3
2.
Step 3: To find the vertical asymptotes, we look for values of xthat make
the denominator equal to zero, but not the numerator. This will create vertical
asymptotes.
2
Step 4: To find the vertical asymptotes of f(x) = 3x2−5x+2
2x2+x−3, we set the
denominator equal to zero and solve for x:2x2+x−3=0. Factoring, we get
(2x−3)(x+ 1) = 0. Thus, x=3
2and x=−1are the vertical asymptotes.
Step 5: To determine if there are any holes in the graph, we need to check
if any factors cancel out in the function. To find any potential holes, set the
function equal to zero and factor.
Step 6: Setting the function equal to zero, we get 3x2−5x+2
2x2+x−3= 0. Factoring
both the numerator and denominator, we get (3x−2)(x−1)
(2x−3)(x+1) = 0. This gives us
two potential holes at x=2
3and x= 1.
Hence, the function f(x) = 3x2−5x+2
2x2+x−3has a horizontal asymptote at y=3
2,
vertical asymptotes at x=3
2and x=−1, and potential holes at x=2
3and
x= 1.
Question 4
Question
Simplify the rational function: 2x2−5x−3
x2−4.
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify by canceling
out common factors. Step 3: Write the simplified form of the rational function.
Step 1: Factor the numerator and denominator. The numerator 2x2−5x−3
can be factored into (2x+ 1)(x−3). The denominator x2−4is a difference of
squares and can be factored into (x+ 2)(x−2).
Step 2: Simplify by canceling out common factors. Therefore, the rational
function can be written as:
2x2−5x−3
x2−4=(2x+ 1)(x−3)
(x+ 2)(x−2)
Step 3: Write the simplified form of the rational function. The simplified
form of the rational function is:
2x+ 1
x+ 2
Question 5
Question
Given the rational function f(x) = x3−2x2−3x
x2−3x−4, find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph of f(x).
3
Solution
Step 1: To find the vertical asymptotes, we need to determine where the de-
nominator of the rational function is equal to zero. Set x2−3x−4 = 0 and
solve for x.
x2−3x−4 = 0
(x−4)(x+ 1) = 0
x= 4 or x=−1
Therefore, the vertical asymptotes are x= 4 and x=−1.
Step 2: To find the horizontal asymptote, consider 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 can be found by divid-
ing the leading coefficients of the numerator and denominator. This means the
horizontal asymptote is y=1
1= 1.
Step 3: To find any holes in the graph, we need to check if any factors cancel
out. Simplify the function f(x)by cancelling out common factors between the
numerator and the denominator.
f(x) = x(x2−2x−3)
(x−4)(x+ 1)
f(x) = x(x−3)(x+ 1)
(x−4)(x+ 1)
f(x) = x(x−3)
x−4
From this simplification, we see that there is a hole in the graph at x=−1since
the factor (x+ 1) canceled out in both the numerator and denominator.
In conclusion, the rational function f(x) = x3−2x2−3x
x2−3x−4has vertical asymp-
totes at x= 4 and x=−1, a horizontal asymptote at y= 1, and a hole at
x=−1.
Question 6
Question
Simplify the rational function:
6x2−5x−6
x2−9∇ · 2x2+ 5x+ 3
x2+ 3x−10
4
Solution
Step 1: To simplify a division of rational functions, we can rewrite it as a mul-
tiplication by the reciprocal of the divisor. Thus, the given expression becomes:
6x2−5x−6
x2−9·x2+ 3x−10
2x2+ 5x+ 3
Step 2: Factor all the quadratic expressions:
6x2−5x−6 = (2x−3)(3x+ 2)
x2−9 = (x−3)(x+ 3)
x2+ 3x−10 = (x+ 5)(x−2)
2x2+ 5x+ 3 = (2x+ 3)(x+ 1)
Step 3: Substitute the factored forms into the expression:
(2x−3)(3x+ 2)
(x−3)(x+ 3) ·(x+ 5)(x−2)
(2x+ 3)(x+ 1)
Step 4: Simplify the expression by canceling out common factors in the
numerators and denominators:
2x−3
x+ 3 ·x+ 5
x+ 1
Step 5: Multiply the remaining terms to get the final simplified expression:
(2x−3)(x+ 5)
(x+ 3)(x+ 1) =2x2+ 7x−15
x2+ 4x+ 3
Therefore, the simplified form of the given rational function is 2x2+7x−15
x2+4x+3 .
Question 7
Question
For the rational function f(x) = x2+3x+2
x2−2x−3, determine the following: (i) Do-
main of f(x). (ii) x-intercepts, if any. (iii) y-intercepts, if any. (iv) Vertical
asymptotes, if any. (v) Horizontal asymptotes, if any.
Solution
(i) To find the domain of f(x), we need to consider where the function is defined.
Since we cannot divide by zero, the denominator x2−2x−3cannot be equal
to zero:
x2−2x−3= 0.
5
This quadratic equation factors as (x−3)(x+ 1) = 0, which means x= 3 and
x=−1. Therefore, the domain of f(x)is all real numbers except x= 3 and
x=−1.
(ii) To find the x-intercepts, we set f(x) = 0 and solve for x:
x2+ 3x+ 2
x2−2x−3= 0.
This gives us x2+ 3x+ 2 = 0, which factors to (x+ 1)(x+ 2) = 0. Therefore,
the x-intercepts are at x=−1and x=−2.
(iii) To find the y-intercept, we set x= 0 in f(x):
f(0) = 02+ 3(0) + 2
02−2(0) −3=2
−3=−2
3.
So, the y-intercept is at y=−2
3.
(iv) To find the vertical asymptotes, we need to find where the denominator
is zero. This occurs at x= 3 and x=−1. Thus, the vertical asymptotes are at
x= 3 and x=−1.
(v) To find the horizontal asymptotes, we compare the degrees of the nu-
merator and denominator of f(x). Since both the numerator and denominator
have the same degree (2), we look at the ratio of the leading coefficients:
lim
x→∞
f(x) = lim
x→∞
x2+ 3x+ 2
x2−2x−3=1
1= 1.
Therefore, there is a horizontal asymptote at y= 1.
Question 8
Question
Let f(x) = x3−4x2−x+ 4
x2−3x+ 2 . Determine the domain of f(x)and find any
vertical asymptotes, horizontal asymptotes, and holes in the graph of f(x).
Solution
Step 1: To find the domain of f(x), we need to identify any values of xthat
would make the denominator x2−3x+2 equal to zero. This will result in division
by zero, which is undefined. So, we need to solve the equation x2−3x+ 2 = 0.
x2−3x+ 2 = 0
(x−2)(x−1) = 0
x= 2 or x= 1
Therefore, the domain of f(x)is all real numbers except x= 1 and x= 2.
6
Step 2: To find the vertical asymptotes, we need to look at the values that
xcannot take on in the domain. Since x= 1 and x= 2 are excluded from the
domain, we have vertical asymptotes at x= 1 and x= 2.
Step 3: The horizontal asymptote of a rational function can be found by
looking at the degrees of the numerator and denominator. Since the degree
of the numerator is greater than the degree of the denominator, there is no
horizontal asymptote.
Step 4: To find any holes in the graph of f(x), we can simplify the function
by factoring the numerator and the denominator.
f(x) = x3−4x2−x+ 4
x2−3x+ 2 =(x−4)(x−1)(x+ 1)
(x−2)(x−1)
We can see that there is a factor of (x−1) in both the numerator and
denominator that cancels out. So, f(x)simplifies to (x−4)(x+ 1)
x−2.
Therefore, there is a hole in the graph at x= 1.
In summary, the domain of f(x)is all real numbers except x= 1 and x= 2.
The graph has vertical asymptotes at x= 1 and x= 2, a hole at x= 1, and no
horizontal asymptote.
Question 9
Question
Simplify the following rational expression:
4x2−16
x2−4x
Solution
Step 1: Factor both the numerator and denominator.
4x2−16 = 4(x2−4)
= 4(x+ 2)(x−2)
x2−4x=x(x−4)
Step 2: Rewrite the expression using the factored forms.
4(x+ 2)(x−2)
x(x−4)
Step 3: Simplify the expression by canceling out common factors.
4(x+ 2)(x−2)
x(x−4)
7
Step 4: The simplified form of the expression is:
4
x
Question 10
Question
Simplify the rational function:
2x2+ 5x−3
x2−4x−5∇ · 4x2−1
2x2+ 7x+ 3
Solution
Step 1: To divide rational expressions, we multiply by the reciprocal of the
divisor. Step 2: Rewrite the division as multiplication and multiply by the
reciprocal. Step 3: Factor all the quadratic expressions and simplify where
possible.
Step 1: Rewrite the division as multiplication and multiply by the recipro-
cal: 2x2+ 5x−3
x2−4x−5·2x2+ 7x+ 3
4x2−1
Step 2: Factor all the quadratic expressions:
(2x−1)(x+ 3)
(x−5)(x+ 1) ·(2x+ 3)(x+ 1)
(2x−1)(2x+ 1)
Step 3: Multiply the numerators and denominators:
(2x−1)(x+ 3)(2x+ 3)(x+ 1)
(x−5)(x+ 1)(2x−1)(2x+ 1)
Therefore, the simplified form of the rational function is:
(2x−1)(x+ 3)(2x+ 3)(x+ 1)
(x−5)(x+ 1)(2x−1)(2x+ 1)
Question 11
Question
Given the rational function f(x) = x3−2x2−5x+6
x2+3x−4, find the vertical asymptotes
of f(x).
8
Solution
Step 1: To find the vertical asymptotes of the rational function f(x), we need
to determine the values of xthat make the denominator equal to zero, but do
not make the numerator zero. Vertical asymptotes occur at these values.
Step 2: Set the denominator x2+ 3x−4equal to zero and solve for x:
x2+ 3x−4 = 0
Step 3: Factor the quadratic equation:
(x+ 4)(x−1) = 0
Step 4: Set each factor equal to zero:
x+ 4 = 0 or x−1 = 0
Step 5: Solve for xin each equation:
x=−4or x= 1
Step 6: Therefore, the vertical asymptotes of the rational function f(x)are
x=−4and x= 1.
Question 12
Question
Simplify the following rational expression:
3x2−9x+ 6
x2−4
Solution
Step 1: Factor out common terms in the numerator and denominator:
3x2−9x+ 6
x2−4=3(x2−3x+ 2)
x2−4
Step 2: Factor the quadratics in the numerator and denominator:
3(x−1)(x−2)
(x+ 2)(x−2)
Step 3: Cancel out common factors:
3(x−1)
x+ 2
Therefore, the simplified form of the rational expression is 3(x−1)
x+2 .
9
Question 13
Question
Simplify the following rational expression:
4x2−7x−12
2x2+ 9x+ 5 ∇ · 2x2+ 5x−3
8x2−7x−12
Solution
Step 1: Re-write the division as multiplication by the reciprocal of the second
fraction: 4x2−7x−12
2x2+ 9x+ 5 ×8x2−7x−12
2x2+ 5x−3
Step 2: Factor all the quadratic expressions to simplify: Factor the numer-
ator of the first fraction:
4x2−7x−12
This expression can be factored as:
(4x+ 3)(x−4)
Factor the denominator of the first fraction:
2x2+ 9x+ 5
This expression can be factored as:
(2x+ 1)(x+ 5)
Factor the numerator of the second fraction:
8x2−7x−12
This expression can be factored as:
(8x−3)(x+ 4)
Factor the denominator of the second fraction:
2x2+ 5x−3
This expression can be factored as:
(2x−1)(x+ 3)
Step 3: Substitute the factored expressions back into the original expression:
(4x+ 3)(x−4)
(2x+ 1)(x+ 5) ×(8x−3)(x+ 4)
(2x−1)(x+ 3)
10
Step 4: Simplify the expression: The factors in the numerator and denomi-
nator can now be cancelled out, resulting in the simplified expression:
4x+ 3
2x+ 1 ×8x−3
2x−1=(4x+ 3)(8x−3)
(2x+ 1)(2x−1)
Therefore, the simplified form of the given rational expression is (4x+3)(8x−3)
(2x+1)(2x−1) .
Question 14
Question
Calculate the domain of the rational function:
f(x) = 3x2−7x−6
x2−4x−5
Solution
Step 1: Find the excluded values by setting the denominator equal to zero and
solving for x. These values are not in the domain of the function.
x2−4x−5 = 0
(x−5)(x+ 1) = 0
x= 5 or x=−1
Step 2: The domain of the function is all real numbers except for the excluded
values. Therefore, the domain of the function is:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 15
Question
Simplify the following rational expression:
3x2−4
x2−x−6∇ · x2−3x−4
x2+ 2x−8
Solution
Step 1: Factor the numerators and denominators of both fractions. Step 2:
Rewrite the division of fractions as a multiplication by the reciprocal. Step 3:
Simplify the expression by multiplying the numerators and denominators of the
rational expression. Step 4: Factor the resulting numerator and denominator
11
if possible. Step 5: Simplify the expression by canceling out common factors if
they exist.
Step 1: Factor the numerators and denominators of both fractions. For the
first fraction: The numerator 3x2−4can be factored as (3x+ 2)(x−2). The
denominator x2−x−6can be factored as (x−3)(x+ 2).
For the second fraction: The numerator x2−3x−4can be factored as
(x−4)(x+ 1). The denominator x2+ 2x−8can be factored as (x−2)(x+ 4).
Step 2: Rewrite the division of fractions as a multiplication by the recipro-
cal. The expression becomes:
(3x+ 2)(x−2)
(x−3)(x+ 2) ×(x+ 4)(x−2)
(x−4)(x+ 1)
Step 3: Simplify the expression by multiplying the numerators and denom-
inators of the rational expression. The expression simplifies to:
(3x+ 2)(x−2)(x+ 4)(x−2)
(x−3)(x+ 2)(x−4)(x+ 1)
Step 4: Factor the resulting numerator and denominator if possible. The
expression becomes:
(3x+ 2)(x+ 4)(x−2)2
(x−3)(x−4)(x+ 1)(x+ 2)
Step 5: Simplify the expression by canceling out common factors if they
exist. There are no common factors to cancel, so the final simplified form of the
rational expression is:
(3x+ 2)(x+ 4)(x−2)2
(x−3)(x−4)(x+ 1)(x+ 2)
Question 16
Question
Consider the rational function given by f(x) = 2x3−5x2−3x
x2−2x−3.
Determine the vertical asymptotes, horizontal asymptotes, x-intercepts, y-
intercepts, and the domain of the function f(x).
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero
and solving for x.
x2−2x−3 = 0
(x−3)(x+ 1) = 0
This gives us vertical asymptotes at x= 3 and x=−1.
12
Step 2: Find the horizontal asymptote by comparing the degrees of the
numerator and the denominator. Since the degree of the numerator (3) is greater
than the degree of the denominator (2), there is no horizontal asymptote.
Step 3: Find the x-intercepts by setting the numerator equal to zero and
solving for x.
2x3−5x2−3x= 0
x(2x2−5x−3) = 0
x(2x+ 1)(x−3) = 0
This gives us x-intercepts at x= 0,x=−1
2, and x= 3.
Step 4: Find the y-intercept by evaluating f(0).
f(0) = 2(0)3−5(0)2−3(0)
02−2(0) −3= 0
Therefore, the y-intercept is at the point (0, 0).
Step 5: Determine the domain of the function. The function is defined for
all real numbers except where the denominator is equal to zero (the vertical
asymptotes). Therefore, the domain of f(x)is all real numbers except x= 3
and x=−1.
Question 17
Question
Find the domain of the rational function:
f(x) = 1
x2−4x+ 3
Solution
Step 1: The domain of a rational function is all real numbers except the values
that would make the denominator equal to zero. Thus, we need to find the
values of xthat make the denominator x2−4x+ 3 equal to zero.
Step 2: To find the values of xthat make the denominator zero, we solve
the quadratic equation:
x2−4x+ 3 = 0
Step 3: Factoring the quadratic equation, we get:
(x−1)(x−3) = 0
Step 4: Setting each factor to zero gives us:
x−1 = 0 or x−3 = 0
13
Step 5: Solving these equations gives us:
x= 1 or x= 3
Step 6: Therefore, the values x= 1 and x= 3 would make the denominator
of the function equal to zero. As a result, the domain of the function f(x)is all
real numbers except x= 1 and x= 3.
Step 7: Thus, the domain of the function f(x) = 1
x2−4x+ 3 is (−∞,1) ∪
(1,3) ∪(3,∞).
Question 18
Question
Solve the following rational inequality:
3
x−2
x−3≥1.
Solution
Step 1: Find a common denominator for the fractions on the left side of the
inequality.
3
x−2
x−3=3(x−3)
x(x−3) −2x
x(x−3) =3x−9−2x
x(x−3) =x−9
x(x−3)
Step 2: Rewrite the inequality with the common denominator.
x−9
x(x−3) ≥1.
Step 3: Solve the inequality for the numerator.
x−9≥x(x−3)
Step 4: Expand the right side and simplify the inequality.
x−9≥x2−3x
Step 5: Rearrange the inequality into a quadratic inequality.
0≥x2−4x+ 9
Step 6: Solve the quadratic inequality by finding the roots of the quadratic
equation.
x=−(−4) ±√(−4)2−4(1)(9)
2(1)
x=4±√16 −36
2=4±√−20
2=4±2i√5
2= 2 ±i√5.
Step 7: Since the inequality is less than or equal to zero, the solution set is
(−∞,2−i√5] ∪[2 + i√5,∞).
14
Question 19
Question
Simplify the following rational expression:
3x2−2x−5
x2−4x−5
Solution
Step 1: Factor both the numerator and the denominator of the rational expres-
sion. 3x2−2x−5
x2−4x−5=(3x+ 1)(x−5)
(x−5)(x+ 1)
Step 2: Simplify by canceling out the common factor in the numerator and
the denominator. (3x+ 1)(x−5)
(x−5)(x+ 1) =3x+ 1
x+ 1
Therefore, the simplified form of the rational expression is 3x+1
x+1 .
Question 20
Question
Find the domain of the rational function:
f(x) = 4x
x2−9.
Solution
Step 1: The domain of a rational function is all real numbers except the values of
xthat make the denominator equal to zero, since division by zero is undefined.
So, we need to find the values of xfor which x2−9= 0.
Step 2: We can rewrite x2−9= 0 as (x+ 3)(x−3) = 0 by factoring the
quadratic polynomial.
Step 3: This inequality is true except when either x+ 3 = 0 or x−3 = 0.
Therefore, we solve for xin both cases:
x+ 3 = 0 ⇒x=−3,
and
x−3 = 0 ⇒x= 3.
Step 4: So, the domain of the rational function f(x) = 4x
x2−9is all real
numbers except x=−3and x= 3. In interval notation, the domain is:
(−∞,−3) ∪(−3,3) ∪(3,∞).
15
Question 21
Question
Find the domain of the rational function:
f(x) = x2−4x+ 3
x2−5x+ 6
Solution
Step 1: Find the values of xthat make the denominator zero, as these values
would result in division by zero which is undefined.
x2−5x+ 6 = 0
Factorizing the quadratic equation gives:
(x−2)(x−3) = 0
So, the values of xthat make the denominator zero are x= 2 and x= 3.
Step 2: The domain of a rational function is all real numbers except those
that make the denominator zero. Thus, the domain of f(x)is all real numbers
except x= 2 and x= 3.
Therefore, the domain of the rational function f(x) = x2−4x+3
x2−5x+6 is (−∞,2) ∪(2,3) ∪(3,∞).
Question 22
Question
Find the domain of the rational function:
f(x) = 7x
x2−5x−6
Solution
Step 1: The domain of a rational function is all real numbers except where the
denominator is equal to zero. So, we need to find the values of xthat make the
denominator x2−5x−6equal to zero.
Step 2: To find these values, we will factor the quadratic equation x2−5x−6.
The factors will be in the form (x+m)(x+n)where mand nare the factors
of −6that add up to −5. The factors are −6and 1as −6 + 1 = −5. Therefore,
the factored form is (x−6)(x+ 1).
Step 3: Setting the factored form equal to zero gives us:
x−6 = 0 =⇒x= 6
x+ 1 = 0 =⇒x=−1
16
Step 4: The domain of the rational function f(x)is all real numbers except
x= 6 and x=−1since they make the denominator equal to zero.
Therefore, the domain of f(x)is (−∞,−1) ∪(−1,6) ∪(6,∞).
Question 23
Question
Solve the rational equation for x:
2
x+ 3 −3
x−2=5
x2+x−6
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation. In this case, 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) =5
x2+x−6
Step 3: Combine the fractions on the left side of the equation.
2x−4−3x−9
(x+ 3)(x−2) =5
x2+x−6
Step 4: Simplify the numerator.
−x−13
(x+ 3)(x−2) =5
x2+x−6
Step 5: Factor the denominators of both sides. The denominator on the
left side is already factored. The denominator on the right side factors as (x+
3)(x−2).
Step 6: Cross multiply to eliminate the denominators.
−x−13 = 5
Step 7: Solve for x.
x=−18
Step 8: Check the solution in the original equation. Substitute x=−18
back into the original equation to check for extraneous solutions.
2
−18 + 3 −3
−18 −2=5
(−18)2+ (−18) −6
2
−15 −3
−20 =5
324 −18 −6=2
−15 +3
−20 =5
300 =1
60
17
The solution x=−18 satisfies the original equation.
Question 24
Question
Simplify the following rational function:
f(x) = 3x2−2x−8
2x2+ 5x−3
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify the rational
function by canceling out any common factors.
Step 1: To factor the numerator and denominator of f(x): Factor the
numerator:
3x2−2x−8 = (3x+ 4)(x−2)
Factor the denominator:
2x2+ 5x−3 = (2x−1)(x+ 3)
Step 2: Now, we can rewrite f(x)with the factored numerator and denom-
inator:
f(x) = (3x+ 4)(x−2)
(2x−1)(x+ 3)
There are no common factors that can be canceled out, so the simplified
form of the rational function is:
f(x) = 3x+ 4
2x−1
Question 25
Question
Let f(x) = x2+4x
x+2 be a rational function. Determine the domain of f(x).
Solution
Step 1: The domain of a rational function is all real numbers except where the
denominator is equal to zero. In this case, the denominator x+ 2 is equal to
zero when x=−2.
Step 2: Therefore, the domain of f(x)is all real numbers except x=−2.
Thus, the domain of f(x)is (−∞,−2) ∪(−2,∞).
18
Question 26
Question
Find the domain of the rational function:
f(x) = 2x2−3x−2
x2−4x−5
Solution
Step 1: To find the domain of the function, we need to determine all the values of
xfor which the function is defined. The function is defined for all real numbers
except those that make the denominator zero, as division by zero is undefined.
Step 2: Find the values of xthat make the denominator zero by setting
x2−4x−5equal to zero and solving for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation, we can factor it or use the quadratic
formula. Let’s use the quadratic formula:
x=−(−4) ±√(−4)2−4(1)(−5)
2(1)
x=4±√16 + 20
2
x=4±√36
2
x=4±6
2
Step 4: This gives us two possible values for x:
x=4+6
2⇒x= 5
x=4−6
2⇒x=−1
Step 5: These two values, x= 5 and x=−1, make the denominator zero.
Therefore, the values -1 and 5 are not in the domain of the function.
Step 6: The domain of the function f(x)is all real numbers except for
x=−1and x= 5. Therefore, the domain is given by:
Domain of f(x) = {x∈R|x=−1,5}
19
Question 27
Question
Simplify the following rational expression:
5x2−3x−2
x2−4x−12∇ · x2+ 2x−8
3x2+ 2x−8
Solution
Step 1: First, rewrite the division as multiplication by the reciprocal of the
second fraction: 5x2−3x−2
x2−4x−12 ×3x2+ 2x−8
x2+ 2x−8
Step 2: Factor the numerators and denominators of each fraction:
(5x+ 1)(x−2)
(x−6)(x+ 2) ×(3x−2)(x+ 4)
(x+ 4)(x−2)
Step 3: Cancel out any common factors in the numerators and denominators:
5x+ 1
x−6×3x−2
1
Step 4: Multiply the remaining factors in the numerators and denominators:
(5x+ 1)(3x−2)
(x−6)(1)
Step 5: Expand the remaining factors:
15x2−4x−2
x−6
Therefore, the simplified form of the given rational expression is 15x2−4x−2
x−6.
Question 28
Question
Find the domain of the rational function:
f(x) = 2x2−5x−3
x2+ 4x−5
20
Solution
Step 1: Determine the excluded values that would make the denominator zero,
as division by zero is undefined. Step 2: Find the domain by excluding these
values from the set of all real numbers.
Step 1: Set the denominator equal to zero and solve for x:
x2+ 4x−5 = 0
Using the quadratic formula, we have:
x=−b±√b2−4ac
2a
Where a= 1,b= 4, and c=−5, we get:
x=−4±√42−4(1)(−5)
2(1)
x=−4±√16 + 20
2
x=−4±√36
2
x=−4±6
2
Therefore, the excluded values are x=−6and x= 1.
Step 2: The domain of a rational function is all real numbers except those
that would make the denominator zero. So, the domain of the function f(x)is:
(−∞,−6) ∪(−6,1) ∪(1,∞)
Question 29
Question
Find the domain of the rational function:
f(x) = 2x−1
x2−5x+ 6.
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the function is defined. In a rational function, the de-
nominator cannot be zero. So, we need to find the values of xthat make the
denominator zero.
21
Step 2: We set the denominator equal to zero and solve for x.
x2−5x+ 6 = 0
(x−2)(x−3) = 0
Step 3: Now, we find the values of xfor which the denominator is zero by
setting each factor equal to zero and solving for x.
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 4: Therefore, the domain of the function f(x) = 2x−1
x2−5x+ 6 is all real
numbers except x= 2 and x= 3.
Step 5: The domain of the function is (−∞,2) ∪(2,3) ∪(3,∞).
Question 30
Question
Simplify the rational function:
f(x) = 3x3−2x2+ 5x−1
x2−2x−3
Solution
Step 1: Factor the numerator and the denominator. Step 2: Express the function
in simplified form.
Step 1: Factor the numerator and the denominator. First, let’s factor the
numerator:
3x3−2x2+ 5x−1 = x2(3x−2) + 1(3x−2) = (x2+ 1)(3x−2)
Next, let’s factor the denominator:
x2−2x−3 = (x−3)(x+ 1)
Step 2: Express the function in simplified form. Now that we have factored
both the numerator and the denominator, we can simplify the expression:
f(x) = 3x3−2x2+ 5x−1
x2−2x−3=(x2+ 1)(3x−2)
(x−3)(x+ 1)
Therefore, the simplified form of the rational function f(x)is (x2+ 1)(3x−2)
(x−3)(x+ 1) .
22
Question 31
Question
Simplify the following rational expression:
4x2−36
x2−16
Solution
Step 1: Factor both the numerator and denominator:
4x2−36 = 4(x2−9) = 4(x+ 3)(x−3)
x2−16 = (x+ 4)(x−4)
Step 2: Rewrite the expression with factored terms:
4(x+ 3)(x−3)
(x+ 4)(x−4)
Step 3: Cancel out common factors:
4(x+ 3)(x−3)
(x+ 4)(x−4)
Step 4: Simplify the expression to its final form:
4
1= 4
Question 32
Question
Given the rational function f(x) = 2x2+3x−2
x2−4x+3 , find the vertical and horizontal
asymptotes, any holes, x-intercepts, and y-intercepts.
Solution
Step 1: Find the vertical asymptotes
To find the vertical asymptotes, we need to determine the values of xthat
make the denominator of the rational function equal to zero. Set the denomi-
nator x2−4x+ 3 equal to zero and solve for x:
x2−4x+ 3 = 0
(x−3)(x−1) = 0
23
So, the vertical asymptotes occur at x= 3 and x= 1.
Step 2: Find the horizontal asymptote
To find the horizontal asymptote, compare the degrees of the numerator
and denominator of the rational function. Since the degrees are the same, the
horizontal asymptote is given by the ratio of the leading coefficients:
Horizontal asymptote: y=2
1= 2
Step 3: Find any holes
To find any holes, factor the numerator 2x2+ 3x−2and the denominator
x2−4x+ 3:
2x2+ 3x−2 = (2x−1)(x+ 2)
x2−4x+ 3 = (x−3)(x−1)
We see that there is a hole at x= 1 where the factor of (x−1) cancels out.
Step 4: Find x-intercepts To find the x-intercepts, set f(x) = 0 and solve
for x:2x2+ 3x−2
x2−4x+ 3 = 0
Since the numerator can only be zero when x=1
2or x=−2, the x-intercepts
are (1
2,0)and (−2,0).
Step 5: Find y-intercept To find the y-intercept, evaluate f(0):
f(0) = 2(0)2+ 3(0) −2
(0)2−4(0) + 3 =−2
3
So, the y-intercept is (0,−2
3).
Question 33
Question
Find the domain of the rational function: 3x2−5x−2
x2−4x−5.
Solution
Step 1: To find the domain of a rational 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
Step 3: To solve the quadratic equation, we can factor it:
(x−5)(x+ 1) = 0
24
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 the rational function is all real numbers
except x= 5 and x=−1.
Step 7: In interval notation, the domain can be expressed as (−∞,−1) ∪
(−1,5) ∪(5,∞).
Question 34
Question
Simplify the following rational expression:
3x2−7x−6
x2−4x−21
Solution
Step 1: Factor the numerator and denominator:
3x2−7x−6 = (3x+ 1)(x−6)
x2−4x−21 = (x−7)(x+ 3)
Step 2: Rewrite the expression with the factored forms of the numerator and
denominator: (3x+ 1)(x−6)
(x−7)(x+ 3)
Step 3: Simplify the expression by canceling out common factors in the
numerator and denominator:
(3x+ 1)(x−6)
(x−7)(x+ 3) =3x+ 1
x+ 3
Therefore, the simplified form of the rational expression is 3x+1
x+3 .
Question 35
Question
Let f(x) = 3x2−5x−2
2x2+7x+3 . Find the domain of the function f(x).
25
Question 2
Question
Let f(x) = 5x3−4x2+2x−1
3x2−x−2. Find the x-intercepts of the function f(x).
Solution
To find the x-intercepts of a function, we need to solve for the values of xwhere
f(x) = 0.
Step 1: Set f(x) = 0 Set 5x3−4x2+2x−1
3x2−x−2= 0.
Step 2: Find Common Denominator To simplify, we find a common
denominator for the fractions in the numerator and denominator:
0 = 5x3−4x2+ 2x−1
3x2−x−2=(5x3−4x2+ 2x−1)(3x2−x−2)
3x2−x−2
Step 3: Use Numerator Setting the numerator equal to zero:
5x3−4x2+ 2x−1 = 0
Step 4: Factorize Quadratic Factorizing the cubic polynomial is challeng-
ing and may involve the Rational Root Theorem or synthetic division. Once we
identify one root, we can divide by the corresponding factor to find the other
roots.
Step 5: Solve for xAfter factoring or using other methods to find the
roots of the equation, we obtain the x-intercepts of the function f(x).
Question 3
Question
Let f(x) = 3x2−5x+2
2x2+x−3be a rational function. Find the horizontal asymptotes,
vertical asymptotes, and any holes in the graph of f(x).
Solution
Step 1: To find the horizontal asymptotes, we compare the degrees of the nu-
merator and denominator. If the degree of the numerator is less than the degree
of the denominator, the x-axis (i.e., y= 0) is a horizontal asymptote. If the de-
grees are equal, the horizontal asymptote is the ratio of the leading coefficients.
If the degree of the numerator is greater, there is no horizontal asymptote.
Step 2: The degree of the numerator is 2 and the degree of the denominator
is 2. Therefore, to find the horizontal asymptote(s), we compare the leading
coefficients: 3
2. Thus, the horizontal asymptote is y=3
2.
Step 3: To find the vertical asymptotes, we look for values of xthat make
the denominator equal to zero, but not the numerator. This will create vertical
asymptotes.
2
Step 4: To find the vertical asymptotes of f(x) = 3x2−5x+2
2x2+x−3, we set the
denominator equal to zero and solve for x:2x2+x−3=0. Factoring, we get
(2x−3)(x+ 1) = 0. Thus, x=3
2and x=−1are the vertical asymptotes.
Step 5: To determine if there are any holes in the graph, we need to check
if any factors cancel out in the function. To find any potential holes, set the
function equal to zero and factor.
Step 6: Setting the function equal to zero, we get 3x2−5x+2
2x2+x−3= 0. Factoring
both the numerator and denominator, we get (3x−2)(x−1)
(2x−3)(x+1) = 0. This gives us
two potential holes at x=2
3and x= 1.
Hence, the function f(x) = 3x2−5x+2
2x2+x−3has a horizontal asymptote at y=3
2,
vertical asymptotes at x=3
2and x=−1, and potential holes at x=2
3and
x= 1.
Question 4
Question
Simplify the rational function: 2x2−5x−3
x2−4.
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify by canceling
out common factors. Step 3: Write the simplified form of the rational function.
Step 1: Factor the numerator and denominator. The numerator 2x2−5x−3
can be factored into (2x+ 1)(x−3). The denominator x2−4is a difference of
squares and can be factored into (x+ 2)(x−2).
Step 2: Simplify by canceling out common factors. Therefore, the rational
function can be written as:
2x2−5x−3
x2−4=(2x+ 1)(x−3)
(x+ 2)(x−2)
Step 3: Write the simplified form of the rational function. The simplified
form of the rational function is:
2x+ 1
x+ 2
Question 5
Question
Given the rational function f(x) = x3−2x2−3x
x2−3x−4, find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph of f(x).
3
Solution
Step 1: To find the vertical asymptotes, we need to determine where the de-
nominator of the rational function is equal to zero. Set x2−3x−4 = 0 and
solve for x.
x2−3x−4 = 0
(x−4)(x+ 1) = 0
x= 4 or x=−1
Therefore, the vertical asymptotes are x= 4 and x=−1.
Step 2: To find the horizontal asymptote, consider 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 can be found by divid-
ing the leading coefficients of the numerator and denominator. This means the
horizontal asymptote is y=1
1= 1.
Step 3: To find any holes in the graph, we need to check if any factors cancel
out. Simplify the function f(x)by cancelling out common factors between the
numerator and the denominator.
f(x) = x(x2−2x−3)
(x−4)(x+ 1)
f(x) = x(x−3)(x+ 1)
(x−4)(x+ 1)
f(x) = x(x−3)
x−4
From this simplification, we see that there is a hole in the graph at x=−1since
the factor (x+ 1) canceled out in both the numerator and denominator.
In conclusion, the rational function f(x) = x3−2x2−3x
x2−3x−4has vertical asymp-
totes at x= 4 and x=−1, a horizontal asymptote at y= 1, and a hole at
x=−1.
Question 6
Question
Simplify the rational function:
6x2−5x−6
x2−9∇ · 2x2+ 5x+ 3
x2+ 3x−10
4
Solution
Step 1: To simplify a division of rational functions, we can rewrite it as a mul-
tiplication by the reciprocal of the divisor. Thus, the given expression becomes:
6x2−5x−6
x2−9·x2+ 3x−10
2x2+ 5x+ 3
Step 2: Factor all the quadratic expressions:
6x2−5x−6 = (2x−3)(3x+ 2)
x2−9 = (x−3)(x+ 3)
x2+ 3x−10 = (x+ 5)(x−2)
2x2+ 5x+ 3 = (2x+ 3)(x+ 1)
Step 3: Substitute the factored forms into the expression:
(2x−3)(3x+ 2)
(x−3)(x+ 3) ·(x+ 5)(x−2)
(2x+ 3)(x+ 1)
Step 4: Simplify the expression by canceling out common factors in the
numerators and denominators:
2x−3
x+ 3 ·x+ 5
x+ 1
Step 5: Multiply the remaining terms to get the final simplified expression:
(2x−3)(x+ 5)
(x+ 3)(x+ 1) =2x2+ 7x−15
x2+ 4x+ 3
Therefore, the simplified form of the given rational function is 2x2+7x−15
x2+4x+3 .
Question 7
Question
For the rational function f(x) = x2+3x+2
x2−2x−3, determine the following: (i) Do-
main of f(x). (ii) x-intercepts, if any. (iii) y-intercepts, if any. (iv) Vertical
asymptotes, if any. (v) Horizontal asymptotes, if any.
Solution
(i) To find the domain of f(x), we need to consider where the function is defined.
Since we cannot divide by zero, the denominator x2−2x−3cannot be equal
to zero:
x2−2x−3= 0.
5
This quadratic equation factors as (x−3)(x+ 1) = 0, which means x= 3 and
x=−1. Therefore, the domain of f(x)is all real numbers except x= 3 and
x=−1.
(ii) To find the x-intercepts, we set f(x) = 0 and solve for x:
x2+ 3x+ 2
x2−2x−3= 0.
This gives us x2+ 3x+ 2 = 0, which factors to (x+ 1)(x+ 2) = 0. Therefore,
the x-intercepts are at x=−1and x=−2.
(iii) To find the y-intercept, we set x= 0 in f(x):
f(0) = 02+ 3(0) + 2
02−2(0) −3=2
−3=−2
3.
So, the y-intercept is at y=−2
3.
(iv) To find the vertical asymptotes, we need to find where the denominator
is zero. This occurs at x= 3 and x=−1. Thus, the vertical asymptotes are at
x= 3 and x=−1.
(v) To find the horizontal asymptotes, we compare the degrees of the nu-
merator and denominator of f(x). Since both the numerator and denominator
have the same degree (2), we look at the ratio of the leading coefficients:
lim
x→∞
f(x) = lim
x→∞
x2+ 3x+ 2
x2−2x−3=1
1= 1.
Therefore, there is a horizontal asymptote at y= 1.
Question 8
Question
Let f(x) = x3−4x2−x+ 4
x2−3x+ 2 . Determine the domain of f(x)and find any
vertical asymptotes, horizontal asymptotes, and holes in the graph of f(x).
Solution
Step 1: To find the domain of f(x), we need to identify any values of xthat
would make the denominator x2−3x+2 equal to zero. This will result in division
by zero, which is undefined. So, we need to solve the equation x2−3x+ 2 = 0.
x2−3x+ 2 = 0
(x−2)(x−1) = 0
x= 2 or x= 1
Therefore, the domain of f(x)is all real numbers except x= 1 and x= 2.
6
Step 2: To find the vertical asymptotes, we need to look at the values that
xcannot take on in the domain. Since x= 1 and x= 2 are excluded from the
domain, we have vertical asymptotes at x= 1 and x= 2.
Step 3: The horizontal asymptote of a rational function can be found by
looking at the degrees of the numerator and denominator. Since the degree
of the numerator is greater than the degree of the denominator, there is no
horizontal asymptote.
Step 4: To find any holes in the graph of f(x), we can simplify the function
by factoring the numerator and the denominator.
f(x) = x3−4x2−x+ 4
x2−3x+ 2 =(x−4)(x−1)(x+ 1)
(x−2)(x−1)
We can see that there is a factor of (x−1) in both the numerator and
denominator that cancels out. So, f(x)simplifies to (x−4)(x+ 1)
x−2.
Therefore, there is a hole in the graph at x= 1.
In summary, the domain of f(x)is all real numbers except x= 1 and x= 2.
The graph has vertical asymptotes at x= 1 and x= 2, a hole at x= 1, and no
horizontal asymptote.
Question 9
Question
Simplify the following rational expression:
4x2−16
x2−4x
Solution
Step 1: Factor both the numerator and denominator.
4x2−16 = 4(x2−4)
= 4(x+ 2)(x−2)
x2−4x=x(x−4)
Step 2: Rewrite the expression using the factored forms.
4(x+ 2)(x−2)
x(x−4)
Step 3: Simplify the expression by canceling out common factors.
4(x+ 2)(x−2)
x(x−4)
7
Step 4: The simplified form of the expression is:
4
x
Question 10
Question
Simplify the rational function:
2x2+ 5x−3
x2−4x−5∇ · 4x2−1
2x2+ 7x+ 3
Solution
Step 1: To divide rational expressions, we multiply by the reciprocal of the
divisor. Step 2: Rewrite the division as multiplication and multiply by the
reciprocal. Step 3: Factor all the quadratic expressions and simplify where
possible.
Step 1: Rewrite the division as multiplication and multiply by the recipro-
cal: 2x2+ 5x−3
x2−4x−5·2x2+ 7x+ 3
4x2−1
Step 2: Factor all the quadratic expressions:
(2x−1)(x+ 3)
(x−5)(x+ 1) ·(2x+ 3)(x+ 1)
(2x−1)(2x+ 1)
Step 3: Multiply the numerators and denominators:
(2x−1)(x+ 3)(2x+ 3)(x+ 1)
(x−5)(x+ 1)(2x−1)(2x+ 1)
Therefore, the simplified form of the rational function is:
(2x−1)(x+ 3)(2x+ 3)(x+ 1)
(x−5)(x+ 1)(2x−1)(2x+ 1)
Question 11
Question
Given the rational function f(x) = x3−2x2−5x+6
x2+3x−4, find the vertical asymptotes
of f(x).
8
Solution
Step 1: To find the vertical asymptotes of the rational function f(x), we need
to determine the values of xthat make the denominator equal to zero, but do
not make the numerator zero. Vertical asymptotes occur at these values.
Step 2: Set the denominator x2+ 3x−4equal to zero and solve for x:
x2+ 3x−4 = 0
Step 3: Factor the quadratic equation:
(x+ 4)(x−1) = 0
Step 4: Set each factor equal to zero:
x+ 4 = 0 or x−1 = 0
Step 5: Solve for xin each equation:
x=−4or x= 1
Step 6: Therefore, the vertical asymptotes of the rational function f(x)are
x=−4and x= 1.
Question 12
Question
Simplify the following rational expression:
3x2−9x+ 6
x2−4
Solution
Step 1: Factor out common terms in the numerator and denominator:
3x2−9x+ 6
x2−4=3(x2−3x+ 2)
x2−4
Step 2: Factor the quadratics in the numerator and denominator:
3(x−1)(x−2)
(x+ 2)(x−2)
Step 3: Cancel out common factors:
3(x−1)
x+ 2
Therefore, the simplified form of the rational expression is 3(x−1)
x+2 .
9
Question 13
Question
Simplify the following rational expression:
4x2−7x−12
2x2+ 9x+ 5 ∇ · 2x2+ 5x−3
8x2−7x−12
Solution
Step 1: Re-write the division as multiplication by the reciprocal of the second
fraction: 4x2−7x−12
2x2+ 9x+ 5 ×8x2−7x−12
2x2+ 5x−3
Step 2: Factor all the quadratic expressions to simplify: Factor the numer-
ator of the first fraction:
4x2−7x−12
This expression can be factored as:
(4x+ 3)(x−4)
Factor the denominator of the first fraction:
2x2+ 9x+ 5
This expression can be factored as:
(2x+ 1)(x+ 5)
Factor the numerator of the second fraction:
8x2−7x−12
This expression can be factored as:
(8x−3)(x+ 4)
Factor the denominator of the second fraction:
2x2+ 5x−3
This expression can be factored as:
(2x−1)(x+ 3)
Step 3: Substitute the factored expressions back into the original expression:
(4x+ 3)(x−4)
(2x+ 1)(x+ 5) ×(8x−3)(x+ 4)
(2x−1)(x+ 3)
10
Step 4: Simplify the expression: The factors in the numerator and denomi-
nator can now be cancelled out, resulting in the simplified expression:
4x+ 3
2x+ 1 ×8x−3
2x−1=(4x+ 3)(8x−3)
(2x+ 1)(2x−1)
Therefore, the simplified form of the given rational expression is (4x+3)(8x−3)
(2x+1)(2x−1) .
Question 14
Question
Calculate the domain of the rational function:
f(x) = 3x2−7x−6
x2−4x−5
Solution
Step 1: Find the excluded values by setting the denominator equal to zero and
solving for x. These values are not in the domain of the function.
x2−4x−5 = 0
(x−5)(x+ 1) = 0
x= 5 or x=−1
Step 2: The domain of the function is all real numbers except for the excluded
values. Therefore, the domain of the function is:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 15
Question
Simplify the following rational expression:
3x2−4
x2−x−6∇ · x2−3x−4
x2+ 2x−8
Solution
Step 1: Factor the numerators and denominators of both fractions. Step 2:
Rewrite the division of fractions as a multiplication by the reciprocal. Step 3:
Simplify the expression by multiplying the numerators and denominators of the
rational expression. Step 4: Factor the resulting numerator and denominator
11
if possible. Step 5: Simplify the expression by canceling out common factors if
they exist.
Step 1: Factor the numerators and denominators of both fractions. For the
first fraction: The numerator 3x2−4can be factored as (3x+ 2)(x−2). The
denominator x2−x−6can be factored as (x−3)(x+ 2).
For the second fraction: The numerator x2−3x−4can be factored as
(x−4)(x+ 1). The denominator x2+ 2x−8can be factored as (x−2)(x+ 4).
Step 2: Rewrite the division of fractions as a multiplication by the recipro-
cal. The expression becomes:
(3x+ 2)(x−2)
(x−3)(x+ 2) ×(x+ 4)(x−2)
(x−4)(x+ 1)
Step 3: Simplify the expression by multiplying the numerators and denom-
inators of the rational expression. The expression simplifies to:
(3x+ 2)(x−2)(x+ 4)(x−2)
(x−3)(x+ 2)(x−4)(x+ 1)
Step 4: Factor the resulting numerator and denominator if possible. The
expression becomes:
(3x+ 2)(x+ 4)(x−2)2
(x−3)(x−4)(x+ 1)(x+ 2)
Step 5: Simplify the expression by canceling out common factors if they
exist. There are no common factors to cancel, so the final simplified form of the
rational expression is:
(3x+ 2)(x+ 4)(x−2)2
(x−3)(x−4)(x+ 1)(x+ 2)
Question 16
Question
Consider the rational function given by f(x) = 2x3−5x2−3x
x2−2x−3.
Determine the vertical asymptotes, horizontal asymptotes, x-intercepts, y-
intercepts, and the domain of the function f(x).
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero
and solving for x.
x2−2x−3 = 0
(x−3)(x+ 1) = 0
This gives us vertical asymptotes at x= 3 and x=−1.
12
Step 2: Find the horizontal asymptote by comparing the degrees of the
numerator and the denominator. Since the degree of the numerator (3) is greater
than the degree of the denominator (2), there is no horizontal asymptote.
Step 3: Find the x-intercepts by setting the numerator equal to zero and
solving for x.
2x3−5x2−3x= 0
x(2x2−5x−3) = 0
x(2x+ 1)(x−3) = 0
This gives us x-intercepts at x= 0,x=−1
2, and x= 3.
Step 4: Find the y-intercept by evaluating f(0).
f(0) = 2(0)3−5(0)2−3(0)
02−2(0) −3= 0
Therefore, the y-intercept is at the point (0, 0).
Step 5: Determine the domain of the function. The function is defined for
all real numbers except where the denominator is equal to zero (the vertical
asymptotes). Therefore, the domain of f(x)is all real numbers except x= 3
and x=−1.
Question 17
Question
Find the domain of the rational function:
f(x) = 1
x2−4x+ 3
Solution
Step 1: The domain of a rational function is all real numbers except the values
that would make the denominator equal to zero. Thus, we need to find the
values of xthat make the denominator x2−4x+ 3 equal to zero.
Step 2: To find the values of xthat make the denominator zero, we solve
the quadratic equation:
x2−4x+ 3 = 0
Step 3: Factoring the quadratic equation, we get:
(x−1)(x−3) = 0
Step 4: Setting each factor to zero gives us:
x−1 = 0 or x−3 = 0
13
Step 5: Solving these equations gives us:
x= 1 or x= 3
Step 6: Therefore, the values x= 1 and x= 3 would make the denominator
of the function equal to zero. As a result, the domain of the function f(x)is all
real numbers except x= 1 and x= 3.
Step 7: Thus, the domain of the function f(x) = 1
x2−4x+ 3 is (−∞,1) ∪
(1,3) ∪(3,∞).
Question 18
Question
Solve the following rational inequality:
3
x−2
x−3≥1.
Solution
Step 1: Find a common denominator for the fractions on the left side of the
inequality.
3
x−2
x−3=3(x−3)
x(x−3) −2x
x(x−3) =3x−9−2x
x(x−3) =x−9
x(x−3)
Step 2: Rewrite the inequality with the common denominator.
x−9
x(x−3) ≥1.
Step 3: Solve the inequality for the numerator.
x−9≥x(x−3)
Step 4: Expand the right side and simplify the inequality.
x−9≥x2−3x
Step 5: Rearrange the inequality into a quadratic inequality.
0≥x2−4x+ 9
Step 6: Solve the quadratic inequality by finding the roots of the quadratic
equation.
x=−(−4) ±√(−4)2−4(1)(9)
2(1)
x=4±√16 −36
2=4±√−20
2=4±2i√5
2= 2 ±i√5.
Step 7: Since the inequality is less than or equal to zero, the solution set is
(−∞,2−i√5] ∪[2 + i√5,∞).
14
Question 19
Question
Simplify the following rational expression:
3x2−2x−5
x2−4x−5
Solution
Step 1: Factor both the numerator and the denominator of the rational expres-
sion. 3x2−2x−5
x2−4x−5=(3x+ 1)(x−5)
(x−5)(x+ 1)
Step 2: Simplify by canceling out the common factor in the numerator and
the denominator. (3x+ 1)(x−5)
(x−5)(x+ 1) =3x+ 1
x+ 1
Therefore, the simplified form of the rational expression is 3x+1
x+1 .
Question 20
Question
Find the domain of the rational function:
f(x) = 4x
x2−9.
Solution
Step 1: The domain of a rational function is all real numbers except the values of
xthat make the denominator equal to zero, since division by zero is undefined.
So, we need to find the values of xfor which x2−9= 0.
Step 2: We can rewrite x2−9= 0 as (x+ 3)(x−3) = 0 by factoring the
quadratic polynomial.
Step 3: This inequality is true except when either x+ 3 = 0 or x−3 = 0.
Therefore, we solve for xin both cases:
x+ 3 = 0 ⇒x=−3,
and
x−3 = 0 ⇒x= 3.
Step 4: So, the domain of the rational function f(x) = 4x
x2−9is all real
numbers except x=−3and x= 3. In interval notation, the domain is:
(−∞,−3) ∪(−3,3) ∪(3,∞).
15
Question 21
Question
Find the domain of the rational function:
f(x) = x2−4x+ 3
x2−5x+ 6
Solution
Step 1: Find the values of xthat make the denominator zero, as these values
would result in division by zero which is undefined.
x2−5x+ 6 = 0
Factorizing the quadratic equation gives:
(x−2)(x−3) = 0
So, the values of xthat make the denominator zero are x= 2 and x= 3.
Step 2: The domain of a rational function is all real numbers except those
that make the denominator zero. Thus, the domain of f(x)is all real numbers
except x= 2 and x= 3.
Therefore, the domain of the rational function f(x) = x2−4x+3
x2−5x+6 is (−∞,2) ∪(2,3) ∪(3,∞).
Question 22
Question
Find the domain of the rational function:
f(x) = 7x
x2−5x−6
Solution
Step 1: The domain of a rational function is all real numbers except where the
denominator is equal to zero. So, we need to find the values of xthat make the
denominator x2−5x−6equal to zero.
Step 2: To find these values, we will factor the quadratic equation x2−5x−6.
The factors will be in the form (x+m)(x+n)where mand nare the factors
of −6that add up to −5. The factors are −6and 1as −6 + 1 = −5. Therefore,
the factored form is (x−6)(x+ 1).
Step 3: Setting the factored form equal to zero gives us:
x−6 = 0 =⇒x= 6
x+ 1 = 0 =⇒x=−1
16
Step 4: The domain of the rational function f(x)is all real numbers except
x= 6 and x=−1since they make the denominator equal to zero.
Therefore, the domain of f(x)is (−∞,−1) ∪(−1,6) ∪(6,∞).
Question 23
Question
Solve the rational equation for x:
2
x+ 3 −3
x−2=5
x2+x−6
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation. In this case, 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) =5
x2+x−6
Step 3: Combine the fractions on the left side of the equation.
2x−4−3x−9
(x+ 3)(x−2) =5
x2+x−6
Step 4: Simplify the numerator.
−x−13
(x+ 3)(x−2) =5
x2+x−6
Step 5: Factor the denominators of both sides. The denominator on the
left side is already factored. The denominator on the right side factors as (x+
3)(x−2).
Step 6: Cross multiply to eliminate the denominators.
−x−13 = 5
Step 7: Solve for x.
x=−18
Step 8: Check the solution in the original equation. Substitute x=−18
back into the original equation to check for extraneous solutions.
2
−18 + 3 −3
−18 −2=5
(−18)2+ (−18) −6
2
−15 −3
−20 =5
324 −18 −6=2
−15 +3
−20 =5
300 =1
60
17
The solution x=−18 satisfies the original equation.
Question 24
Question
Simplify the following rational function:
f(x) = 3x2−2x−8
2x2+ 5x−3
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify the rational
function by canceling out any common factors.
Step 1: To factor the numerator and denominator of f(x): Factor the
numerator:
3x2−2x−8 = (3x+ 4)(x−2)
Factor the denominator:
2x2+ 5x−3 = (2x−1)(x+ 3)
Step 2: Now, we can rewrite f(x)with the factored numerator and denom-
inator:
f(x) = (3x+ 4)(x−2)
(2x−1)(x+ 3)
There are no common factors that can be canceled out, so the simplified
form of the rational function is:
f(x) = 3x+ 4
2x−1
Question 25
Question
Let f(x) = x2+4x
x+2 be a rational function. Determine the domain of f(x).
Solution
Step 1: The domain of a rational function is all real numbers except where the
denominator is equal to zero. In this case, the denominator x+ 2 is equal to
zero when x=−2.
Step 2: Therefore, the domain of f(x)is all real numbers except x=−2.
Thus, the domain of f(x)is (−∞,−2) ∪(−2,∞).
18
Question 26
Question
Find the domain of the rational function:
f(x) = 2x2−3x−2
x2−4x−5
Solution
Step 1: To find the domain of the function, we need to determine all the values of
xfor which the function is defined. The function is defined for all real numbers
except those that make the denominator zero, as division by zero is undefined.
Step 2: Find the values of xthat make the denominator zero by setting
x2−4x−5equal to zero and solving for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation, we can factor it or use the quadratic
formula. Let’s use the quadratic formula:
x=−(−4) ±√(−4)2−4(1)(−5)
2(1)
x=4±√16 + 20
2
x=4±√36
2
x=4±6
2
Step 4: This gives us two possible values for x:
x=4+6
2⇒x= 5
x=4−6
2⇒x=−1
Step 5: These two values, x= 5 and x=−1, make the denominator zero.
Therefore, the values -1 and 5 are not in the domain of the function.
Step 6: The domain of the function f(x)is all real numbers except for
x=−1and x= 5. Therefore, the domain is given by:
Domain of f(x) = {x∈R|x=−1,5}
19
Question 27
Question
Simplify the following rational expression:
5x2−3x−2
x2−4x−12∇ · x2+ 2x−8
3x2+ 2x−8
Solution
Step 1: First, rewrite the division as multiplication by the reciprocal of the
second fraction: 5x2−3x−2
x2−4x−12 ×3x2+ 2x−8
x2+ 2x−8
Step 2: Factor the numerators and denominators of each fraction:
(5x+ 1)(x−2)
(x−6)(x+ 2) ×(3x−2)(x+ 4)
(x+ 4)(x−2)
Step 3: Cancel out any common factors in the numerators and denominators:
5x+ 1
x−6×3x−2
1
Step 4: Multiply the remaining factors in the numerators and denominators:
(5x+ 1)(3x−2)
(x−6)(1)
Step 5: Expand the remaining factors:
15x2−4x−2
x−6
Therefore, the simplified form of the given rational expression is 15x2−4x−2
x−6.
Question 28
Question
Find the domain of the rational function:
f(x) = 2x2−5x−3
x2+ 4x−5
20
Solution
Step 1: Determine the excluded values that would make the denominator zero,
as division by zero is undefined. Step 2: Find the domain by excluding these
values from the set of all real numbers.
Step 1: Set the denominator equal to zero and solve for x:
x2+ 4x−5 = 0
Using the quadratic formula, we have:
x=−b±√b2−4ac
2a
Where a= 1,b= 4, and c=−5, we get:
x=−4±√42−4(1)(−5)
2(1)
x=−4±√16 + 20
2
x=−4±√36
2
x=−4±6
2
Therefore, the excluded values are x=−6and x= 1.
Step 2: The domain of a rational function is all real numbers except those
that would make the denominator zero. So, the domain of the function f(x)is:
(−∞,−6) ∪(−6,1) ∪(1,∞)
Question 29
Question
Find the domain of the rational function:
f(x) = 2x−1
x2−5x+ 6.
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the function is defined. In a rational function, the de-
nominator cannot be zero. So, we need to find the values of xthat make the
denominator zero.
21
Step 2: We set the denominator equal to zero and solve for x.
x2−5x+ 6 = 0
(x−2)(x−3) = 0
Step 3: Now, we find the values of xfor which the denominator is zero by
setting each factor equal to zero and solving for x.
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 4: Therefore, the domain of the function f(x) = 2x−1
x2−5x+ 6 is all real
numbers except x= 2 and x= 3.
Step 5: The domain of the function is (−∞,2) ∪(2,3) ∪(3,∞).
Question 30
Question
Simplify the rational function:
f(x) = 3x3−2x2+ 5x−1
x2−2x−3
Solution
Step 1: Factor the numerator and the denominator. Step 2: Express the function
in simplified form.
Step 1: Factor the numerator and the denominator. First, let’s factor the
numerator:
3x3−2x2+ 5x−1 = x2(3x−2) + 1(3x−2) = (x2+ 1)(3x−2)
Next, let’s factor the denominator:
x2−2x−3 = (x−3)(x+ 1)
Step 2: Express the function in simplified form. Now that we have factored
both the numerator and the denominator, we can simplify the expression:
f(x) = 3x3−2x2+ 5x−1
x2−2x−3=(x2+ 1)(3x−2)
(x−3)(x+ 1)
Therefore, the simplified form of the rational function f(x)is (x2+ 1)(3x−2)
(x−3)(x+ 1) .
22
Question 31
Question
Simplify the following rational expression:
4x2−36
x2−16
Solution
Step 1: Factor both the numerator and denominator:
4x2−36 = 4(x2−9) = 4(x+ 3)(x−3)
x2−16 = (x+ 4)(x−4)
Step 2: Rewrite the expression with factored terms:
4(x+ 3)(x−3)
(x+ 4)(x−4)
Step 3: Cancel out common factors:
4(x+ 3)(x−3)
(x+ 4)(x−4)
Step 4: Simplify the expression to its final form:
4
1= 4
Question 32
Question
Given the rational function f(x) = 2x2+3x−2
x2−4x+3 , find the vertical and horizontal
asymptotes, any holes, x-intercepts, and y-intercepts.
Solution
Step 1: Find the vertical asymptotes
To find the vertical asymptotes, we need to determine the values of xthat
make the denominator of the rational function equal to zero. Set the denomi-
nator x2−4x+ 3 equal to zero and solve for x:
x2−4x+ 3 = 0
(x−3)(x−1) = 0
23
So, the vertical asymptotes occur at x= 3 and x= 1.
Step 2: Find the horizontal asymptote
To find the horizontal asymptote, compare the degrees of the numerator
and denominator of the rational function. Since the degrees are the same, the
horizontal asymptote is given by the ratio of the leading coefficients:
Horizontal asymptote: y=2
1= 2
Step 3: Find any holes
To find any holes, factor the numerator 2x2+ 3x−2and the denominator
x2−4x+ 3:
2x2+ 3x−2 = (2x−1)(x+ 2)
x2−4x+ 3 = (x−3)(x−1)
We see that there is a hole at x= 1 where the factor of (x−1) cancels out.
Step 4: Find x-intercepts To find the x-intercepts, set f(x) = 0 and solve
for x:2x2+ 3x−2
x2−4x+ 3 = 0
Since the numerator can only be zero when x=1
2or x=−2, the x-intercepts
are (1
2,0)and (−2,0).
Step 5: Find y-intercept To find the y-intercept, evaluate f(0):
f(0) = 2(0)2+ 3(0) −2
(0)2−4(0) + 3 =−2
3
So, the y-intercept is (0,−2
3).
Question 33
Question
Find the domain of the rational function: 3x2−5x−2
x2−4x−5.
Solution
Step 1: To find the domain of a rational 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
Step 3: To solve the quadratic equation, we can factor it:
(x−5)(x+ 1) = 0
24
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 the rational function is all real numbers
except x= 5 and x=−1.
Step 7: In interval notation, the domain can be expressed as (−∞,−1) ∪
(−1,5) ∪(5,∞).
Question 34
Question
Simplify the following rational expression:
3x2−7x−6
x2−4x−21
Solution
Step 1: Factor the numerator and denominator:
3x2−7x−6 = (3x+ 1)(x−6)
x2−4x−21 = (x−7)(x+ 3)
Step 2: Rewrite the expression with the factored forms of the numerator and
denominator: (3x+ 1)(x−6)
(x−7)(x+ 3)
Step 3: Simplify the expression by canceling out common factors in the
numerator and denominator:
(3x+ 1)(x−6)
(x−7)(x+ 3) =3x+ 1
x+ 3
Therefore, the simplified form of the rational expression is 3x+1
x+3 .
Question 35
Question
Let f(x) = 3x2−5x−2
2x2+7x+3 . Find the domain of the function f(x).
25
Solution
Step 1: To find the domain of the rational function f(x), we need to identify
the values of xthat make the denominator equal to zero, since division by zero
is undefined. Step 2: Set the denominator equal to zero and solve for x.
2x2+ 7x+ 3 = 0
Step 3: Factor the quadratic equation or use the quadratic formula to solve for
x.
(2x+ 1)(x+ 3) = 0
This gives us x=−1
2and x=−3. Step 4: The domain of the function f(x)
will be all real numbers except those that make the denominator zero.
Domain: {x∈R|x=−1
2,−3}
Therefore, the domain of the function f(x)is (−∞,−3) ∪(−3,−1
2)∪(−1
2,∞).
26
