MATH 121 - COLLEGE ALGEBRA -
Rational Functions
Question Bank - Set 3
Liberty University
Question 1
Question
Given the rational function f(x) = 2x2−3x−2
x2−1, find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph of f(x).
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes of a rational function, we need to determine
where the denominator becomes zero, but the numerator does not. In this case,
the denominator x2−1becomes zero when x=±1. We check if the numerator
2x2−3x−2also becomes zero at x=±1. Evaluating the numerator at x= 1:
f(1) = 2(1)2−3(1) −2
(1)2−1=−3
0
Since the numerator is also zero at x= 1, there is a hole at x= 1. At x=−1,
the function is undefined, so x=−1is a vertical asymptote.
Step 2: Horizontal Asymptotes
To find the horizontal asymptotes of a rational function, we look at the de-
grees of the numerator and denominator. Since both the numerator and denom-
inator have the same degree, we compare the leading coefficients to determine
the horizontal asymptote. The horizontal asymptote is given by the ratio of the
leading coefficients:
y=2
1= 2
Therefore, the horizontal asymptote of the function is y= 2.
Step 3: Hole
As we found earlier, there is a hole in the graph at x= 1. To find the
y-coordinate of the hole, simplify the function at x= 1:
f(x) = 2x2−3x−2
x2−1=(2x+ 1)(x−2)
(x−1)(x+ 1)
At x= 1, the simplified function is:
f(1) = (2(1) + 1)(1 −2)
(1 −1)(1 + 1) =−3
0
Hence, the y-coordinate of the hole is −3, so the hole is at the point (1,−3).
In summary, the rational function f(x) = 2x2−3x−2
x2−1has a vertical asymptote
at x=−1, a horizontal asymptote at y= 2, and a hole at (1,−3).
Question 2
Question
Find the domain of the rational function:
f(x) = x2−4x+ 3
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 zero. Step 2: Factor the denominator x2−5x+ 6 = 0 to
find the values of xthat make the denominator zero. Step 3: x2−5x+ 6 =
(x−2)(x−3), so the values of xthat make the denominator zero are x= 2 and
x= 3. Step 4: Therefore, the domain of the function f(x) = x2−4x+3
x2−5x+6 is all real
numbers except x= 2 and x= 3. Step 5: The domain in interval notation is
(−∞,2) ∪(2,3) ∪(3,∞).
Question 3
Question
Let f(x) = x2+ 3x−10
x2−4x+ 3 . Find any vertical asymptotes, horizontal asymptotes,
x-intercepts, and y-intercepts of the rational function f(x).
Solution
Step 1: To find the vertical asymptotes of the rational function f(x), we need
to determine where the denominator x2−4x+ 3 is equal to zero. This occurs
at x= 1 and x= 3, so the vertical asymptotes are x= 1 and x= 3.
2
Step 2: To find the horizontal asymptote of the rational function f(x), we
compare the degrees of the numerator and denominator. Since both have the
same degree (2), we look at the leading coefficients. The horizontal asymptote
is the ratio of the leading coefficients, which is 1
1= 1. Therefore, the horizontal
asymptote is y= 1.
Step 3: To find the x-intercepts, we set the numerator equal to zero and
solve for x.
x2+ 3x−10 = 0
(x+ 5)(x−2) = 0
So, x=−5and x= 2. Therefore, the x-intercepts are (−5,0) and (2,0).
Step 4: To find the y-intercept, we substitute x= 0 into the function f(x).
f(0) = 02+ 3(0) −10
02−4(0) + 3 =−10
3=−10
3
Therefore, the y-intercept is (0,−10
3).
Question 4
Question
Determine the domain of the rational function:
f(x) = x2−4
x2−5x+ 6
Solution
Step 1: To find the domain of a rational function, we need to identify 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:
x2−5x+ 6 = 0
Step 3: Factor the quadratic equation:
(x−2)(x−3) = 0
Step 4: Find the roots of the quadratic equation:
x= 2 or x= 3
Step 5: Therefore, the values x= 2 and x= 3 make the denominator zero.
Step 6: As division by zero is undefined, the domain of the function f(x) =
x2−4
x2−5x+6 is all real numbers except x= 2 and x= 3, so the domain is:
(−∞,2) ∪(2,3) ∪(3,∞)
3
Question 5
Question
Simplify the following rational function:
6x2−5x−6
3x2+ 3x−18
Solution
Step 1: Factor both the numerator and denominator.
Step 2: Simplify the rational function by canceling out common factors, if pos-
sible.
Step 1: Factor both the numerator and denominator.
The numerator 6x2−5x−6can be factored as (2x+ 1)(3x−6).
The denominator 3x2+ 3x−18 can be factored as 3(x+ 3)(x−2).
Step 2: Simplify the rational function by canceling out common factors, if
possible.
Therefore, the rational function simplifies as follows:
6x2−5x−6
3x2+ 3x−18 =(2x+ 1)(3x−6)
3(x+ 3)(x−2) =2x+ 1
x+ 3
Question 6
Question
Find the domain of the rational function: f(x) = x+3
x2+x−6.
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the denominator is not equal to zero.
Step 2: First, let’s find the factors of the denominator x2+x−6. The factors
are (x+ 3)(x−2).
Step 3: Now, we know that the denominator cannot be equal to zero. There-
fore, we need to find when x2+x−6= 0.
Step 4: Setting the denominator not equal to zero, we have (x+3)(x−2) = 0.
Step 5: This implies that x+ 3 = 0 and x−2= 0.
Step 6: Solving x+ 3 = 0, we get x=−3.
Step 7: Solving x−2= 0, we get x= 2.
Step 8: Therefore, the domain of the rational function f(x) = x+3
x2+x−6is all
real numbers except x=−3and x= 2. In interval notation, the domain is
(−∞,−3) ∪(−3,2) ∪(2,∞).
4
Question 7
Question
Let f(x) = 4x2+ 7x+ 3
x2−x−6. Find the domain of f(x).
Solution
Step 1: We need to determine the values of xfor which f(x)is defined. f(x)
will be undefined if the denominator of the rational function is equal to 0. Step
2: Set the denominator equal to 0and solve for x:
x2−x−6 = 0
Step 3: Factor the quadratic equation:
(x−3)(x+ 2) = 0
Step 4: Set each factor to zero and solve for x:
x−3 = 0 =⇒x= 3
x+ 2 = 0 =⇒x=−2
Step 5: The values x= 3 and x=−2make the denominator of the function
equal to 0, which would make the function undefined. Therefore, xcannot be
equal to 3or −2. Step 6: The domain of f(x)is all real numbers except x= 3
and x=−2. Therefore, the domain of f(x)is:
(−∞,−2) ∪(−2,3) ∪(3,∞)
Question 8
Question
Let f(x) = 3x2−5x+ 2
x2−4. Determine the vertical asymptotes, horizontal asymp-
totes, and the x-intercepts of the rational function f(x).
Solution
Step 1: To find the vertical asymptotes of the rational function, set the denom-
inator equal to zero and solve for x:
x2−4 = 0
(x−2)(x+ 2) = 0
5
x= 2 or x=−2
Therefore, the vertical asymptotes are x= 2 and x=−2.
Step 2: To find the horizontal asymptotes of the rational function, compare
the degrees of the numerator and denominator. Since the degree of the numer-
ator is equal to the degree of the denominator, divide the leading coefficients to
find the horizontal asymptote:
lim
x→∞
3x2
x2= lim
x→∞ 3 = 3
Therefore, the horizontal asymptote is y= 3.
Step 3: To find the x-intercepts of the rational function, set the numerator
equal to zero and solve for x:
3x2−5x+ 2 = 0
(3x−2)(x−1) = 0
x=2
3or x= 1
Therefore, the x-intercepts are (2
3,0)and (1,0).
Question 9
Question
Find the vertical asymptotes of the rational function:
f(x) = 3x2−5x−2
x2−4x−5
Solution
Step 1: First, factor the numerator and denominator to simplify the expression:
f(x) = 3x2−5x−2
x2−4x−5=(3x+ 1)(x−2)
(x−5)(x+ 1)
Step 2: Next, identify the values of xthat make the denominator equal to
zero, as these are potential vertical asymptotes. Set the denominator equal to
zero and solve for x:
x2−4x−5 = 0
(x−5)(x+ 1) = 0
x= 5 or x=−1
Step 3: Therefore, the vertical asymptotes of the rational function f(x)are
x= 5 and x=−1.
6
Question 10
Question
Given the rational function f(x) = 2x3−7x2−2x+8
x2−5x−6, determine the vertical asymp-
totes (if any) of the function.
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero.
x2−5x−6 = 0
(x−6)(x+ 1) = 0
So, the values x= 6 and x=−1make the denominator zero. These are the
potential vertical asymptotes.
Step 2: Check if any of these potential vertical asymptotes are canceled out
by factors in the numerator.
f(x) = 2x3−7x2−2x+ 8
x2−5x−6
=(x−4)(2x+ 1)(x−2)
(x−6)(x+ 1)
=(x−4)(2x+ 1)(x−2)
(x−6)(x+ 1)
Step 3: From the factored form of f(x), we can see that canceling terms
does not eliminate any of the vertical asymptotes x= 6 and x=−1.
Therefore, the vertical asymptotes of the function f(x) = 2x3−7x2−2x+8
x2−5x−6are
x= 6 and x=−1.
Question 11
Question
Find the domain of the rational function
f(x) = 4x2−9
2x2+ 7x−4
.
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xfor which the function is defined. The function is defined everywhere
except where the denominator is equal to zero.
7
Step 2: Set the denominator equal to zero to find where the function is
undefined.
2x2+ 7x−4 = 0
Step 3: Factor the quadratic expression.
2x2+ 7x−4 = (2x−1)(x+ 4)
Step 4: Set each factor equal to zero and solve for x.
2x−1 = 0 =⇒x=1
2
x+ 4 = 0 =⇒x=−4
Step 5: Therefore, the domain of the rational function f(x)is all real num-
bers except x=−4, x =1
2.
Domain = (−∞,−4) ∪(−4,1
2)∪(1
2,∞)
Question 12
Question
Simplify the following rational expression:
5x2−3x−2
2x2−5x−3
Solution
Step 1: Factor the numerator and the denominator.
5x2−3x−2 = (5x+ 1)(x−2)
2x2−5x−3 = (2x+ 1)(x−3)
Step 2: Rewrite the expression with the factored forms.
(5x+ 1)(x−2)
(2x+ 1)(x−3)
Step 3: Simplify by cancelling common factors.
5x+ 1
2x+ 1 ·x−2
x−3
Therefore, the simplified form of the given rational expression is:
5x+ 1
2x+ 1 ·x−2
x−3
8
Question 13
Question
Find the domain of the rational function:
f(x) = x2+ 5x+ 6
x2−x−6
Solution
To find the domain of the rational function, we need to determine all values of
xthat make the denominator x2−x−6non-zero.
x2−x−6= 0
(x−3)(x+ 2) = 0
From this, we can see that the function is undefined when x= 3 and x=−2
(since these values would make the denominator zero), so the domain is all real
numbers except x= 3 and x=−2.
Therefore, the domain of the rational function f(x) = x2+5x+6
x2−x−6is (−∞,−2) ∪(−2,3) ∪(3,∞).
Question 14
Question
Find the domain of the rational function: 3x2−4x−1
x2−9.
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the denominator is not equal to zero, because division by
zero is undefined. In this case, the denominator is x2−9. Set the denominator
equal to zero and solve for x:
x2−9 = 0
x2= 9
x=±3
So, the denominator is equal to zero when x= 3 or x=−3.
Step 2: The domain of the rational function is all real numbers except the
values of xthat make the denominator zero. So, the domain is:
Domain: {x∈R|x= 3, x =−3}
Domain: {x∈R|x= 3 and x=−3}
9
Question 15
Question
Find the domain of the rational function f(x) = x2−5x+6
x2−4.
Solution
Step 1: The domain of a rational function is all real numbers except those values
of xthat make the denominator equal to zero, since division by zero is undefined.
So, to find the domain of f(x), we need to solve the equation x2−4 = 0.
Step 2: Factor the denominator by recognizing it as a difference of squares.
We have (x+ 2)(x−2) = 0.
Step 3: Set each factor equal to zero and solve for x. We get x+ 2 = 0 or
x−2 = 0, which gives x=−2or x= 2.
Step 4: The values x=−2and x= 2 are the values that make the denom-
inator equal to zero. Therefore, the domain of the function f(x) = x2−5x+6
x2−4is
all real numbers except x=−2and x= 2.
Step 5: So, the domain of the function is (−∞,−2) ∪(−2,2) ∪(2,∞).
Question 16
Question
Simplify the rational function:
4x3−6x2+ 3x
2x2−4x
Solution
Step 1: Factor out common terms in the numerator and denominator. Step 2:
Simplify the rational function by canceling out common factors. Step 3: Write
the simplified rational function.
Step 1: Factor out common terms in the numerator and denominator.
4x3−6x2+ 3x
2x2−4x=x(4x2−6x+ 3)
2x(x−2)
Step 2: Simplify the rational function by canceling out common factors.
x(4x2−6x+ 3)
2x(x−2) =x(4x2−6x+ 3)
2x(x−2) =4x2−6x+ 3
2(x−2)
Step 3: Write the simplified rational function.
4x2−6x+ 3
2(x−2)
Therefore, the simplified form of the given rational function is 4x2−6x+3
2(x−2) .
10
Question 17
Question
Express the rational function in partial fractions:
4x3+ 10x2−3x+ 2
x4+ 3x3+ 3x2
Solution
Step 1: Factor the denominator. Step 2: Write the given rational function as a
sum of partial fractions. Step 3: Equate coefficients and solve for the unknown
constants. Step 4: Combine the partial fractions to simplify the expression.
Step 1: Factor the denominator x4+ 3x3+ 3x2.
x4+ 3x3+ 3x2=x2(x2+ 3x+ 3)
Step 2: Write the given rational function as a sum of partial fractions.
4x3+ 10x2−3x+ 2
x2(x2+ 3x+ 3) =A
x+B
x2+Cx +D
x2+ 3x+ 3
Step 3: Equate coefficients and solve for the unknown constants. Multiply-
ing by the common denominator x2(x2+ 3x+ 3), we get:
4x3+ 10x2−3x+ 2 = A(x2+ 3x+ 3) + Bx(x2+ 3x+3)+(Cx +D)x2
4x3+ 10x2−3x+ 2 = (A+B)x3+ (3A+C)x2+ (3A+D)x+ 3A
Equating coefficients, we have the system of equations:
A+B= 4
3A+C= 10
3A+D=−3
3A= 2
Solving this system, we find A=2
3,B=10
3,C= 4, and D=−11.
Step 4: Combine the partial fractions. Substitute the values of A,B,C,
and Dback into the partial fractions:
4x3+ 10x2−3x+ 2
x2(x2+ 3x+ 3) =2
3x+10
3x2+4x−11
x2+ 3x+ 3
Therefore, the given rational function can be expressed as the sum of partial
fractions.
11
Question 18
Question
Solve the rational inequality: x+ 5
x−2>1.
Solution
To solve the rational inequality x+ 5
x−2>1, we will first find the critical points
where the expression equals 1 and then test the intervals between these points.
Step 1: Find the critical points
Set the inequality equal to 1 and solve for x:
x+ 5
x−2= 1
x+ 5 = x−2
0 = −7
The equation has no solution, which means there are no critical points to con-
sider.
Step 2: Test intervals
Choose a test point from each interval separated by the critical points. We will
test points from the intervals (−∞,2),(2,∞).
Test x= 0:0+5
0−2=5
−2<1
Test x= 3:3+5
3−2=8
1>1
Step 3: Determine the solution
Since the inequality is true for x > 2, the solution to the inequality x+ 5
x−2>1
is x∈(2,∞).
Question 19
Question
Simplify the following rational function:
f(x) = 4x2+ 6x−5
2x2−3x−2
12
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify by canceling
common factors.
Step 1: To factor the numerator and denominator, we need to find two
numbers that multiply to the constant term and add up to the coefficient of the
linear term.
For the numerator 4x2+ 6x−5: The product is 4×−5 = −20 and the sum
is 4+(−5) = −1. So, we can factor it as (4x−5)(x+ 1).
For the denominator 2x2−3x−2: The product is 2×−2 = −4and the sum
is 2+(−4) = −2. So, we can factor it as (2x+ 1)(x−2).
Therefore, the rational function can be rewritten as:
f(x) = (4x−5)(x+ 1)
(2x+ 1)(x−2)
Step 2: Now, we simplify by canceling common factors:
f(x) = 4x−5
2x+ 1 ·x+ 1
x−2
So, the simplified form of the rational function is:
f(x) = 4x−5
2x+ 1 ·x+ 1
x−2
Question 20
Question
Determine the domain of the following rational function:
f(x) = 3x+ 1
x2−4x−5
Solution
Step 1: To find the domain of the rational function, we need to identify the
values of xthat would make the denominator equal to zero, since division by
zero is undefined.
Step 2: Set the denominator x2−4x−5equal to zero and solve for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation x2−4x−5=0, we can use the
quadratic formula:
x=−b±√b2−4ac
2a
where a= 1,b=−4, and c=−5.
13
Step 4: Plug in a= 1,b=−4, and c=−5into 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 5: This gives us two possible values for x:
x=4+6
2=10
2= 5
or
x=4−6
2=−2
2=−1
Step 6: Therefore, the domain of the function f(x)is all real numbers ex-
cept x= 5 and x=−1. In interval notation, the domain can be written as
(−∞,−1) ∪(−1,5) ∪(5,∞).
Question 21
Question
Let f(x) = 2x2−5x−3
x2−x−6. Find the following:
1. Domain of 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 f(x):
The domain of a rational function is all real numbers except the values
that make the denominator equal to zero. So, we find the values of xsuch
that x2−x−6 = 0.
Factoring the quadratic, we get (x−3)(x+ 2) = 0, so x= 3 or x=−2.
Therefore, the domain of f(x)is all real numbers except x= 3 and x=−2,
or in interval notation, (−∞,−2) ∪(−2,3) ∪(3,∞).
14
2. x-intercepts:
To find the x-intercept(s), we set f(x) = 0 and solve for x:2x2−5x−3
x2−x−6=
0.
Simplifying further, we get 2x2−5x−3 = 0. Factoring the quadratic
gives (2x+ 1)(x−3) = 0, so x=−1
2or x= 3. Thus, the x-intercepts are
(−1
2,0) and (3,0).
3. y-intercept:
To find the y-intercept, we set x= 0 in f(x):f(0) = 2(0)2−5(0) −3
(0)2−(0) −6=
−3
−6=1
2.
Therefore, the y-intercept is (0,1
2).
4. Vertical asymptotes:
Vertical asymptotes occur where the denominator of a rational function is
zero. Thus, the vertical asymptotes are at x=−2and x= 3.
5. Horizontal asymptotes:
To find the horizontal asymptote(s), we look at the degrees of the numer-
ator and denominator. Since they have the same degree (2), we compare
the leading coefficients of the polynomial terms.
The horizontal asymptote is the ratio of the leading coefficients, which is
2
1= 2. Therefore, the horizontal asymptote of this function is y= 2.
Question 22
Question
Find the domain of the rational function:
f(x) = x+ 3
x2−4x−5
Solution
Step 1: We first need to find the values of xthat make the denominator zero,
since division by zero is undefined. Set the denominator equal to zero and solve
for x:
x2−4x−5 = 0
Step 2: This is a quadratic equation, so we can factor it or use the quadratic
formula. Factoring, we get:
(x−5)(x+ 1) = 0
15
Step 3: Set each factor to zero and solve for x:
x−5 = 0 or x+ 1 = 0
x= 5 or x=−1
Step 4: Therefore, the values x= 5 and x=−1make the denominator zero.
These values are not in the domain of the function. The domain of the function
f(x)is all real numbers except x= 5 and x=−1.
So, the domain of the function is:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 23
Question
Simplify the rational function:
3x2−6x
x2−x−6∇ · 2x2−6x
2x2+ 3x−2
Solution
Step 1: Simplify the complex fraction by multiplying the numerator by the
reciprocal of the denominator. Step 2: Factorize the numerators and denomina-
tors where possible. Step 3: Simplify the rational function by canceling factors
in the numerator and denominator. Step 4: Write the simplified form of the
rational function.
Step 1:
3x2−6x
x2−x−6∇ · 2x2−6x
2x2+ 3x−2=3x2−6x
x2−x−6·2x2+ 3x−2
2x2−6x
Step 2: Factorize where possible. For the first fraction: 3x2−6x= 3x(x−2)
x2−x−6 = (x−3)(x+ 2)
For the second fraction: 2x2+ 3x−2 = (2x−1)(x+ 2) 2x2−6x= 2x(x−3)
Step 3: Now, simplify the rational function:
3x(x−2) ·(2x−1)(x+ 2)
(x−3)(x+ 2) ·2x(x−3)
Cancel out common factors in the numerator and denominator:
3(2x−1)
2x
Step 4: The simplified form of the rational function is:
6x−3
2x
16
Question 24
Question
Solve the following rational equation for x:3x
x2−4−1
x−2=2
x+2 .
Solution
Step 1: First, let’s get rid of the fractions in the equation by finding a common
denominator. The common denominator in this case is (x2−4)(x−2)(x+ 2).
Multiply each term of the equation by this common denominator.
Step 2: After multiplying by the common denominator, the equation be-
comes: 3x(x+ 2) −(x2−4)(x+ 2) = 2(x2−4). Simplify this equation.
Step 3: Expand and simplify each term in the equation: 3x2+ 6x−x3−
2x2−4x+ 8 = 2x2−8. This simplifies to −x3+x2+ 2x+ 8 = 2x2−8.
Step 4: Rearrange the equation by moving all terms to one side to set it
equal to zero. This gives us −x3−x2−4x+ 16 = 0.
Step 5: Factor out a negative sign to make the equation easier to factor:
x3+x2+ 4x−16 = 0.
Step 6: Now, we need to factor the polynomial on the left side of the equa-
tion. By inspection, we find that x= 2 is a root of the polynomial. Use synthetic
division or polynomial division to factor the polynomial completely.
Step 7: After factoring, we get (x−2)(x2+ 3x+ 8) = 0. Setting each factor
equal to zero gives us solutions x= 2,x=−3
2+√7i
2, and x=−3
2−√7i
2.
Therefore, the solutions to the rational equation are x= 2,x=−3
2+√7i
2,
and x=−3
2−√7i
2.
Question 25
Question
Simplify the rational function:
R(x) = x2−5x+ 6
x2−4
Solution
To simplify R(x) = x2−5x+6
x2−4, we first factor the numerator and denominator.
Step 1: Factor the numerator: x2−5x+ 6 = (x−3)(x−2).
Step 2: Factor the denominator: x2−4 = (x+ 2)(x−2).
So, the rational function R(x)becomes:
R(x) = (x−3)(x−2)
(x+ 2)(x−2)
17
Step 3: Simplify by canceling out common factors in the numerator and
denominator.
Since there is a common factor of (x−2) in both the numerator and denom-
inator, we can simplify further:
R(x) = x−3
x+ 2
Therefore, the simplified form of the rational function R(x)is x−3
x+2 .
Question 26
Question
Simplify the rational expression:
2x2−7x−30
3x2+ 2x−8.
Solution
Step 1: Factor both the numerator and the denominator.
For the numerator 2x2−7x−30, we need to find two numbers that multiply
to 2×−30 = −60 and add up to −7. These numbers are −12 and 5, so we can
rewrite the numerator as 2x2−12x+ 5x−30 and factor by grouping:
(2x2−12x) + (5x−30) = 2x(x−6) + 5(x−6) = (2x+ 5)(x−6).
For the denominator 3x2+2x−8, we need to find two numbers that multiply
to 3× −8 = −24 and add up to 2. These numbers are −4and 6, so we can
rewrite the denominator as 3x2−4x+ 6x−8and factor by grouping:
(3x2−4x) + (6x−8) = x(3x−4) + 2(3x−4) = (x+ 2)(3x−4).
Therefore, the rational expression simplifies to:
2x2−7x−30
3x2+ 2x−8=(2x+ 5)(x−6)
(x+ 2)(3x−4).
So the simplified form of the rational expression is 2x+ 5
x+ 2 ·x−6
3x−4.
Question 27
Question
Let f(x) = 2x2+x−3
x2−4x+3 . Find the domain of f(x).
18
Solution
Step 1: The function f(x)is defined for all xvalues except those that make the
denominator equal to zero. Thus, we need to find the values of xthat satisfy
x2−4x+ 3 = 0.
Step 2: Factor the quadratic expression x2−4x+ 3 = 0 to find the values
of x.
(x−1)(x−3) = 0
Step 3: Set each factor equal to zero and solve for x.
x−1 = 0 =⇒x= 1
x−3 = 0 =⇒x= 3
Step 4: The solutions to the equation x2−4x+ 3 = 0 are x= 1 and x= 3.
These values make the denominator of f(x)equal to zero, so they are not in the
domain of f(x).
Step 5: Therefore, the domain of the function f(x)is all real numbers except
x= 1 and x= 3.
Step 6: In interval notation, the domain of f(x)can be expressed as (−∞,1)∪
(1,3) ∪(3,∞).
Question 28
Question
Simplify the following rational expression:
3x2−2x−5
x2+ 4x+ 3
Solution
Step 1: Factor both the numerator and the denominator.
Step 2: Factor the quadratic expressions.
Step 3: Simplify the expression by canceling out common factors.
Step 1: Factor the numerator: 3x2−2x−5We need to find two numbers
that multiply to 3·−5 = −15 and add up to −2. These numbers are −5and 3.
So, we can write 3x2−2x−5as (3x+ 3)(x−5).
Factor the denominator: x2+ 4x+ 3 We need to find two numbers that
multiply to 1·3 = 3 and add up to 4. These numbers are 3and 1. So, we can
write x2+ 4x+ 3 as (x+ 3)(x+ 1).
Step 2: Our expression becomes (3x+3)(x−5)
(x+3)(x+1) .
Step 3: Now, we can cancel out the common factor of 3 in the numerator
and the denominator. This simplifies the expression to x−5
x+1 .
Therefore, the simplified form of the given rational expression is x−5
x+ 1 .
19
Question 29
Question
Simplify the following rational function:
f(x) = 4x2−12x−16
x2−4x−5
Solution
Step 1: Factor the numerator and denominator of the rational function f(x):
f(x) = 4x2−12x−16
x2−4x−5=4(x2−3x−4)
(x−5)(x+ 1)
Step 2: Further simplify the factored numerator:
f(x) = 4(x2−3x−4)
(x−5)(x+ 1) =4(x−4)(x+ 1)
(x−5)(x+ 1)
Step 3: Cancel out common factors in the numerator and denominator:
f(x) = 4(x−4)(x+ 1)
(x−5)(x+ 1) =4(x−4)
x−5
Therefore, the simplified form of the rational function f(x)is:
f(x) = 4(x−4)
x−5
Question 30
Question
Simplify the following rational expression:
3x2−7x−20
2x2−5x−3
Solution
To simplify the rational expression, we can factor both the numerator and de-
nominator and then cancel out any common factors.
Step 1: Factor the numerator and denominator:
3x2−7x−20 = (3x+ 5)(x−4)
2x2−5x−3 = (2x+ 1)(x−3)
20
Step 2: Rewrite the expression with the factored forms:
(3x+ 5)(x−4)
(2x+ 1)(x−3)
Step 3: Simplify the expression by canceling out common factors:
(3x+ 5)(x−4)
(2x+ 1)(x−3) =3x+ 5
2x+ 1 ·x−4
x−3
So, the simplified form of the given rational expression is 3x+ 5
2x+ 1 ·x−4
x−3.
Question 31
Question
Simplify the rational function:
f(x) = x2−4
x2+ 3x−10
Solution
Step 1: Factor the numerator and the denominator. Step 2: Simplify by can-
celing out common factors.
Step 1: We factor the numerator and denominator:
f(x) = x2−4
x2+ 3x−10 =(x+ 2)(x−2)
(x+ 5)(x−2)
Step 2: Now, we simplify by canceling out the common factor (x−2) in
the numerator and denominator:
f(x) = (x+ 2)(x−2)
(x+ 5)(x−2) =x+ 2
x+ 5
Therefore, the simplified form of the rational function is f(x) = x+ 2
x+ 5.
Question 32
Question
Simplify the rational function:
4x3+ 12x2−7x−21
2x2+ 5x−3
21
Solution
Step 1: Factor both the numerator and denominator.
4x3+ 12x2−7x−21 = 4x2(x+ 3) −7(x+ 3) = (4x2−7)(x+ 3)
2x2+ 5x−3 = (2x−1)(x+ 3)
Step 2: Rewrite the original rational function with the factored forms.
4x2−7
2x−1
Thus, the simplified form of the given rational function is 4x2−7
2x−1.
Question 33
Question
Find the domain of the rational function: f(x) = x2−4
x2−5x+ 6.
Solution
Step 1: To find the domain of a rational function, we need to identify all the
values of xfor which the function is defined. The function will not be defined
where the denominator is equal to zero (since division by zero is undefined).
Step 2: Set the denominator equal to zero and solve for x:
x2−5x+ 6 = 0
Step 3: Factor the quadratic:
(x−2)(x−3) = 0
Step 4: Set each factor to zero:
x−2 = 0 or x−3 = 0
Step 5: Solve for xin each case:
x= 2 or x= 3
Step 6: Therefore, the rational function is undefined at x= 2 and x= 3.
Step 7: The domain of the rational function f(x)is all real numbers except
x= 2 and x= 3. It can be written in interval notation as:
(−∞,2) ∪(2,3) ∪(3,∞)
22
Question 34
Question
Given the rational function f(x) = 4x2−3x−4
x2+1 , find: (a) the domain of f(x),
(b) the x- and y-intercepts of the graph of f(x), (c) the vertical and horizontal
asymptotes (if any) of the graph of f(x).
Solution
(a) To find the domain of f(x), we need to identify any values of xthat make
the denominator of the rational function equal to zero, since division by zero is
undefined. In this case, the denominator x2+ 1 is never equal to zero for any
real number x. Hence, the domain of f(x)is all real numbers: (−∞,∞).
(b) To find the x-intercepts of the graph of f(x), we set f(x) = 0 and solve
for x:4x2−3x−4
x2+ 1 = 0
4x2−3x−4 = 0
This quadratic equation does not factor easily, so we can use the quadratic
formula:
x=−(−3) ±√(−3)2−4(4)(−4)
2(4)
x=3±√9 + 64
8
x=3±√73
8
Therefore, the x-intercepts are (3+√73
8,0)and (3−√73
8,0).
To find the y-intercept, we evaluate f(0):
f(0) = 4(0)2−3(0) −4
(0)2+ 1 =−4
1=−4
Hence, the y-intercept is (0,−4).
(c) To find the vertical asymptotes of the graph of f(x), we look for values
of xthat make the denominator of f(x)equal to zero. In this case, x2+ 1 = 0
has no real solutions, so there are no vertical asymptotes.
To find the horizontal asymptote, we compare the degrees of the numerator
and denominator of f(x). Since the degree of the numerator is equal to the
degree of the denominator, we divide the leading coefficients:
lim
x→∞
4x2−3x−4
x2+ 1 = lim
x→∞
4
1= 4
Therefore, the horizontal asymptote is y= 4.
23
Question 35
Question
Solve the following rational inequality for x:
3x+ 2
x−5≤4.
Solution
To solve the rational inequality, we first find the critical points by setting the
numerator equal to zero. Then we determine the sign of the rational function
in the intervals created by the critical points.
Step 1: Find the critical points Setting the numerator equal to zero
gives:
3x+ 2 = 0.
Solving for x, we get:
x=−2
3.
The critical point is x=−2
3.
Step 2: Determine the sign of the rational function Choose test
points in each interval to determine the sign of the rational function.
• Test point x=−1:
3(−1) + 2
(−1) −5=−1
−6=1
6<4.
This means the function is negative in this interval.
• Test point x= 0:
3(0) + 2
(0) −5=2
−5=−2
5<4.
This means the function is negative in this interval.
• Test point x=−2
3:
3(−2
3)+ 2
(−2
3)−5=−6+2
−2
3−5=−4
−17
3
=12
17 <4.
This means the function is negative in this interval.
• Test point x= 1:
3(1) + 2
(1) −5=5
−4=−5
4<4.
This means the function is negative in this interval.
24
Question 5
Question
Simplify the following rational function:
6x2−5x−6
3x2+ 3x−18
Solution
Step 1: Factor both the numerator and denominator.
Step 2: Simplify the rational function by canceling out common factors, if pos-
sible.
Step 1: Factor both the numerator and denominator.
The numerator 6x2−5x−6can be factored as (2x+ 1)(3x−6).
The denominator 3x2+ 3x−18 can be factored as 3(x+ 3)(x−2).
Step 2: Simplify the rational function by canceling out common factors, if
possible.
Therefore, the rational function simplifies as follows:
6x2−5x−6
3x2+ 3x−18 =(2x+ 1)(3x−6)
3(x+ 3)(x−2) =2x+ 1
x+ 3
Question 6
Question
Find the domain of the rational function: f(x) = x+3
x2+x−6.
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the denominator is not equal to zero.
Step 2: First, let’s find the factors of the denominator x2+x−6. The factors
are (x+ 3)(x−2).
Step 3: Now, we know that the denominator cannot be equal to zero. There-
fore, we need to find when x2+x−6= 0.
Step 4: Setting the denominator not equal to zero, we have (x+3)(x−2) = 0.
Step 5: This implies that x+ 3 = 0 and x−2= 0.
Step 6: Solving x+ 3 = 0, we get x=−3.
Step 7: Solving x−2= 0, we get x= 2.
Step 8: Therefore, the domain of the rational function f(x) = x+3
x2+x−6is all
real numbers except x=−3and x= 2. In interval notation, the domain is
(−∞,−3) ∪(−3,2) ∪(2,∞).
4
Question 7
Question
Let f(x) = 4x2+ 7x+ 3
x2−x−6. Find the domain of f(x).
Solution
Step 1: We need to determine the values of xfor which f(x)is defined. f(x)
will be undefined if the denominator of the rational function is equal to 0. Step
2: Set the denominator equal to 0and solve for x:
x2−x−6 = 0
Step 3: Factor the quadratic equation:
(x−3)(x+ 2) = 0
Step 4: Set each factor to zero and solve for x:
x−3 = 0 =⇒x= 3
x+ 2 = 0 =⇒x=−2
Step 5: The values x= 3 and x=−2make the denominator of the function
equal to 0, which would make the function undefined. Therefore, xcannot be
equal to 3or −2. Step 6: The domain of f(x)is all real numbers except x= 3
and x=−2. Therefore, the domain of f(x)is:
(−∞,−2) ∪(−2,3) ∪(3,∞)
Question 8
Question
Let f(x) = 3x2−5x+ 2
x2−4. Determine the vertical asymptotes, horizontal asymp-
totes, and the x-intercepts of the rational function f(x).
Solution
Step 1: To find the vertical asymptotes of the rational function, set the denom-
inator equal to zero and solve for x:
x2−4 = 0
(x−2)(x+ 2) = 0
5
x= 2 or x=−2
Therefore, the vertical asymptotes are x= 2 and x=−2.
Step 2: To find the horizontal asymptotes of the rational function, compare
the degrees of the numerator and denominator. Since the degree of the numer-
ator is equal to the degree of the denominator, divide the leading coefficients to
find the horizontal asymptote:
lim
x→∞
3x2
x2= lim
x→∞ 3 = 3
Therefore, the horizontal asymptote is y= 3.
Step 3: To find the x-intercepts of the rational function, set the numerator
equal to zero and solve for x:
3x2−5x+ 2 = 0
(3x−2)(x−1) = 0
x=2
3or x= 1
Therefore, the x-intercepts are (2
3,0)and (1,0).
Question 9
Question
Find the vertical asymptotes of the rational function:
f(x) = 3x2−5x−2
x2−4x−5
Solution
Step 1: First, factor the numerator and denominator to simplify the expression:
f(x) = 3x2−5x−2
x2−4x−5=(3x+ 1)(x−2)
(x−5)(x+ 1)
Step 2: Next, identify the values of xthat make the denominator equal to
zero, as these are potential vertical asymptotes. Set the denominator equal to
zero and solve for x:
x2−4x−5 = 0
(x−5)(x+ 1) = 0
x= 5 or x=−1
Step 3: Therefore, the vertical asymptotes of the rational function f(x)are
x= 5 and x=−1.
6
Question 10
Question
Given the rational function f(x) = 2x3−7x2−2x+8
x2−5x−6, determine the vertical asymp-
totes (if any) of the function.
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero.
x2−5x−6 = 0
(x−6)(x+ 1) = 0
So, the values x= 6 and x=−1make the denominator zero. These are the
potential vertical asymptotes.
Step 2: Check if any of these potential vertical asymptotes are canceled out
by factors in the numerator.
f(x) = 2x3−7x2−2x+ 8
x2−5x−6
=(x−4)(2x+ 1)(x−2)
(x−6)(x+ 1)
=(x−4)(2x+ 1)(x−2)
(x−6)(x+ 1)
Step 3: From the factored form of f(x), we can see that canceling terms
does not eliminate any of the vertical asymptotes x= 6 and x=−1.
Therefore, the vertical asymptotes of the function f(x) = 2x3−7x2−2x+8
x2−5x−6are
x= 6 and x=−1.
Question 11
Question
Find the domain of the rational function
f(x) = 4x2−9
2x2+ 7x−4
.
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xfor which the function is defined. The function is defined everywhere
except where the denominator is equal to zero.
7
Step 2: Set the denominator equal to zero to find where the function is
undefined.
2x2+ 7x−4 = 0
Step 3: Factor the quadratic expression.
2x2+ 7x−4 = (2x−1)(x+ 4)
Step 4: Set each factor equal to zero and solve for x.
2x−1 = 0 =⇒x=1
2
x+ 4 = 0 =⇒x=−4
Step 5: Therefore, the domain of the rational function f(x)is all real num-
bers except x=−4, x =1
2.
Domain = (−∞,−4) ∪(−4,1
2)∪(1
2,∞)
Question 12
Question
Simplify the following rational expression:
5x2−3x−2
2x2−5x−3
Solution
Step 1: Factor the numerator and the denominator.
5x2−3x−2 = (5x+ 1)(x−2)
2x2−5x−3 = (2x+ 1)(x−3)
Step 2: Rewrite the expression with the factored forms.
(5x+ 1)(x−2)
(2x+ 1)(x−3)
Step 3: Simplify by cancelling common factors.
5x+ 1
2x+ 1 ·x−2
x−3
Therefore, the simplified form of the given rational expression is:
5x+ 1
2x+ 1 ·x−2
x−3
8
Question 13
Question
Find the domain of the rational function:
f(x) = x2+ 5x+ 6
x2−x−6
Solution
To find the domain of the rational function, we need to determine all values of
xthat make the denominator x2−x−6non-zero.
x2−x−6= 0
(x−3)(x+ 2) = 0
From this, we can see that the function is undefined when x= 3 and x=−2
(since these values would make the denominator zero), so the domain is all real
numbers except x= 3 and x=−2.
Therefore, the domain of the rational function f(x) = x2+5x+6
x2−x−6is (−∞,−2) ∪(−2,3) ∪(3,∞).
Question 14
Question
Find the domain of the rational function: 3x2−4x−1
x2−9.
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the denominator is not equal to zero, because division by
zero is undefined. In this case, the denominator is x2−9. Set the denominator
equal to zero and solve for x:
x2−9 = 0
x2= 9
x=±3
So, the denominator is equal to zero when x= 3 or x=−3.
Step 2: The domain of the rational function is all real numbers except the
values of xthat make the denominator zero. So, the domain is:
Domain: {x∈R|x= 3, x =−3}
Domain: {x∈R|x= 3 and x=−3}
9
Question 15
Question
Find the domain of the rational function f(x) = x2−5x+6
x2−4.
Solution
Step 1: The domain of a rational function is all real numbers except those values
of xthat make the denominator equal to zero, since division by zero is undefined.
So, to find the domain of f(x), we need to solve the equation x2−4 = 0.
Step 2: Factor the denominator by recognizing it as a difference of squares.
We have (x+ 2)(x−2) = 0.
Step 3: Set each factor equal to zero and solve for x. We get x+ 2 = 0 or
x−2 = 0, which gives x=−2or x= 2.
Step 4: The values x=−2and x= 2 are the values that make the denom-
inator equal to zero. Therefore, the domain of the function f(x) = x2−5x+6
x2−4is
all real numbers except x=−2and x= 2.
Step 5: So, the domain of the function is (−∞,−2) ∪(−2,2) ∪(2,∞).
Question 16
Question
Simplify the rational function:
4x3−6x2+ 3x
2x2−4x
Solution
Step 1: Factor out common terms in the numerator and denominator. Step 2:
Simplify the rational function by canceling out common factors. Step 3: Write
the simplified rational function.
Step 1: Factor out common terms in the numerator and denominator.
4x3−6x2+ 3x
2x2−4x=x(4x2−6x+ 3)
2x(x−2)
Step 2: Simplify the rational function by canceling out common factors.
x(4x2−6x+ 3)
2x(x−2) =x(4x2−6x+ 3)
2x(x−2) =4x2−6x+ 3
2(x−2)
Step 3: Write the simplified rational function.
4x2−6x+ 3
2(x−2)
Therefore, the simplified form of the given rational function is 4x2−6x+3
2(x−2) .
10
Question 17
Question
Express the rational function in partial fractions:
4x3+ 10x2−3x+ 2
x4+ 3x3+ 3x2
Solution
Step 1: Factor the denominator. Step 2: Write the given rational function as a
sum of partial fractions. Step 3: Equate coefficients and solve for the unknown
constants. Step 4: Combine the partial fractions to simplify the expression.
Step 1: Factor the denominator x4+ 3x3+ 3x2.
x4+ 3x3+ 3x2=x2(x2+ 3x+ 3)
Step 2: Write the given rational function as a sum of partial fractions.
4x3+ 10x2−3x+ 2
x2(x2+ 3x+ 3) =A
x+B
x2+Cx +D
x2+ 3x+ 3
Step 3: Equate coefficients and solve for the unknown constants. Multiply-
ing by the common denominator x2(x2+ 3x+ 3), we get:
4x3+ 10x2−3x+ 2 = A(x2+ 3x+ 3) + Bx(x2+ 3x+3)+(Cx +D)x2
4x3+ 10x2−3x+ 2 = (A+B)x3+ (3A+C)x2+ (3A+D)x+ 3A
Equating coefficients, we have the system of equations:
A+B= 4
3A+C= 10
3A+D=−3
3A= 2
Solving this system, we find A=2
3,B=10
3,C= 4, and D=−11.
Step 4: Combine the partial fractions. Substitute the values of A,B,C,
and Dback into the partial fractions:
4x3+ 10x2−3x+ 2
x2(x2+ 3x+ 3) =2
3x+10
3x2+4x−11
x2+ 3x+ 3
Therefore, the given rational function can be expressed as the sum of partial
fractions.
11
Question 18
Question
Solve the rational inequality: x+ 5
x−2>1.
Solution
To solve the rational inequality x+ 5
x−2>1, we will first find the critical points
where the expression equals 1 and then test the intervals between these points.
Step 1: Find the critical points
Set the inequality equal to 1 and solve for x:
x+ 5
x−2= 1
x+ 5 = x−2
0 = −7
The equation has no solution, which means there are no critical points to con-
sider.
Step 2: Test intervals
Choose a test point from each interval separated by the critical points. We will
test points from the intervals (−∞,2),(2,∞).
Test x= 0:0+5
0−2=5
−2<1
Test x= 3:3+5
3−2=8
1>1
Step 3: Determine the solution
Since the inequality is true for x > 2, the solution to the inequality x+ 5
x−2>1
is x∈(2,∞).
Question 19
Question
Simplify the following rational function:
f(x) = 4x2+ 6x−5
2x2−3x−2
12
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify by canceling
common factors.
Step 1: To factor the numerator and denominator, we need to find two
numbers that multiply to the constant term and add up to the coefficient of the
linear term.
For the numerator 4x2+ 6x−5: The product is 4×−5 = −20 and the sum
is 4+(−5) = −1. So, we can factor it as (4x−5)(x+ 1).
For the denominator 2x2−3x−2: The product is 2×−2 = −4and the sum
is 2+(−4) = −2. So, we can factor it as (2x+ 1)(x−2).
Therefore, the rational function can be rewritten as:
f(x) = (4x−5)(x+ 1)
(2x+ 1)(x−2)
Step 2: Now, we simplify by canceling common factors:
f(x) = 4x−5
2x+ 1 ·x+ 1
x−2
So, the simplified form of the rational function is:
f(x) = 4x−5
2x+ 1 ·x+ 1
x−2
Question 20
Question
Determine the domain of the following rational function:
f(x) = 3x+ 1
x2−4x−5
Solution
Step 1: To find the domain of the rational function, we need to identify the
values of xthat would make the denominator equal to zero, since division by
zero is undefined.
Step 2: Set the denominator x2−4x−5equal to zero and solve for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation x2−4x−5=0, we can use the
quadratic formula:
x=−b±√b2−4ac
2a
where a= 1,b=−4, and c=−5.
13
Step 4: Plug in a= 1,b=−4, and c=−5into 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 5: This gives us two possible values for x:
x=4+6
2=10
2= 5
or
x=4−6
2=−2
2=−1
Step 6: Therefore, the domain of the function f(x)is all real numbers ex-
cept x= 5 and x=−1. In interval notation, the domain can be written as
(−∞,−1) ∪(−1,5) ∪(5,∞).
Question 21
Question
Let f(x) = 2x2−5x−3
x2−x−6. Find the following:
1. Domain of 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 f(x):
The domain of a rational function is all real numbers except the values
that make the denominator equal to zero. So, we find the values of xsuch
that x2−x−6 = 0.
Factoring the quadratic, we get (x−3)(x+ 2) = 0, so x= 3 or x=−2.
Therefore, the domain of f(x)is all real numbers except x= 3 and x=−2,
or in interval notation, (−∞,−2) ∪(−2,3) ∪(3,∞).
14
2. x-intercepts:
To find the x-intercept(s), we set f(x) = 0 and solve for x:2x2−5x−3
x2−x−6=
0.
Simplifying further, we get 2x2−5x−3 = 0. Factoring the quadratic
gives (2x+ 1)(x−3) = 0, so x=−1
2or x= 3. Thus, the x-intercepts are
(−1
2,0) and (3,0).
3. y-intercept:
To find the y-intercept, we set x= 0 in f(x):f(0) = 2(0)2−5(0) −3
(0)2−(0) −6=
−3
−6=1
2.
Therefore, the y-intercept is (0,1
2).
4. Vertical asymptotes:
Vertical asymptotes occur where the denominator of a rational function is
zero. Thus, the vertical asymptotes are at x=−2and x= 3.
5. Horizontal asymptotes:
To find the horizontal asymptote(s), we look at the degrees of the numer-
ator and denominator. Since they have the same degree (2), we compare
the leading coefficients of the polynomial terms.
The horizontal asymptote is the ratio of the leading coefficients, which is
2
1= 2. Therefore, the horizontal asymptote of this function is y= 2.
Question 22
Question
Find the domain of the rational function:
f(x) = x+ 3
x2−4x−5
Solution
Step 1: We first need to find the values of xthat make the denominator zero,
since division by zero is undefined. Set the denominator equal to zero and solve
for x:
x2−4x−5 = 0
Step 2: This is a quadratic equation, so we can factor it or use the quadratic
formula. Factoring, we get:
(x−5)(x+ 1) = 0
15
Step 3: Set each factor to zero and solve for x:
x−5 = 0 or x+ 1 = 0
x= 5 or x=−1
Step 4: Therefore, the values x= 5 and x=−1make the denominator zero.
These values are not in the domain of the function. The domain of the function
f(x)is all real numbers except x= 5 and x=−1.
So, the domain of the function is:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 23
Question
Simplify the rational function:
3x2−6x
x2−x−6∇ · 2x2−6x
2x2+ 3x−2
Solution
Step 1: Simplify the complex fraction by multiplying the numerator by the
reciprocal of the denominator. Step 2: Factorize the numerators and denomina-
tors where possible. Step 3: Simplify the rational function by canceling factors
in the numerator and denominator. Step 4: Write the simplified form of the
rational function.
Step 1:
3x2−6x
x2−x−6∇ · 2x2−6x
2x2+ 3x−2=3x2−6x
x2−x−6·2x2+ 3x−2
2x2−6x
Step 2: Factorize where possible. For the first fraction: 3x2−6x= 3x(x−2)
x2−x−6 = (x−3)(x+ 2)
For the second fraction: 2x2+ 3x−2 = (2x−1)(x+ 2) 2x2−6x= 2x(x−3)
Step 3: Now, simplify the rational function:
3x(x−2) ·(2x−1)(x+ 2)
(x−3)(x+ 2) ·2x(x−3)
Cancel out common factors in the numerator and denominator:
3(2x−1)
2x
Step 4: The simplified form of the rational function is:
6x−3
2x
16
Question 24
Question
Solve the following rational equation for x:3x
x2−4−1
x−2=2
x+2 .
Solution
Step 1: First, let’s get rid of the fractions in the equation by finding a common
denominator. The common denominator in this case is (x2−4)(x−2)(x+ 2).
Multiply each term of the equation by this common denominator.
Step 2: After multiplying by the common denominator, the equation be-
comes: 3x(x+ 2) −(x2−4)(x+ 2) = 2(x2−4). Simplify this equation.
Step 3: Expand and simplify each term in the equation: 3x2+ 6x−x3−
2x2−4x+ 8 = 2x2−8. This simplifies to −x3+x2+ 2x+ 8 = 2x2−8.
Step 4: Rearrange the equation by moving all terms to one side to set it
equal to zero. This gives us −x3−x2−4x+ 16 = 0.
Step 5: Factor out a negative sign to make the equation easier to factor:
x3+x2+ 4x−16 = 0.
Step 6: Now, we need to factor the polynomial on the left side of the equa-
tion. By inspection, we find that x= 2 is a root of the polynomial. Use synthetic
division or polynomial division to factor the polynomial completely.
Step 7: After factoring, we get (x−2)(x2+ 3x+ 8) = 0. Setting each factor
equal to zero gives us solutions x= 2,x=−3
2+√7i
2, and x=−3
2−√7i
2.
Therefore, the solutions to the rational equation are x= 2,x=−3
2+√7i
2,
and x=−3
2−√7i
2.
Question 25
Question
Simplify the rational function:
R(x) = x2−5x+ 6
x2−4
Solution
To simplify R(x) = x2−5x+6
x2−4, we first factor the numerator and denominator.
Step 1: Factor the numerator: x2−5x+ 6 = (x−3)(x−2).
Step 2: Factor the denominator: x2−4 = (x+ 2)(x−2).
So, the rational function R(x)becomes:
R(x) = (x−3)(x−2)
(x+ 2)(x−2)
17
Step 3: Simplify by canceling out common factors in the numerator and
denominator.
Since there is a common factor of (x−2) in both the numerator and denom-
inator, we can simplify further:
R(x) = x−3
x+ 2
Therefore, the simplified form of the rational function R(x)is x−3
x+2 .
Question 26
Question
Simplify the rational expression:
2x2−7x−30
3x2+ 2x−8.
Solution
Step 1: Factor both the numerator and the denominator.
For the numerator 2x2−7x−30, we need to find two numbers that multiply
to 2×−30 = −60 and add up to −7. These numbers are −12 and 5, so we can
rewrite the numerator as 2x2−12x+ 5x−30 and factor by grouping:
(2x2−12x) + (5x−30) = 2x(x−6) + 5(x−6) = (2x+ 5)(x−6).
For the denominator 3x2+2x−8, we need to find two numbers that multiply
to 3× −8 = −24 and add up to 2. These numbers are −4and 6, so we can
rewrite the denominator as 3x2−4x+ 6x−8and factor by grouping:
(3x2−4x) + (6x−8) = x(3x−4) + 2(3x−4) = (x+ 2)(3x−4).
Therefore, the rational expression simplifies to:
2x2−7x−30
3x2+ 2x−8=(2x+ 5)(x−6)
(x+ 2)(3x−4).
So the simplified form of the rational expression is 2x+ 5
x+ 2 ·x−6
3x−4.
Question 27
Question
Let f(x) = 2x2+x−3
x2−4x+3 . Find the domain of f(x).
18
Solution
Step 1: The function f(x)is defined for all xvalues except those that make the
denominator equal to zero. Thus, we need to find the values of xthat satisfy
x2−4x+ 3 = 0.
Step 2: Factor the quadratic expression x2−4x+ 3 = 0 to find the values
of x.
(x−1)(x−3) = 0
Step 3: Set each factor equal to zero and solve for x.
x−1 = 0 =⇒x= 1
x−3 = 0 =⇒x= 3
Step 4: The solutions to the equation x2−4x+ 3 = 0 are x= 1 and x= 3.
These values make the denominator of f(x)equal to zero, so they are not in the
domain of f(x).
Step 5: Therefore, the domain of the function f(x)is all real numbers except
x= 1 and x= 3.
Step 6: In interval notation, the domain of f(x)can be expressed as (−∞,1)∪
(1,3) ∪(3,∞).
Question 28
Question
Simplify the following rational expression:
3x2−2x−5
x2+ 4x+ 3
Solution
Step 1: Factor both the numerator and the denominator.
Step 2: Factor the quadratic expressions.
Step 3: Simplify the expression by canceling out common factors.
Step 1: Factor the numerator: 3x2−2x−5We need to find two numbers
that multiply to 3·−5 = −15 and add up to −2. These numbers are −5and 3.
So, we can write 3x2−2x−5as (3x+ 3)(x−5).
Factor the denominator: x2+ 4x+ 3 We need to find two numbers that
multiply to 1·3 = 3 and add up to 4. These numbers are 3and 1. So, we can
write x2+ 4x+ 3 as (x+ 3)(x+ 1).
Step 2: Our expression becomes (3x+3)(x−5)
(x+3)(x+1) .
Step 3: Now, we can cancel out the common factor of 3 in the numerator
and the denominator. This simplifies the expression to x−5
x+1 .
Therefore, the simplified form of the given rational expression is x−5
x+ 1 .
19
Question 29
Question
Simplify the following rational function:
f(x) = 4x2−12x−16
x2−4x−5
Solution
Step 1: Factor the numerator and denominator of the rational function f(x):
f(x) = 4x2−12x−16
x2−4x−5=4(x2−3x−4)
(x−5)(x+ 1)
Step 2: Further simplify the factored numerator:
f(x) = 4(x2−3x−4)
(x−5)(x+ 1) =4(x−4)(x+ 1)
(x−5)(x+ 1)
Step 3: Cancel out common factors in the numerator and denominator:
f(x) = 4(x−4)(x+ 1)
(x−5)(x+ 1) =4(x−4)
x−5
Therefore, the simplified form of the rational function f(x)is:
f(x) = 4(x−4)
x−5
Question 30
Question
Simplify the following rational expression:
3x2−7x−20
2x2−5x−3
Solution
To simplify the rational expression, we can factor both the numerator and de-
nominator and then cancel out any common factors.
Step 1: Factor the numerator and denominator:
3x2−7x−20 = (3x+ 5)(x−4)
2x2−5x−3 = (2x+ 1)(x−3)
20
Step 2: Rewrite the expression with the factored forms:
(3x+ 5)(x−4)
(2x+ 1)(x−3)
Step 3: Simplify the expression by canceling out common factors:
(3x+ 5)(x−4)
(2x+ 1)(x−3) =3x+ 5
2x+ 1 ·x−4
x−3
So, the simplified form of the given rational expression is 3x+ 5
2x+ 1 ·x−4
x−3.
Question 31
Question
Simplify the rational function:
f(x) = x2−4
x2+ 3x−10
Solution
Step 1: Factor the numerator and the denominator. Step 2: Simplify by can-
celing out common factors.
Step 1: We factor the numerator and denominator:
f(x) = x2−4
x2+ 3x−10 =(x+ 2)(x−2)
(x+ 5)(x−2)
Step 2: Now, we simplify by canceling out the common factor (x−2) in
the numerator and denominator:
f(x) = (x+ 2)(x−2)
(x+ 5)(x−2) =x+ 2
x+ 5
Therefore, the simplified form of the rational function is f(x) = x+ 2
x+ 5.
Question 32
Question
Simplify the rational function:
4x3+ 12x2−7x−21
2x2+ 5x−3
21
Solution
Step 1: Factor both the numerator and denominator.
4x3+ 12x2−7x−21 = 4x2(x+ 3) −7(x+ 3) = (4x2−7)(x+ 3)
2x2+ 5x−3 = (2x−1)(x+ 3)
Step 2: Rewrite the original rational function with the factored forms.
4x2−7
2x−1
Thus, the simplified form of the given rational function is 4x2−7
2x−1.
Question 33
Question
Find the domain of the rational function: f(x) = x2−4
x2−5x+ 6.
Solution
Step 1: To find the domain of a rational function, we need to identify all the
values of xfor which the function is defined. The function will not be defined
where the denominator is equal to zero (since division by zero is undefined).
Step 2: Set the denominator equal to zero and solve for x:
x2−5x+ 6 = 0
Step 3: Factor the quadratic:
(x−2)(x−3) = 0
Step 4: Set each factor to zero:
x−2 = 0 or x−3 = 0
Step 5: Solve for xin each case:
x= 2 or x= 3
Step 6: Therefore, the rational function is undefined at x= 2 and x= 3.
Step 7: The domain of the rational function f(x)is all real numbers except
x= 2 and x= 3. It can be written in interval notation as:
(−∞,2) ∪(2,3) ∪(3,∞)
22
Question 34
Question
Given the rational function f(x) = 4x2−3x−4
x2+1 , find: (a) the domain of f(x),
(b) the x- and y-intercepts of the graph of f(x), (c) the vertical and horizontal
asymptotes (if any) of the graph of f(x).
Solution
(a) To find the domain of f(x), we need to identify any values of xthat make
the denominator of the rational function equal to zero, since division by zero is
undefined. In this case, the denominator x2+ 1 is never equal to zero for any
real number x. Hence, the domain of f(x)is all real numbers: (−∞,∞).
(b) To find the x-intercepts of the graph of f(x), we set f(x) = 0 and solve
for x:4x2−3x−4
x2+ 1 = 0
4x2−3x−4 = 0
This quadratic equation does not factor easily, so we can use the quadratic
formula:
x=−(−3) ±√(−3)2−4(4)(−4)
2(4)
x=3±√9 + 64
8
x=3±√73
8
Therefore, the x-intercepts are (3+√73
8,0)and (3−√73
8,0).
To find the y-intercept, we evaluate f(0):
f(0) = 4(0)2−3(0) −4
(0)2+ 1 =−4
1=−4
Hence, the y-intercept is (0,−4).
(c) To find the vertical asymptotes of the graph of f(x), we look for values
of xthat make the denominator of f(x)equal to zero. In this case, x2+ 1 = 0
has no real solutions, so there are no vertical asymptotes.
To find the horizontal asymptote, we compare the degrees of the numerator
and denominator of f(x). Since the degree of the numerator is equal to the
degree of the denominator, we divide the leading coefficients:
lim
x→∞
4x2−3x−4
x2+ 1 = lim
x→∞
4
1= 4
Therefore, the horizontal asymptote is y= 4.
23
Question 35
Question
Solve the following rational inequality for x:
3x+ 2
x−5≤4.
Solution
To solve the rational inequality, we first find the critical points by setting the
numerator equal to zero. Then we determine the sign of the rational function
in the intervals created by the critical points.
Step 1: Find the critical points Setting the numerator equal to zero
gives:
3x+ 2 = 0.
Solving for x, we get:
x=−2
3.
The critical point is x=−2
3.
Step 2: Determine the sign of the rational function Choose test
points in each interval to determine the sign of the rational function.
• Test point x=−1:
3(−1) + 2
(−1) −5=−1
−6=1
6<4.
This means the function is negative in this interval.
• Test point x= 0:
3(0) + 2
(0) −5=2
−5=−2
5<4.
This means the function is negative in this interval.
• Test point x=−2
3:
3(−2
3)+ 2
(−2
3)−5=−6+2
−2
3−5=−4
−17
3
=12
17 <4.
This means the function is negative in this interval.
• Test point x= 1:
3(1) + 2
(1) −5=5
−4=−5
4<4.
This means the function is negative in this interval.
24
Question 5
Question
Simplify the following rational function:
6x2−5x−6
3x2+ 3x−18
Solution
Step 1: Factor both the numerator and denominator.
Step 2: Simplify the rational function by canceling out common factors, if pos-
sible.
Step 1: Factor both the numerator and denominator.
The numerator 6x2−5x−6can be factored as (2x+ 1)(3x−6).
The denominator 3x2+ 3x−18 can be factored as 3(x+ 3)(x−2).
Step 2: Simplify the rational function by canceling out common factors, if
possible.
Therefore, the rational function simplifies as follows:
6x2−5x−6
3x2+ 3x−18 =(2x+ 1)(3x−6)
3(x+ 3)(x−2) =2x+ 1
x+ 3
Question 6
Question
Find the domain of the rational function: f(x) = x+3
x2+x−6.
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the denominator is not equal to zero.
Step 2: First, let’s find the factors of the denominator x2+x−6. The factors
are (x+ 3)(x−2).
Step 3: Now, we know that the denominator cannot be equal to zero. There-
fore, we need to find when x2+x−6= 0.
Step 4: Setting the denominator not equal to zero, we have (x+3)(x−2) = 0.
Step 5: This implies that x+ 3 = 0 and x−2= 0.
Step 6: Solving x+ 3 = 0, we get x=−3.
Step 7: Solving x−2= 0, we get x= 2.
Step 8: Therefore, the domain of the rational function f(x) = x+3
x2+x−6is all
real numbers except x=−3and x= 2. In interval notation, the domain is
(−∞,−3) ∪(−3,2) ∪(2,∞).
4
Question 7
Question
Let f(x) = 4x2+ 7x+ 3
x2−x−6. Find the domain of f(x).
Solution
Step 1: We need to determine the values of xfor which f(x)is defined. f(x)
will be undefined if the denominator of the rational function is equal to 0. Step
2: Set the denominator equal to 0and solve for x:
x2−x−6 = 0
Step 3: Factor the quadratic equation:
(x−3)(x+ 2) = 0
Step 4: Set each factor to zero and solve for x:
x−3 = 0 =⇒x= 3
x+ 2 = 0 =⇒x=−2
Step 5: The values x= 3 and x=−2make the denominator of the function
equal to 0, which would make the function undefined. Therefore, xcannot be
equal to 3or −2. Step 6: The domain of f(x)is all real numbers except x= 3
and x=−2. Therefore, the domain of f(x)is:
(−∞,−2) ∪(−2,3) ∪(3,∞)
Question 8
Question
Let f(x) = 3x2−5x+ 2
x2−4. Determine the vertical asymptotes, horizontal asymp-
totes, and the x-intercepts of the rational function f(x).
Solution
Step 1: To find the vertical asymptotes of the rational function, set the denom-
inator equal to zero and solve for x:
x2−4 = 0
(x−2)(x+ 2) = 0
5
x= 2 or x=−2
Therefore, the vertical asymptotes are x= 2 and x=−2.
Step 2: To find the horizontal asymptotes of the rational function, compare
the degrees of the numerator and denominator. Since the degree of the numer-
ator is equal to the degree of the denominator, divide the leading coefficients to
find the horizontal asymptote:
lim
x→∞
3x2
x2= lim
x→∞ 3 = 3
Therefore, the horizontal asymptote is y= 3.
Step 3: To find the x-intercepts of the rational function, set the numerator
equal to zero and solve for x:
3x2−5x+ 2 = 0
(3x−2)(x−1) = 0
x=2
3or x= 1
Therefore, the x-intercepts are (2
3,0)and (1,0).
Question 9
Question
Find the vertical asymptotes of the rational function:
f(x) = 3x2−5x−2
x2−4x−5
Solution
Step 1: First, factor the numerator and denominator to simplify the expression:
f(x) = 3x2−5x−2
x2−4x−5=(3x+ 1)(x−2)
(x−5)(x+ 1)
Step 2: Next, identify the values of xthat make the denominator equal to
zero, as these are potential vertical asymptotes. Set the denominator equal to
zero and solve for x:
x2−4x−5 = 0
(x−5)(x+ 1) = 0
x= 5 or x=−1
Step 3: Therefore, the vertical asymptotes of the rational function f(x)are
x= 5 and x=−1.
6
Question 10
Question
Given the rational function f(x) = 2x3−7x2−2x+8
x2−5x−6, determine the vertical asymp-
totes (if any) of the function.
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero.
x2−5x−6 = 0
(x−6)(x+ 1) = 0
So, the values x= 6 and x=−1make the denominator zero. These are the
potential vertical asymptotes.
Step 2: Check if any of these potential vertical asymptotes are canceled out
by factors in the numerator.
f(x) = 2x3−7x2−2x+ 8
x2−5x−6
=(x−4)(2x+ 1)(x−2)
(x−6)(x+ 1)
=(x−4)(2x+ 1)(x−2)
(x−6)(x+ 1)
Step 3: From the factored form of f(x), we can see that canceling terms
does not eliminate any of the vertical asymptotes x= 6 and x=−1.
Therefore, the vertical asymptotes of the function f(x) = 2x3−7x2−2x+8
x2−5x−6are
x= 6 and x=−1.
Question 11
Question
Find the domain of the rational function
f(x) = 4x2−9
2x2+ 7x−4
.
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xfor which the function is defined. The function is defined everywhere
except where the denominator is equal to zero.
7
Step 2: Set the denominator equal to zero to find where the function is
undefined.
2x2+ 7x−4 = 0
Step 3: Factor the quadratic expression.
2x2+ 7x−4 = (2x−1)(x+ 4)
Step 4: Set each factor equal to zero and solve for x.
2x−1 = 0 =⇒x=1
2
x+ 4 = 0 =⇒x=−4
Step 5: Therefore, the domain of the rational function f(x)is all real num-
bers except x=−4, x =1
2.
Domain = (−∞,−4) ∪(−4,1
2)∪(1
2,∞)
Question 12
Question
Simplify the following rational expression:
5x2−3x−2
2x2−5x−3
Solution
Step 1: Factor the numerator and the denominator.
5x2−3x−2 = (5x+ 1)(x−2)
2x2−5x−3 = (2x+ 1)(x−3)
Step 2: Rewrite the expression with the factored forms.
(5x+ 1)(x−2)
(2x+ 1)(x−3)
Step 3: Simplify by cancelling common factors.
5x+ 1
2x+ 1 ·x−2
x−3
Therefore, the simplified form of the given rational expression is:
5x+ 1
2x+ 1 ·x−2
x−3
8
Question 13
Question
Find the domain of the rational function:
f(x) = x2+ 5x+ 6
x2−x−6
Solution
To find the domain of the rational function, we need to determine all values of
xthat make the denominator x2−x−6non-zero.
x2−x−6= 0
(x−3)(x+ 2) = 0
From this, we can see that the function is undefined when x= 3 and x=−2
(since these values would make the denominator zero), so the domain is all real
numbers except x= 3 and x=−2.
Therefore, the domain of the rational function f(x) = x2+5x+6
x2−x−6is (−∞,−2) ∪(−2,3) ∪(3,∞).
Question 14
Question
Find the domain of the rational function: 3x2−4x−1
x2−9.
Solution
Step 1: To find the domain of a rational function, we need to determine the
values of xfor which the denominator is not equal to zero, because division by
zero is undefined. In this case, the denominator is x2−9. Set the denominator
equal to zero and solve for x:
x2−9 = 0
x2= 9
x=±3
So, the denominator is equal to zero when x= 3 or x=−3.
Step 2: The domain of the rational function is all real numbers except the
values of xthat make the denominator zero. So, the domain is:
Domain: {x∈R|x= 3, x =−3}
Domain: {x∈R|x= 3 and x=−3}
9
Question 15
Question
Find the domain of the rational function f(x) = x2−5x+6
x2−4.
Solution
Step 1: The domain of a rational function is all real numbers except those values
of xthat make the denominator equal to zero, since division by zero is undefined.
So, to find the domain of f(x), we need to solve the equation x2−4 = 0.
Step 2: Factor the denominator by recognizing it as a difference of squares.
We have (x+ 2)(x−2) = 0.
Step 3: Set each factor equal to zero and solve for x. We get x+ 2 = 0 or
x−2 = 0, which gives x=−2or x= 2.
Step 4: The values x=−2and x= 2 are the values that make the denom-
inator equal to zero. Therefore, the domain of the function f(x) = x2−5x+6
x2−4is
all real numbers except x=−2and x= 2.
Step 5: So, the domain of the function is (−∞,−2) ∪(−2,2) ∪(2,∞).
Question 16
Question
Simplify the rational function:
4x3−6x2+ 3x
2x2−4x
Solution
Step 1: Factor out common terms in the numerator and denominator. Step 2:
Simplify the rational function by canceling out common factors. Step 3: Write
the simplified rational function.
Step 1: Factor out common terms in the numerator and denominator.
4x3−6x2+ 3x
2x2−4x=x(4x2−6x+ 3)
2x(x−2)
Step 2: Simplify the rational function by canceling out common factors.
x(4x2−6x+ 3)
2x(x−2) =x(4x2−6x+ 3)
2x(x−2) =4x2−6x+ 3
2(x−2)
Step 3: Write the simplified rational function.
4x2−6x+ 3
2(x−2)
Therefore, the simplified form of the given rational function is 4x2−6x+3
2(x−2) .
10
Question 17
Question
Express the rational function in partial fractions:
4x3+ 10x2−3x+ 2
x4+ 3x3+ 3x2
Solution
Step 1: Factor the denominator. Step 2: Write the given rational function as a
sum of partial fractions. Step 3: Equate coefficients and solve for the unknown
constants. Step 4: Combine the partial fractions to simplify the expression.
Step 1: Factor the denominator x4+ 3x3+ 3x2.
x4+ 3x3+ 3x2=x2(x2+ 3x+ 3)
Step 2: Write the given rational function as a sum of partial fractions.
4x3+ 10x2−3x+ 2
x2(x2+ 3x+ 3) =A
x+B
x2+Cx +D
x2+ 3x+ 3
Step 3: Equate coefficients and solve for the unknown constants. Multiply-
ing by the common denominator x2(x2+ 3x+ 3), we get:
4x3+ 10x2−3x+ 2 = A(x2+ 3x+ 3) + Bx(x2+ 3x+3)+(Cx +D)x2
4x3+ 10x2−3x+ 2 = (A+B)x3+ (3A+C)x2+ (3A+D)x+ 3A
Equating coefficients, we have the system of equations:
A+B= 4
3A+C= 10
3A+D=−3
3A= 2
Solving this system, we find A=2
3,B=10
3,C= 4, and D=−11.
Step 4: Combine the partial fractions. Substitute the values of A,B,C,
and Dback into the partial fractions:
4x3+ 10x2−3x+ 2
x2(x2+ 3x+ 3) =2
3x+10
3x2+4x−11
x2+ 3x+ 3
Therefore, the given rational function can be expressed as the sum of partial
fractions.
11
Question 18
Question
Solve the rational inequality: x+ 5
x−2>1.
Solution
To solve the rational inequality x+ 5
x−2>1, we will first find the critical points
where the expression equals 1 and then test the intervals between these points.
Step 1: Find the critical points
Set the inequality equal to 1 and solve for x:
x+ 5
x−2= 1
x+ 5 = x−2
0 = −7
The equation has no solution, which means there are no critical points to con-
sider.
Step 2: Test intervals
Choose a test point from each interval separated by the critical points. We will
test points from the intervals (−∞,2),(2,∞).
Test x= 0:0+5
0−2=5
−2<1
Test x= 3:3+5
3−2=8
1>1
Step 3: Determine the solution
Since the inequality is true for x > 2, the solution to the inequality x+ 5
x−2>1
is x∈(2,∞).
Question 19
Question
Simplify the following rational function:
f(x) = 4x2+ 6x−5
2x2−3x−2
12
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify by canceling
common factors.
Step 1: To factor the numerator and denominator, we need to find two
numbers that multiply to the constant term and add up to the coefficient of the
linear term.
For the numerator 4x2+ 6x−5: The product is 4×−5 = −20 and the sum
is 4+(−5) = −1. So, we can factor it as (4x−5)(x+ 1).
For the denominator 2x2−3x−2: The product is 2×−2 = −4and the sum
is 2+(−4) = −2. So, we can factor it as (2x+ 1)(x−2).
Therefore, the rational function can be rewritten as:
f(x) = (4x−5)(x+ 1)
(2x+ 1)(x−2)
Step 2: Now, we simplify by canceling common factors:
f(x) = 4x−5
2x+ 1 ·x+ 1
x−2
So, the simplified form of the rational function is:
f(x) = 4x−5
2x+ 1 ·x+ 1
x−2
Question 20
Question
Determine the domain of the following rational function:
f(x) = 3x+ 1
x2−4x−5
Solution
Step 1: To find the domain of the rational function, we need to identify the
values of xthat would make the denominator equal to zero, since division by
zero is undefined.
Step 2: Set the denominator x2−4x−5equal to zero and solve for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation x2−4x−5=0, we can use the
quadratic formula:
x=−b±√b2−4ac
2a
where a= 1,b=−4, and c=−5.
13
Step 4: Plug in a= 1,b=−4, and c=−5into 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 5: This gives us two possible values for x:
x=4+6
2=10
2= 5
or
x=4−6
2=−2
2=−1
Step 6: Therefore, the domain of the function f(x)is all real numbers ex-
cept x= 5 and x=−1. In interval notation, the domain can be written as
(−∞,−1) ∪(−1,5) ∪(5,∞).
Question 21
Question
Let f(x) = 2x2−5x−3
x2−x−6. Find the following:
1. Domain of 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 f(x):
The domain of a rational function is all real numbers except the values
that make the denominator equal to zero. So, we find the values of xsuch
that x2−x−6 = 0.
Factoring the quadratic, we get (x−3)(x+ 2) = 0, so x= 3 or x=−2.
Therefore, the domain of f(x)is all real numbers except x= 3 and x=−2,
or in interval notation, (−∞,−2) ∪(−2,3) ∪(3,∞).
14
2. x-intercepts:
To find the x-intercept(s), we set f(x) = 0 and solve for x:2x2−5x−3
x2−x−6=
0.
Simplifying further, we get 2x2−5x−3 = 0. Factoring the quadratic
gives (2x+ 1)(x−3) = 0, so x=−1
2or x= 3. Thus, the x-intercepts are
(−1
2,0) and (3,0).
3. y-intercept:
To find the y-intercept, we set x= 0 in f(x):f(0) = 2(0)2−5(0) −3
(0)2−(0) −6=
−3
−6=1
2.
Therefore, the y-intercept is (0,1
2).
4. Vertical asymptotes:
Vertical asymptotes occur where the denominator of a rational function is
zero. Thus, the vertical asymptotes are at x=−2and x= 3.
5. Horizontal asymptotes:
To find the horizontal asymptote(s), we look at the degrees of the numer-
ator and denominator. Since they have the same degree (2), we compare
the leading coefficients of the polynomial terms.
The horizontal asymptote is the ratio of the leading coefficients, which is
2
1= 2. Therefore, the horizontal asymptote of this function is y= 2.
Question 22
Question
Find the domain of the rational function:
f(x) = x+ 3
x2−4x−5
Solution
Step 1: We first need to find the values of xthat make the denominator zero,
since division by zero is undefined. Set the denominator equal to zero and solve
for x:
x2−4x−5 = 0
Step 2: This is a quadratic equation, so we can factor it or use the quadratic
formula. Factoring, we get:
(x−5)(x+ 1) = 0
15
Step 3: Set each factor to zero and solve for x:
x−5 = 0 or x+ 1 = 0
x= 5 or x=−1
Step 4: Therefore, the values x= 5 and x=−1make the denominator zero.
These values are not in the domain of the function. The domain of the function
f(x)is all real numbers except x= 5 and x=−1.
So, the domain of the function is:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 23
Question
Simplify the rational function:
3x2−6x
x2−x−6∇ · 2x2−6x
2x2+ 3x−2
Solution
Step 1: Simplify the complex fraction by multiplying the numerator by the
reciprocal of the denominator. Step 2: Factorize the numerators and denomina-
tors where possible. Step 3: Simplify the rational function by canceling factors
in the numerator and denominator. Step 4: Write the simplified form of the
rational function.
Step 1:
3x2−6x
x2−x−6∇ · 2x2−6x
2x2+ 3x−2=3x2−6x
x2−x−6·2x2+ 3x−2
2x2−6x
Step 2: Factorize where possible. For the first fraction: 3x2−6x= 3x(x−2)
x2−x−6 = (x−3)(x+ 2)
For the second fraction: 2x2+ 3x−2 = (2x−1)(x+ 2) 2x2−6x= 2x(x−3)
Step 3: Now, simplify the rational function:
3x(x−2) ·(2x−1)(x+ 2)
(x−3)(x+ 2) ·2x(x−3)
Cancel out common factors in the numerator and denominator:
3(2x−1)
2x
Step 4: The simplified form of the rational function is:
6x−3
2x
16
Question 24
Question
Solve the following rational equation for x:3x
x2−4−1
x−2=2
x+2 .
Solution
Step 1: First, let’s get rid of the fractions in the equation by finding a common
denominator. The common denominator in this case is (x2−4)(x−2)(x+ 2).
Multiply each term of the equation by this common denominator.
Step 2: After multiplying by the common denominator, the equation be-
comes: 3x(x+ 2) −(x2−4)(x+ 2) = 2(x2−4). Simplify this equation.
Step 3: Expand and simplify each term in the equation: 3x2+ 6x−x3−
2x2−4x+ 8 = 2x2−8. This simplifies to −x3+x2+ 2x+ 8 = 2x2−8.
Step 4: Rearrange the equation by moving all terms to one side to set it
equal to zero. This gives us −x3−x2−4x+ 16 = 0.
Step 5: Factor out a negative sign to make the equation easier to factor:
x3+x2+ 4x−16 = 0.
Step 6: Now, we need to factor the polynomial on the left side of the equa-
tion. By inspection, we find that x= 2 is a root of the polynomial. Use synthetic
division or polynomial division to factor the polynomial completely.
Step 7: After factoring, we get (x−2)(x2+ 3x+ 8) = 0. Setting each factor
equal to zero gives us solutions x= 2,x=−3
2+√7i
2, and x=−3
2−√7i
2.
Therefore, the solutions to the rational equation are x= 2,x=−3
2+√7i
2,
and x=−3
2−√7i
2.
Question 25
Question
Simplify the rational function:
R(x) = x2−5x+ 6
x2−4
Solution
To simplify R(x) = x2−5x+6
x2−4, we first factor the numerator and denominator.
Step 1: Factor the numerator: x2−5x+ 6 = (x−3)(x−2).
Step 2: Factor the denominator: x2−4 = (x+ 2)(x−2).
So, the rational function R(x)becomes:
R(x) = (x−3)(x−2)
(x+ 2)(x−2)
17
Step 3: Simplify by canceling out common factors in the numerator and
denominator.
Since there is a common factor of (x−2) in both the numerator and denom-
inator, we can simplify further:
R(x) = x−3
x+ 2
Therefore, the simplified form of the rational function R(x)is x−3
x+2 .
Question 26
Question
Simplify the rational expression:
2x2−7x−30
3x2+ 2x−8.
Solution
Step 1: Factor both the numerator and the denominator.
For the numerator 2x2−7x−30, we need to find two numbers that multiply
to 2×−30 = −60 and add up to −7. These numbers are −12 and 5, so we can
rewrite the numerator as 2x2−12x+ 5x−30 and factor by grouping:
(2x2−12x) + (5x−30) = 2x(x−6) + 5(x−6) = (2x+ 5)(x−6).
For the denominator 3x2+2x−8, we need to find two numbers that multiply
to 3× −8 = −24 and add up to 2. These numbers are −4and 6, so we can
rewrite the denominator as 3x2−4x+ 6x−8and factor by grouping:
(3x2−4x) + (6x−8) = x(3x−4) + 2(3x−4) = (x+ 2)(3x−4).
Therefore, the rational expression simplifies to:
2x2−7x−30
3x2+ 2x−8=(2x+ 5)(x−6)
(x+ 2)(3x−4).
So the simplified form of the rational expression is 2x+ 5
x+ 2 ·x−6
3x−4.
Question 27
Question
Let f(x) = 2x2+x−3
x2−4x+3 . Find the domain of f(x).
18
Solution
Step 1: The function f(x)is defined for all xvalues except those that make the
denominator equal to zero. Thus, we need to find the values of xthat satisfy
x2−4x+ 3 = 0.
Step 2: Factor the quadratic expression x2−4x+ 3 = 0 to find the values
of x.
(x−1)(x−3) = 0
Step 3: Set each factor equal to zero and solve for x.
x−1 = 0 =⇒x= 1
x−3 = 0 =⇒x= 3
Step 4: The solutions to the equation x2−4x+ 3 = 0 are x= 1 and x= 3.
These values make the denominator of f(x)equal to zero, so they are not in the
domain of f(x).
Step 5: Therefore, the domain of the function f(x)is all real numbers except
x= 1 and x= 3.
Step 6: In interval notation, the domain of f(x)can be expressed as (−∞,1)∪
(1,3) ∪(3,∞).
Question 28
Question
Simplify the following rational expression:
3x2−2x−5
x2+ 4x+ 3
Solution
Step 1: Factor both the numerator and the denominator.
Step 2: Factor the quadratic expressions.
Step 3: Simplify the expression by canceling out common factors.
Step 1: Factor the numerator: 3x2−2x−5We need to find two numbers
that multiply to 3·−5 = −15 and add up to −2. These numbers are −5and 3.
So, we can write 3x2−2x−5as (3x+ 3)(x−5).
Factor the denominator: x2+ 4x+ 3 We need to find two numbers that
multiply to 1·3 = 3 and add up to 4. These numbers are 3and 1. So, we can
write x2+ 4x+ 3 as (x+ 3)(x+ 1).
Step 2: Our expression becomes (3x+3)(x−5)
(x+3)(x+1) .
Step 3: Now, we can cancel out the common factor of 3 in the numerator
and the denominator. This simplifies the expression to x−5
x+1 .
Therefore, the simplified form of the given rational expression is x−5
x+ 1 .
19
Question 29
Question
Simplify the following rational function:
f(x) = 4x2−12x−16
x2−4x−5
Solution
Step 1: Factor the numerator and denominator of the rational function f(x):
f(x) = 4x2−12x−16
x2−4x−5=4(x2−3x−4)
(x−5)(x+ 1)
Step 2: Further simplify the factored numerator:
f(x) = 4(x2−3x−4)
(x−5)(x+ 1) =4(x−4)(x+ 1)
(x−5)(x+ 1)
Step 3: Cancel out common factors in the numerator and denominator:
f(x) = 4(x−4)(x+ 1)
(x−5)(x+ 1) =4(x−4)
x−5
Therefore, the simplified form of the rational function f(x)is:
f(x) = 4(x−4)
x−5
Question 30
Question
Simplify the following rational expression:
3x2−7x−20
2x2−5x−3
Solution
To simplify the rational expression, we can factor both the numerator and de-
nominator and then cancel out any common factors.
Step 1: Factor the numerator and denominator:
3x2−7x−20 = (3x+ 5)(x−4)
2x2−5x−3 = (2x+ 1)(x−3)
20
Step 2: Rewrite the expression with the factored forms:
(3x+ 5)(x−4)
(2x+ 1)(x−3)
Step 3: Simplify the expression by canceling out common factors:
(3x+ 5)(x−4)
(2x+ 1)(x−3) =3x+ 5
2x+ 1 ·x−4
x−3
So, the simplified form of the given rational expression is 3x+ 5
2x+ 1 ·x−4
x−3.
Question 31
Question
Simplify the rational function:
f(x) = x2−4
x2+ 3x−10
Solution
Step 1: Factor the numerator and the denominator. Step 2: Simplify by can-
celing out common factors.
Step 1: We factor the numerator and denominator:
f(x) = x2−4
x2+ 3x−10 =(x+ 2)(x−2)
(x+ 5)(x−2)
Step 2: Now, we simplify by canceling out the common factor (x−2) in
the numerator and denominator:
f(x) = (x+ 2)(x−2)
(x+ 5)(x−2) =x+ 2
x+ 5
Therefore, the simplified form of the rational function is f(x) = x+ 2
x+ 5.
Question 32
Question
Simplify the rational function:
4x3+ 12x2−7x−21
2x2+ 5x−3
21
Solution
Step 1: Factor both the numerator and denominator.
4x3+ 12x2−7x−21 = 4x2(x+ 3) −7(x+ 3) = (4x2−7)(x+ 3)
2x2+ 5x−3 = (2x−1)(x+ 3)
Step 2: Rewrite the original rational function with the factored forms.
4x2−7
2x−1
Thus, the simplified form of the given rational function is 4x2−7
2x−1.
Question 33
Question
Find the domain of the rational function: f(x) = x2−4
x2−5x+ 6.
Solution
Step 1: To find the domain of a rational function, we need to identify all the
values of xfor which the function is defined. The function will not be defined
where the denominator is equal to zero (since division by zero is undefined).
Step 2: Set the denominator equal to zero and solve for x:
x2−5x+ 6 = 0
Step 3: Factor the quadratic:
(x−2)(x−3) = 0
Step 4: Set each factor to zero:
x−2 = 0 or x−3 = 0
Step 5: Solve for xin each case:
x= 2 or x= 3
Step 6: Therefore, the rational function is undefined at x= 2 and x= 3.
Step 7: The domain of the rational function f(x)is all real numbers except
x= 2 and x= 3. It can be written in interval notation as:
(−∞,2) ∪(2,3) ∪(3,∞)
22
Question 34
Question
Given the rational function f(x) = 4x2−3x−4
x2+1 , find: (a) the domain of f(x),
(b) the x- and y-intercepts of the graph of f(x), (c) the vertical and horizontal
asymptotes (if any) of the graph of f(x).
Solution
(a) To find the domain of f(x), we need to identify any values of xthat make
the denominator of the rational function equal to zero, since division by zero is
undefined. In this case, the denominator x2+ 1 is never equal to zero for any
real number x. Hence, the domain of f(x)is all real numbers: (−∞,∞).
(b) To find the x-intercepts of the graph of f(x), we set f(x) = 0 and solve
for x:4x2−3x−4
x2+ 1 = 0
4x2−3x−4 = 0
This quadratic equation does not factor easily, so we can use the quadratic
formula:
x=−(−3) ±√(−3)2−4(4)(−4)
2(4)
x=3±√9 + 64
8
x=3±√73
8
Therefore, the x-intercepts are (3+√73
8,0)and (3−√73
8,0).
To find the y-intercept, we evaluate f(0):
f(0) = 4(0)2−3(0) −4
(0)2+ 1 =−4
1=−4
Hence, the y-intercept is (0,−4).
(c) To find the vertical asymptotes of the graph of f(x), we look for values
of xthat make the denominator of f(x)equal to zero. In this case, x2+ 1 = 0
has no real solutions, so there are no vertical asymptotes.
To find the horizontal asymptote, we compare the degrees of the numerator
and denominator of f(x). Since the degree of the numerator is equal to the
degree of the denominator, we divide the leading coefficients:
lim
x→∞
4x2−3x−4
x2+ 1 = lim
x→∞
4
1= 4
Therefore, the horizontal asymptote is y= 4.
23
Question 35
Question
Solve the following rational inequality for x:
3x+ 2
x−5≤4.
Solution
To solve the rational inequality, we first find the critical points by setting the
numerator equal to zero. Then we determine the sign of the rational function
in the intervals created by the critical points.
Step 1: Find the critical points Setting the numerator equal to zero
gives:
3x+ 2 = 0.
Solving for x, we get:
x=−2
3.
The critical point is x=−2
3.
Step 2: Determine the sign of the rational function Choose test
points in each interval to determine the sign of the rational function.
• Test point x=−1:
3(−1) + 2
(−1) −5=−1
−6=1
6<4.
This means the function is negative in this interval.
• Test point x= 0:
3(0) + 2
(0) −5=2
−5=−2
5<4.
This means the function is negative in this interval.
• Test point x=−2
3:
3(−2
3)+ 2
(−2
3)−5=−6+2
−2
3−5=−4
−17
3
=12
17 <4.
This means the function is negative in this interval.
• Test point x= 1:
3(1) + 2
(1) −5=5
−4=−5
4<4.
This means the function is negative in this interval.
24
• Test point x= 6:
3(6) + 2
6−5=20
1= 20 <4.
This means the function is positive in this interval.
Step 3: Write the solution The solution to the inequality is:
x∈(−∞,−2
3]∪(−2
3,5) .
25
