MATH 121 - COLLEGE ALGEBRA Rational Functions Question Bank Set 6

MATH 121 - COLLEGE ALGEBRA -

Rational Functions

Question Bank - Set 6

Liberty University

Question 1

Question

Simplify the rational function: 4x2−9

2x2+11x+3 .

Solution

Step 1: Factor the numerator and denominator as much as possible. Step 2:

Simplify the expression by canceling out common factors.

Step 1: Factor the numerator and denominator. The numerator 4x2−9can

be factored as the difference of squares (2x)2−32, which gives us (2x+3)(2x−3).

The denominator 2x2+ 11x+ 3 can be factored as (2x+ 1)(x+ 3).

Step 2: Simplify the expression. Substitute the factored forms back into

the original expression:

4x2−9

2x2+ 11x+ 3 =(2x+ 3)(2x−3)

(2x+ 1)(x+ 3) .

Now, we can cancel out common factors:

(2x+ 3)(2x−3)

(2x+ 1)(x+ 3) =2x−3

2x+ 1.

Therefore, the simplified form of the rational function is 2x−3

2x+1 .

Question 2

Question

Let f(x) = 2x3−x2−6x+3

x2−4be a rational function. Determine the vertical asymp-

totes, horizontal asymptotes, and any holes (if they exist) of the graph of f(x).

Solution

Step 1: To find the vertical asymptotes, we look for values of xthat make the

denominator of f(x)equal to zero. The denominator x2−4is equal to zero

when x=±2. Therefore, the vertical asymptotes of the graph of f(x)are x= 2

and x=−2.

Step 2: To find the horizontal asymptote, we compare the degrees of the

numerator and denominator of f(x). The degree of the numerator is 3, and the

degree of the denominator is 2. Since the degree of the numerator is greater,

there is no horizontal asymptote.

Step 3: To determine if there are any holes in the graph of f(x), we factor

the numerator and denominator and see if there are any common factors that

cancel out. Factoring, we have f(x) = (2x−1)(x2+3)

(x+2)(x−2) . There are no common

factors that cancel out, so there are no holes in the graph of f(x).

Therefore, the vertical asymptotes of the graph of f(x)are x= 2 and x=−2,

there is no horizontal asymptote, and there are no holes in the graph of f(x).

Question 3

Question

Find the domain of the rational function:

f(x) = 3x2−5x+ 2

x2−4x+ 3

Solution

Step 1: We need to find the values of xfor which the denominator is not equal

to zero, since division by zero is undefined.

Step 2: Set the denominator x2−4x+ 3 to zero and solve for x:

x2−4x+ 3 = 0

Step 3: Factor the quadratic equation:

(x−3)(x−1) = 0

Step 4: Set each factor to zero and solve for x:

x−3 = 0 ⇒x= 3

x−1 = 0 ⇒x= 1

Step 5: The values of xthat make the denominator zero are x= 1 and

x= 3.

Step 6: Therefore, the domain of the function f(x)is all real numbers except

x= 1 and x= 3, or in interval notation:

(−∞,1) ∪(1,3) ∪(3,∞)

Question 4

Question

Find the domain of the rational function:

f(x) = 1

x2−4x

Solution

To find the domain of a rational function, we need to identify the values of x

that would make the denominator zero, as division by zero is undefined.

Step 1: Determine the values of xthat make the denominator zero by

setting the denominator equal to zero and solving for x:

x2−4x= 0

Step 2: Factor out an xfrom the equation:

x(x−4) = 0

Step 3: Use the zero-product property to solve for x:

Setting x= 0:

x= 0

Setting x−4 = 0:

x= 4

Step 4: The values x= 0 and x= 4 make the denominator x2−4xzero.

Step 5: The domain of the function is all real numbers except x= 0 and

x= 4.

Therefore, the domain of the given rational function is (−∞,0) ∪(0,4) ∪(4,∞).

Question 5

Question

Find the domain of the rational function:

f(x) = x+ 5

x2−9.

Solution

Step 1: Recall that the domain of a rational function is all real numbers except

for the values that would make the denominator zero, since division by zero is

undefined. So, we need to find the values of xthat would make the denominator

x2−9equal to zero.

Step 2: Factor the denominator to find its zeros:

x2−9 = 0

(x+ 3)(x−3) = 0

Step 3: Set each factor to zero and solve for x:

x+ 3 = 0 ⇒x=−3

x−3 = 0 ⇒x= 3

Step 4: The domain of the function f(x) = x+5

x2−9is all real numbers except

x=−3and x= 3. Therefore, the domain is x∈(−∞,−3) ∪(−3,3) ∪(3,∞).

Question 6

Question

Given the rational function f(x) = 2x2+x−1

x2−4x+3 , determine the following:

1. Vertical asymptotes, if any.

2. Horizontal asymptotes, if any.

3. Holes, if any.

Solution

1. To find the vertical asymptotes, we need to identify any values of xthat

make the denominator of the function equal to zero.

Step 1: Set the denominator equal to zero and solve for x:

x2−4x+ 3 = 0

Step 2: Factor the quadratic equation:

(x−3)(x−1) = 0

Step 3: Set each factor to zero and solve for x:

x−3 = 0 or x−1 = 0

x= 3 or x= 1

Therefore, the vertical asymptotes are x= 3 and x= 1.

2. To find the horizontal asymptotes, we need to compare the degrees of the

numerator and denominator of the rational function.

Step 1: Compare the degrees of the numerator and denominator: The degree

of the numerator is 2 and the degree of the denominator is also 2.

Step 2: Divide the leading coeﬀicients of the numerator and denominator:

The leading coeﬀicients are both 2.

Step 3: Since the degrees and leading coeﬀicients are equal, the horizontal

asymptote is:

y=2

1= 2

Therefore, the horizontal asymptote is y= 2.

3. To find the holes, we need to look for common factors between the nu-

merator and denominator that can be cancelled out.

Step 1: Factor the numerator and denominator: Numerator: 2x2+x−1 =

(2x−1)(x+ 1) Denominator: x2−4x+ 3 = (x−3)(x−1)

Step 2: Simplify the rational function:

f(x) = (2x−1)(x+ 1)

(x−3)(x−1)

Step 3: From the simplified form, we see that the factor (x−1) in both the

numerator and denominator can be cancelled out. This creates a hole at x= 1.

Therefore, the hole in the graph of the function occurs at x= 1.

Question 7

Question

Let f(x) = 2x2−5x+3

x2−4and g(x) = x2−4

x2−5x+6 . Determine the domain of the com-

posite function h(x) = f(g(x)).

Solution

Step 1: Find the domain of g(x): Since g(x) = x2−4

x2−5x+6 , we need to find the

values of xthat make the denominator (x2−5x+ 6) equal to zero. This will

give us the values that xcannot take on.

Setting the denominator equal to zero:

x2−5x+ 6 = 0

Factoring the quadratic:

(x−2)(x−3) = 0

So, x= 2 and x= 3 are the values that xcannot be in order for the

denominator to not be zero. Therefore, the domain of g(x)is all real numbers

except x= 2 and x= 3.

Step 2: Find the domain of h(x): Since h(x) = f(g(x)), we need to find the

values of xsuch that both f(x)and g(x)are defined.

For f(x)to be defined, the denominator of f(x)should not be zero. There-

fore, the domain of f(x)is all real numbers except x= 2 and x=−2(since

x2−4 = 0 when x=±2).

To determine the domain of h(x) = f(g(x)), we need to consider the values

that g(x)can take on which will make f(g(x)) undefined.

Since the domain of g(x)is all real numbers except x= 2 and x= 3, the

value g(x)takes on when x= 2 is 22−4

22−5(2)+6 =0

0which is undefined. Therefore,

x= 2 is not in the domain of h(x).

Finally, we determine the domain of h(x)by combining the restrictions from

f(x)and g(x). Thus, the domain of the composite function h(x) = f(g(x)) is

all real numbers except x= 2 and x= 3.

Question 8

Question

Given the rational function f(x) = 5x2−4x−3

x2−3x+2 , find the horizontal asymptotes,

vertical asymptotes (if any), and any holes in the graph of f(x).

Solution

Step 1: To find the horizontal asymptotes of f(x), we examine the degrees of

the numerator and denominator. If the degree of the numerator is equal to

the degree of the denominator, the horizontal asymptote is the ratio of the

leading coeﬀicients. If the degree of the numerator is less than the degree of the

denominator, the horizontal asymptote is y= 0. If the degree of the numerator

is greater than the degree of the denominator, there is no horizontal asymptote.

The degrees of the numerator and denominator are both 2. Therefore, the

horizontal asymptote is the ratio of the leading coeﬀicients, which is y=5

1= 5.

Step 2: To find the vertical asymptotes of f(x), we set the denominator equal

to zero and solve for x. These values of xare the vertical asymptotes. Setting

x2−3x+ 2 = 0, we get (x−1)(x−2) = 0. Therefore, x= 1 and x= 2 are the

vertical asymptotes of the function f(x). Step 3: To find any holes in the graph

of f(x), we check for common factors in the numerator and the denominator.

Factoring 5x2−4x−3and x2−3x+ 2 gives us (5x+3)(x−1) and (x−1)(x−2),

respectively. We see that there is a factor of (x−1) that cancels out, leaving

f(x) = 5x+3

x−2. Since the factor (x−1) canceled out, x= 1 is a hole in the graph

of f(x). Therefore, the horizontal asymptote of f(x)is y= 5, the vertical

asymptotes are x= 1 and x= 2, and there is a hole at (1,8

3).

Question 9

Question

Simplify the following rational expression:

3x2−5x−2

x2+ 4x−5

Solution

Step 1: Factor the numerator and denominator.

Step 2: Factor the numerator: To factor the numerator 3x2−5x−2, we

need to find two numbers that multiply to 3(−2) = −6and add up to −5. The

numbers are −6and 1. So we can rewrite the expression as:

(3x+ 1)(x−2)

x2+ 4x−5

Step 3: Factor the denominator: To factor the denominator x2+ 4x−5, we

need to find two numbers that multiply to (−5) and add up to 4. The numbers

are 5and −1. So we can rewrite the expression as:

(3x+ 1)(x−2)

(x+ 5)(x−1)

Step 4: Cancel out common factors: Now that we have factored the numer-

ator and the denominator, we can cancel out the common factors of (3x+ 1)

and (x−2) from the numerator and denominator:

(3x+ 1)(x−2)

(x+ 5)(x−1)

Step 5: Simplify the expression: After canceling out the common factors,

the simplified expression is: 1

x+ 5

Therefore, the simplified form of the given rational expression is 1

x+5 .

Question 10

Question

Simplify the rational function:

f(x) = 4x2−15x+ 18

x2−3x−10

Solution

Step 1: Factor the numerator and denominator.

f(x) = 4x2−15x+ 18

x2−3x−10 =(4x−3)(x−6)

(x−5)(x+ 2)

Step 2: Simplify the fraction by canceling out any common factors in the

numerator and denominator.

f(x) = (4x−3)(x−6)

(x−5)(x+ 2) =4x−3

x+ 2

Thus, the simplified form of the rational function f(x) = 4x2−15x+18

x2−3x−10 is

f(x) = 4x−3

x+2 .

Question 11

Question

For a certain rational function, the numerator has degree 3 while the denomi-

nator has degree 2. One of the vertical asymptotes of this rational function is

x= 4. If the graph of the function also passes through the point (2,−3), find

the equation of the rational function.

Solution

Step 1: Let’s denote the rational function as f(x) = ax3+bx2+cx+d

ex2+fx+g. Given that

x= 4 is a vertical asymptote, we know that the denominator must equal 0 at

x= 4:e(4)2+f(4) + g= 0. This gives us 16e+ 4f+g= 0.

Step 2: Since x= 4 is a vertical asymptote, the factor (x−4) must appear

in the denominator but not in the numerator. This means that emust equal 1.

Thus, the equation 16e+ 4f+g= 0 simplifies to 16 + 4f+g= 0.

Step 3: Now we can substitute the point (2,−3) into the function to find

values of a,b,c, and d. Substituting x= 2 and f(x) = −3into the function

gives us: 8a+4b+2c+d

f(2) =−3. This simplifies to 8a+ 4b+ 2c+d=−3f(2).

Step 4: We also know that the graph passes through the point (2,−3), thus

the function must satisfy f(2) = −3. Substituting x= 2 into the function gives:

8a+4b+2c+d

4+2f+g=−3. Simplifying this yields 8a+ 4b+ 2c+d=−3(4 + 2f+g).

Step 5: Now we have a system of equations:











16 + 4f+g= 0

8a+ 4b+ 2c+d=−3f(2)

8a+ 4b+ 2c+d=−3(4 + 2f+g)

Step 6: Solving the system of equations, we find a=−5

4,b=9

4,c=−11

and d=−3.

Step 7: Therefore, the equation of the rational function is f(x) = −5

4x3+9

4x2−11

4x−3

x2+fx+g.

Substituting e= 1 and g=−16−4f, we have the final answer: f(x) = −5

4x3+9

4x2−11

4x−3

x2+fx −16 −4f.

Question 12

Question

Simplify the following rational expression:

3x2−5x+ 2

2x2+ 3x−2∇ · 2x2+ 3x−2

3x2−x−2.

Solution

To simplify the given expression, we will first rewrite the division as multiplica-

tion by the reciprocal of the second fraction. Then, we will factor the numerator

and denominator of each fraction if possible, and simplify by canceling out com-

mon factors.

Step 1: Begin by rewriting the expression as multiplication by the recipro-

cal: 3x2−5x+ 2

2x2+ 3x−2·3x2−x−2

2x2+ 3x−2.

Step 2: Factor the numerators and denominators of each fraction:

(3x−2)(x−1)

(2x−1)(x+ 2) ·(3x+ 2)(x−1)

(2x−1)(x+ 2).

Step 3: Simplify by canceling out common factors in the numerator and

denominator: 3x−2

x+ 2 ·3x+ 2

x+ 2 .

Step 4: Combine the fractions by multiplying the numerators and denomi-

nators: (3x−2)(3x+ 2)

(x+ 2)2.

Step 5: Expand the numerators:

9x2−4

x2+ 4x+ 4.

Therefore, the simplified form of the given expression is 9x2−4

x2+4x+4 .

Question 13

Question

Solve the rational equation:

x−1−1

x+ 2 =2

x2+x−2

Solution

Step 1: Find a common denominator for all the fractions. In this case, the least

common denominator is (x−1)(x+ 2).

Step 2: Rewrite each fraction with the common denominator.

3(x+ 2)

(x−1)(x+ 2) −(x−1)

(x−1)(x+ 2) =2

x2+x−2

Step 3: Simplify the fractions.

3x+ 6

(x−1)(x+ 2) −x−1

(x−1)(x+ 2) =2

x2+x−2

Step 4: Combine the fractions on the left side of the equation.

3x+ 6 −(x−1)

(x−1)(x+ 2) =2

x2+x−2

Step 5: Simplify the left side of the equation.

3x+ 6 −x+ 1

(x−1)(x+ 2) =2

x2+x−2

2x+ 7

(x−1)(x+ 2) =2

(x+ 2)(x−1)

Step 6: Cross multiply to eliminate denominators.

2(x+ 2)(x−1) = (2x+ 7)

2x2+ 2x−4 = 2x+ 7

Step 7: Simplify the equation.

2x2+ 2x−4−2x−7 = 0

2x2−11 = 0

Step 8: Solve the quadratic equation.

2x2−11 = 0

2x2= 11

x2=11

x=±√11

x=±√22

Therefore, the solutions to the rational equation are x=√22

2and x=−√22

Question 14

Question

Solve the rational equation for x:

x−1+2

x+ 2 =5

x2+x−2

Solution

Step 1: Find a common denominator for the fractions on the left side of the

equation. The common denominator here is (x−1)(x+ 2). Step 2: Rewrite the

equation with the common denominator:

3(x+ 2)

(x−1)(x+ 2) +2(x−1)

(x−1)(x+ 2) =5

x2+x−2

Step 3: Combine the fractions on the left side of the equation:

3(x+ 2) + 2(x−1)

(x−1)(x+ 2) =5

x2+x−2

3x+6+2x−2

(x−1)(x+ 2) =5

x2+x−2

5x+ 4

(x−1)(x+ 2) =5

x2+x−2

Step 4: Cross multiply to solve for x:

5(x2+x−2) = (5x+ 4)

5x2+ 5x−10 = 5x+ 4

5x2+ 5x−5x−10 −4 = 0

5x2−14 = 0

Step 5: Factor the quadratic equation:

(5x+ 2)(x−7) = 0

Step 6: Set each factor to zero and solve for x: For 5x+ 2 = 0:

5x=−2

x=−2

For x−7 = 0:

x= 7

Therefore, the solutions to the equation are x=−2

5and x= 7.

Question 15

Question

Simplify the following rational expression:

(3x2−5x−2

2x2+ 9x+ 5)∇ · (4x2−1

6x2−5x−6)

Solution

Step 1: To simplify the given expression, first rewrite it as a multiplication of

the reciprocal of the second fraction:

3x2−5x−2

2x2+ 9x+ 5 ×6x2−5x−6

4x2−1

Step 2: Factor each quadratic expression: For the first expression, 3x2−

5x−2, we have: 3x2−6x+x−2 = 3x(x−2) + 1(x−2) = (3x+ 1)(x−2)

For the second expression, 2x2+ 9x+ 5, we have: 2x2+ 2x+ 7x+ 5 =

2x(x+ 2) + 5(x+ 1) = (2x+ 5)(x+ 1)

For the third expression, 6x2−5x−6, we have: 6x2−9x+ 4x−6 =

6x(x−1) + 4(x−1) = (6x+ 4)(x−1)

For the fourth expression, 4x2−1, we have: 4x2−2x+x−1 = 4x(x−1) +

1(x−1) = (4x+ 1)(x−1)

Step 3: Rewrite the expression with the factored forms:

(3x+ 1)(x−2)

(2x+ 5)(x+ 1) ×(6x+ 4)(x−1)

(4x+ 1)(x−1)

Step 4: Cancel out any common factors in the numerator and denominator:

3x+ 1

2x+ 5 ×6x+ 4

4x+ 1

Step 5: Multiply the remaining factors to get the simplified expression:

3x+ 1

2x+ 5 ×6x+ 4

4x+ 1 =18x2+ 12x+ 6x+ 4

8x2+ 2x+ 20x+ 5

=18x2+ 18x+ 4

8x2+ 22x+ 5

Therefore, the simplified form of the given expression is 18x2+18x+4

8x2+22x+5 .

Question 16

Question

Let f(x) = 2x2−5x+1

x+3 . Find all values of xfor which f(x)is undefined.

Solution

Step 1: To find the values of xfor which f(x)is undefined, we need to iden-

tify where the denominator of f(x)is equal to zero, since division by zero is

undefined.

Step 2: Set the denominator x+ 3 equal to zero and solve for x.

x+ 3 = 0

Step 3: Subtract 3 from both sides to solve for x.

x=−3

Step 4: Therefore, f(x)is undefined at x=−3.

Question 17

Question

Simplify the following rational expression:

3x2−6x+ 3

x2−2x

Solution

Step 1: Factor out the common terms in the numerator and denominator:

3(x2−2x+ 1)

x(x−2)

Step 2: Simplify the expression inside the parentheses in the numerator:

3(x−1)2

x(x−2)

Step 3: The expression is now simplified and cannot be reduced any further.

Therefore, the final simplified form of the rational expression is:

3(x−1)2

x(x−2)

Question 18

Question

Simplify the following rational expression:

3x2−9x

x2−4

Solution

Step 1: Factor out the common factor in the numerator:

3x2−9x

x2−4=3x(x−3)

x2−4

Step 2: Factor the denominator as the difference of squares:

3x(x−3)

x2−4=3x(x−3)

(x+ 2)(x−2)

Step 3: Cancel out common factors in the numerator and denominator:

3x(x−3)

(x+ 2)(x−2) =3(x−3)

x+ 2

Therefore, the simplified form of the rational expression is 3(x−3)

x+2 .

Question 19

Question

Find the domain of the rational function:

f(x) = 3x2−6x

x2−4x−32

Solution

Step 1: We start by identifying any values of xthat would make the denominator

equal to zero, as these values would make the function undefined. So, we solve

the equation x2−4x−32 = 0 to find the values that would make the denominator

equal to zero.

Step 2: Factoring the quadratic equation, we have:

x2−4x−32 = 0

(x−8)(x+ 4) = 0

So, the values that would make the denominator zero are x= 8 and x=−4.

Step 3: The domain of the rational function is all real numbers except the

values that make the denominator zero. Therefore, the domain of f(x)is:

(−∞,−4) ∪(−4,8) ∪(8,∞)

Question 20

Question

Solve the rational equation: x

x−2+4

x+2 =5

x2−4.

Solution

Step 1: Find a common denominator for all fractions on both sides of the

equation. In this case, the common denominator is (x−2)(x+ 2). Step 2:

Rewrite each fraction with the common denominator. Step 3: Simplify the

equation by multiplying both sides by the common denominator to eliminate

fractions. Step 4: Solve the resulting polynomial equation by moving all terms

to one side and factoring. Step 5: Check for extraneous solutions by ensuring

the solutions do not make any denominators equal to zero. Step 6: State the

final solution set.

Step 1: The common denominator is (x−2)(x+ 2).

Step 2: Rewrite each fraction with the common denominator:

x−2·x+ 2

x+ 2 +4

x+ 2 ·x−2

x−2=5

x2−4

x(x+ 2)

(x−2)(x+ 2) +4(x−2)

(x−2)(x+ 2) =5

x2−4

x2+ 2x+ 4x−8

(x−2)(x+ 2) =5

x2−4

x2+ 6x−8

(x−2)(x+ 2) =5

x2−4

Step 3: Multiply both sides by the common denominator (x−2)(x+ 2):

(x2+ 6x−8) = 5

x2+ 6x−8 = 5

x2+ 6x−13 = 0

Step 4: Solve the resulting polynomial equation by factoring: The equation

x2+ 6x−13 = 0 does not factor easily, so we can use the quadratic formula:

x=−b±√b2−4ac

where a= 1,b= 6, and c=−13.

x=−6±√62−4(1)(−13)

2(1)

x=−6±√36 + 52

x=−6±√88

x=−6±2√22

x=−3±√22

Step 5: Check for extraneous solutions by ensuring the solutions do not make

any denominators equal to zero. Since the original equation had no excluded

values, both solutions x=−3 + √22 and x=−3−√22 are valid.

Step 6: The solution set is x=−3 + √22,−3−√22.

Question 21

Question

Let f(x) = 3x2−2x−5

x2−4x−5. Find the vertical asymptotes of the function f(x).

Solution

Step 1: Find the vertical asymptotes by determining the values of xthat make

the denominator of the rational function equal to zero.

Step 2: Set the denominator equal to zero and solve for x.

x2−4x−5 = 0

(x−5)(x+ 1) = 0

Step 3: Set each factor equal to zero.

x−5 = 0 =⇒x= 5

x+ 1 = 0 =⇒x=−1

Step 4: Therefore, the vertical asymptotes of the function f(x)are x= 5

and x=−1.

Question 22

Question

Find the domain of the rational function: f(x) = 2x2+ 5x−3

x2−4x−5.

Solution

Step 1: We need to find the values of xfor which the denominator is not equal

to zero, since division by zero is undefined. Therefore, we set the denominator

x2−4x−5not equal to zero and solve for x.

x2−4x−5= 0

Step 2: To find the values of xthat satisfy the inequality, we can factor the

quadratic expression on the left side.

(x−5)(x+ 1) = 0

Step 3: Now, set each factor not equal to zero to find the values of x.

x−5= 0 and x+ 1 = 0

Step 4: Solve for xin each equation.

x= 5 and x=−1

Step 5: Combining the solutions, we find that the domain of f(x)is all real

numbers except x= 5 and x=−1. Therefore, the domain of the rational

function f(x) = 2x2+ 5x−3

x2−4x−5is (−∞,−1) ∪(−1,5) ∪(5,∞).

Question 23

Question

Find the domain of the rational function:

f(x) = x2−4

x3−8x

Solution

Step 1: Determine the values of xthat make the denominator zero, since division

by zero is undefined. Set the denominator equal to zero and solve for x:

x3−8x= 0

Step 2: Factor out an xfrom the equation:

x(x2−8) = 0

Step 3: Further factor the quadratic term:

x(x−2)(x+ 2) = 0

Step 4: Set each factor equal to zero and solve for x:

x= 0, x −2 = 0 or x+ 2 = 0

x= 0, x = 2 or x=−2

Step 5: So, the values x= 0,x= 2, and x=−2make the denominator

zero. Therefore, the domain of the function is all real numbers except for x= 0,

x= 2, and x=−2.

Step 6: Therefore, the domain of the rational function f(x) = x2−4

x3−8xis

x∈R|x= 0,2,−2.

Question 24

Question

Let f(x) = 3x2−5x−2

x2−4x−5. Determine the equations of the vertical and horizontal

asymptotes of the graph of f(x).

Solution

Step 1: To find the equations of the vertical asymptotes, we first need to find

the values at which the denominator of f(x)equals 0 (since division by zero is

undefined).

x2−4x−5 = 0

Step 2: We can factor the quadratic equation x2−4x−5 = 0 as (x−5)(x+1) =

0. Therefore, x= 5 or x=−1.

So, the equations of the vertical asymptotes are x= 5 and x=−1.

Step 3: To find the equation of the horizontal asymptote, we compare the

degrees of the numerator and denominator of f(x).

Step 4: Since the degree of the numerator is equal to the degree of the denom-

inator, we can divide the leading coeﬀicients to find the horizontal asymptote:

lim

x→∞

f(x) = lim

x→∞

3x2

x2= lim

x→∞ 3 = 3

Therefore, the equation of the horizontal asymptote is y= 3.

Question 25

Question

Solve the rational inequality: 4x−3

x−2>2.

Solution

To solve the rational inequality 4x−3

x−2>2, we will first find the critical points

by setting the numerator equal to the denominator.

Step 1: Find the critical point(s).

4x−3 = 2(x−2)

4x−3 = 2x−4

4x−2x=−4+3

2x=−1

x=−1

So the critical point is x=−1

Step 2: Test the intervals. We will test the intervals (−∞,−1

2),(−1

2,2),

and (2,∞)by choosing test points.

1. Test x=−1:

4(−1) −3

(−1) −2=−7

−3=7

3≮2

So x=−1is not in the solution set.

2. Test x= 0:

4(0) −3

(0) −2=−3

−2=3

2≮2

So x= 0 is not in the solution set.

3. Test x= 1:

4(1) −3

(1) −2=1

−1=−1<2

So x= 1 is in the solution set.

Step 3: Write the solution set. The solution set to the given rational

inequality is x∈(2,∞).

Question 26

Question

Simplify the following rational expression:

1−2

1 + 2

Solution

Step 1: To simplify the expression, we first need to find a common denominator

for the terms in both the numerator and denominator. Step 2: The common

denominator for 1and 2

xis simply x, so we rewrite the expression as:

x−2

x+2

Step 3: Simplify the expression inside the fraction:

x−2

x+2

Step 4: When dividing by a fraction, we can multiply by the reciprocal of the

denominator: x−2

x·x

x+ 2

Step 5: Multiply the numerators and denominators:

x(x−2)

x(x+ 2)

Step 6: Further simplify the expression by expanding both the numerator and

denominator: x2−2x

x2+ 2x

Step 7: Factor out an xfrom both the numerator and denominator:

x(x−2)

x(x+ 2)

Step 8: Finally, cancel out the common factor of xin the numerator and de-

nominator to obtain the simplified expression:

x−2

x+ 2

Question 27

Question

Find the horizontal and vertical asymptotes of the rational function f(x) =

3x2−5x+2

x2−4.

Solution

Step 1: Identify the Vertical Asymptotes To find the vertical asymptotes

of a rational function, we need to determine the values of xthat make the

denominator equal to zero. The denominator x2−4equals zero when x= 2 or

x=−2. Thus, the vertical asymptotes are x= 2 and x=−2.

Step 2: Identify the Horizontal Asymptote To find the horizontal

asymptote of a rational function, we compare the degrees of the numerator

and denominator. If the degree of the numerator is less than the degree of the

denominator, the horizontal asymptote is y= 0. If the degree of the numerator

is equal to the degree of the denominator, we divide the leading coeﬀicients. If

the degree of the numerator is greater, there is no horizontal asymptote.

In this case, both the numerator and denominator have a degree of 2. Thus,

we divide the leading coeﬀicients: 3

1= 3.

Therefore, the horizontal asymptote is y= 3.

Therefore, the vertical asymptotes are x= 2 and x=−2, and the horizontal

asymptote is y= 3.

Question 28

Question

Find the domain of the rational function: f(x) = x2+ 2x−8

x2−4x.

Solution

Step 1: We start by identifying the values of xthat will make the denominator

of the rational function equal to zero since division by zero is undefined. The

denominator x2−4x=x(x−4). Setting x(x−4) = 0, we find that x= 0 and

x= 4 are the values that make the denominator zero.

Step 2: Now, we determine the values of xfor which the rational function

f(x)is defined. The domain of f(x)consists of all real numbers except those

that make the denominator zero. Thus, the domain of f(x)is all real numbers

minus the values that make the denominator zero. Therefore, the domain of the

rational function f(x) = x2+ 2x−8

x2−4xis (−∞,0) ∪(0,4) ∪(4,∞).

Question 29

Question

Find the domain of the following rational function:

f(x) = x2−5x+ 6

x2−9

Solution

Step 1: The denominator cannot be equal to zero. Thus, we need to solve the

equation x2−9 = 0. Step 2: Factor the denominator to find the solutions:

x2−9=(x+ 3)(x−3) = 0. Step 3: Set each factor to zero to solve for x:

x+ 3 = 0 =⇒x=−3, and x−3 = 0 =⇒x= 3. Step 4: The domain of the

function is all real numbers except for x=−3and x= 3. Step 5: Therefore,

the domain of the function f(x) = x2−5x+6

x2−9is (−∞,−3) ∪(−3,3) ∪(3,∞).

Question 30

Question

Simplify the rational expression:

5x3−7x2+ 11x−3

x2−4x+ 3

Solution

Step 1: Factor the numerator and the denominator.

Step 2: Simplify the rational expression by cancelling out common factors.

Step 1:

To factor the numerator of the rational expression 5x3−7x2+ 11x−3, we

first look for any common factors. In this case, there are no common factors, so

we proceed by using the grouping method.

5x3−7x2+ 11x−3 = (5x3−7x2) + (11x−3)

=x2(5x−7) + 1(11x−3)

Now, looking at x2(5x−7) + 1(11x−3), we have no further factorization

we can do. So, the factored form of the numerator is 5x3−7x2+ 11x−3 =

x2(5x−7) + 1(11x−3).

Similarly, to factor the denominator x2−4x+ 3, we look for numbers that

multiply to 3 (the constant term) and add to -4 (the coeﬀicient of the linear

term). The numbers that fit this criteria are -3 and -1.

Therefore, x2−4x+ 3 = (x−3)(x−1).

Step 2:

Now that we have factored the numerator and denominator, we can rewrite

the rational expression as follows:

5x3−7x2+ 11x−3

x2−4x+ 3 =x2(5x−7) + 1(11x−3)

(x−3)(x−1)

We have no common factors in the numerator and denominator that can be

cancelled out, so the simplified form of the rational expression is:

5x3−7x2+ 11x−3

x2−4x+ 3 =x2(5x−7) + 1(11x−3)

(x−3)(x−1)

Question 31

Question

Find the domain of the rational function:

f(x) = 5x2−3x−2

x2−9

Solution

To find the domain of a rational function, we need to identify all the values

of xthat would make the denominator equal to zero, since division by zero is

undefined.

Step 1: Find the values of xthat make the denominator zero. The

denominator of the rational function is x2−9. Setting this equal to zero gives

us:

x2−9 = 0

(x−3)(x+ 3) = 0

This gives us solutions x= 3 and x=−3.

Step 2: State the domain of the rational function. The domain of the

rational function f(x) = 5x2−3x−2

x2−9is all real numbers except for the values that

make the denominator zero. Thus, the domain of f(x)is (−∞,−3) ∪(−3,3) ∪(3,∞).

Question 32

Question

Given the rational function

f(x) = x3−2x2−x+ 2

x2−3x+ 2 ,

determine the following:

• Vertical asymptotes, if any

• Horizontal asymptotes, if any

• Holes, if any

• x-intercepts, if any

• y-intercepts, if any

Solution

To determine the characteristics of the rational function, we need to analyze

both the numerator and the denominator for any common factors and then

simplify the function if necessary.

Step 1: Factor the numerator and denominator

Factor the numerator x3−2x2−x+ 2 and denominator x2−3x+ 2:

x3−2x2−x+ 2 = x2(x−2) −1(x−2)

= (x2−1)(x−2)

= (x+ 1)(x−1)(x−2)

x2−3x+ 2 = (x−2)(x−1)

So, the rational function can be expressed as

f(x) = (x+ 1)(x−1)(x−2)

(x−2)(x−1) .

Step 2: Determine vertical asymptotes

Vertical asymptotes occur where the denominator is equal to zero but the nu-

merator is not zero. In this case, x= 2 is a vertical asymptote since it makes

the denominator zero.

Step 3: Determine horizontal asymptotes

To find the horizontal asymptote, compare the degrees of the numerator and

the denominator. Since the degree of the numerator is equal to the degree of

the denominator, the horizontal asymptote can be found by dividing the leading

coeﬀicient of the numerator by the leading coeﬀicient of the denominator. Thus,

there is a horizontal asymptote at y= 1.

Step 4: Find holes

Since there are common factors in the numerator and denominator that can be

reduced, we must analyze where these factors are undefined. Both x= 1 and

x= 2 result in zeros in both the numerator and denominator, thus resulting in

holes at x= 1 and x= 2.

Step 5: Find x-intercepts

To find the x-intercepts, set f(x) = 0 and solve for x:

(x+ 1)(x−1)(x−2)

(x−2)(x−1) = 0

(x+ 1) = 0

x=−1

So, the x-intercept is (−1,0).

Step 6: Find y-intercepts

To find the y-intercept, set x= 0 in the function:

(0 + 1)(0 −1)(0 −2)

0−2)(0 −1) =−2

So, the y-intercept is (0,−2).

Therefore, the vertical asymptote is x= 2, the horizontal asymptote is y= 1,

there are holes at x= 1 and x= 2, the x-intercept is (−1,0), and the y-intercept

is (0,−2).

Question 33

Question

Simplify the following rational function:

f(x) = 4x2−16x

2x2−8x

Solution

Step 1: Factor out a 4xfrom the numerator and a 2xfrom the denominator:

f(x) = 4x(x−4)

2x(x−4)

Step 2: Cancel out the common factor (x−4) in the numerator and denom-

inator:

f(x) = 4x

Step 3: Simplify the expression:

f(x) = 2

Therefore, the simplified form of the rational function is f(x) = 2.

Question 34

Question

Simplify the rational function: 2x3+ 6x2−4x

4x2−16 .

Solution

Step 1: Factor out the greatest common factor from the numerator and the

denominator. 2x3+ 6x2−4x

4x2−16 =2x(x2+ 3x−2)

4(x2−4)

Step 2: Factor the quadratic expressions inside the parentheses.

2x(x2+ 3x−2)

4(x2−4) =2x(x+ 2)(x−1)

4(x+ 2)(x−2)

Step 3: Cancel out common factors.

2x(x+ 2)(x−1)

4(x+ 2)(x−2) =x(x−1)

2(x−2)

Step 4: Simplify the resulting expression.

x(x−1)

2(x−2) =x2−x

2x−4

Therefore, the simplified form of the rational function is x2−x

2x−4.

Question 35

Question

Simplify the following rational expression:

2x2−3x−2

x2+x−6

Solution

Step 1: Factor both the numerator and the denominator:

2x2−3x−2

x2+x−6=(2x+ 1)(x−2)

(x+ 3)(x−2)

Step 2: Cancel out any common factors in the numerator and denominator:

(2x+ 1)(x−2)

(x+ 3)(x−2) =2x+ 1

x+ 3

Therefore, the simplified form of the rational expression is 2x+1

x+3 .

Question 4

Question

Find the domain of the rational function:

f(x) = 1

x2−4x

Solution

To find the domain of a rational function, we need to identify the values of x

that would make the denominator zero, as division by zero is undefined.

Step 1: Determine the values of xthat make the denominator zero by

setting the denominator equal to zero and solving for x:

x2−4x= 0

Step 2: Factor out an xfrom the equation:

x(x−4) = 0

Step 3: Use the zero-product property to solve for x:

Setting x= 0:

x= 0

Setting x−4 = 0:

x= 4

Step 4: The values x= 0 and x= 4 make the denominator x2−4xzero.

Step 5: The domain of the function is all real numbers except x= 0 and

x= 4.

Therefore, the domain of the given rational function is (−∞,0) ∪(0,4) ∪(4,∞).

Question 5

Question

Find the domain of the rational function:

f(x) = x+ 5

x2−9.

Solution

Step 1: Recall that the domain of a rational function is all real numbers except

for the values that would make the denominator zero, since division by zero is

undefined. So, we need to find the values of xthat would make the denominator

x2−9equal to zero.

Step 2: Factor the denominator to find its zeros:

x2−9 = 0

(x+ 3)(x−3) = 0

Step 3: Set each factor to zero and solve for x:

x+ 3 = 0 ⇒x=−3

x−3 = 0 ⇒x= 3

Step 4: The domain of the function f(x) = x+5

x2−9is all real numbers except

x=−3and x= 3. Therefore, the domain is x∈(−∞,−3) ∪(−3,3) ∪(3,∞).

Question 6

Question

Given the rational function f(x) = 2x2+x−1

x2−4x+3 , determine the following:

1. Vertical asymptotes, if any.

2. Horizontal asymptotes, if any.

3. Holes, if any.

Solution

1. To find the vertical asymptotes, we need to identify any values of xthat

make the denominator of the function equal to zero.

Step 1: Set the denominator equal to zero and solve for x:

x2−4x+ 3 = 0

Step 2: Factor the quadratic equation:

(x−3)(x−1) = 0

Step 3: Set each factor to zero and solve for x:

x−3 = 0 or x−1 = 0

x= 3 or x= 1

Therefore, the vertical asymptotes are x= 3 and x= 1.

2. To find the horizontal asymptotes, we need to compare the degrees of the

numerator and denominator of the rational function.

Step 1: Compare the degrees of the numerator and denominator: The degree

of the numerator is 2 and the degree of the denominator is also 2.

Step 2: Divide the leading coeﬀicients of the numerator and denominator:

The leading coeﬀicients are both 2.

Step 3: Since the degrees and leading coeﬀicients are equal, the horizontal

asymptote is:

y=2

1= 2

Therefore, the horizontal asymptote is y= 2.

3. To find the holes, we need to look for common factors between the nu-

merator and denominator that can be cancelled out.

Step 1: Factor the numerator and denominator: Numerator: 2x2+x−1 =

(2x−1)(x+ 1) Denominator: x2−4x+ 3 = (x−3)(x−1)

Step 2: Simplify the rational function:

f(x) = (2x−1)(x+ 1)

(x−3)(x−1)

Step 3: From the simplified form, we see that the factor (x−1) in both the

numerator and denominator can be cancelled out. This creates a hole at x= 1.

Therefore, the hole in the graph of the function occurs at x= 1.

Question 7

Question

Let f(x) = 2x2−5x+3

x2−4and g(x) = x2−4

x2−5x+6 . Determine the domain of the com-

posite function h(x) = f(g(x)).

Solution

Step 1: Find the domain of g(x): Since g(x) = x2−4

x2−5x+6 , we need to find the

values of xthat make the denominator (x2−5x+ 6) equal to zero. This will

give us the values that xcannot take on.

Setting the denominator equal to zero:

x2−5x+ 6 = 0

Factoring the quadratic:

(x−2)(x−3) = 0

So, x= 2 and x= 3 are the values that xcannot be in order for the

denominator to not be zero. Therefore, the domain of g(x)is all real numbers

except x= 2 and x= 3.

Step 2: Find the domain of h(x): Since h(x) = f(g(x)), we need to find the

values of xsuch that both f(x)and g(x)are defined.