MATH 121 - COLLEGE ALGEBRA -
Rational Functions
Question Bank - Set 5
Liberty University
Question 1
Question
Solve the rational function for x:
3x
2x+ 5 =4
x−1
Solution
Step 1: Multiply both sides by (2x+ 5)(x−1) to clear the denominators:
3x
2x+ 5 ·(2x+ 5)(x−1) = 4
x−1·(2x+ 5)(x−1)
Step 2: Simplify both sides by canceling out common factors:
3x(x−1) = 4(2x+ 5)
Step 3: Expand both sides to get rid of the parentheses:
3x2−3x= 8x+ 20
Step 4: Rearrange the equation by moving all terms to one side:
3x2−11x−20 = 0
Step 5: Factor the quadratic equation:
(3x+ 4)(x−5) = 0
Step 6: Set each factor to zero and solve for x:
3x+ 4 = 0 or x−5 = 0
Step 7: Solve for xin each equation:
3x=−4⇒x=−4
3
x= 5
Therefore, the solutions to the rational function are x=−4
3and x= 5.
Question 2
Question
Let f(x) = x2−5x−6
x2−3x−10 . Find the domain of the function f(x).
Solution
Step 1: The domain of a rational function is all real numbers except for the
values that make the denominator equal to zero, since division by zero is un-
defined. Therefore, we need to find the values of xthat make the denominator
x2−3x−10 equal to zero.
Step 2: Let’s solve the equation x2−3x−10 = 0 to find the values that
make the denominator zero. This is a quadratic equation and we can factor it
as (x−5)(x+ 2) = 0.
Step 3: Setting each factor to zero gives us x−5 = 0 and x+ 2 = 0. Solving
these equations gives us x= 5 and x=−2.
Step 4: Since x= 5 and x=−2make the denominator zero, the domain of
f(x)is all real numbers except x= 5 and x=−2.
Step 5: Therefore, the domain of the function f(x) = x2−5x−6
x2−3x−10 is (−∞,−2) ∪(−2,5) ∪(5,∞).
Question 3
Question
Solve the rational inequality: x+ 1
x−4≤2.
Solution
Step 1: Find the critical points by setting the numerator equal to zero.
x+ 1 = 0
x=−1
Step 2: Find the critical points by setting the denominator equal to zero and
checking the validities of each.
x−4 = 0
x= 4
2
Because x= 4 is excluded from the domain, it is also a critical point.
Step 3: Rewrite the inequality with the critical points. We have x+ 1
x−4≤2.
Now, we rewrite the inequality as two separate inequalities: x+ 1 ≤2(x−4)
and x−4>0.
Step 4: Solve the first inequality x+ 1 ≤2(x−4).
x+ 1 ≤2x−8
1+8≤2x−x
9≤x
Step 5: Solve the second inequality x−4>0.
x > 4
Step 6: Combine the solutions from Step 4 and Step 5. The solution to the
original inequality x+ 1
x−4≤2is 9≤x < 4.
Question 4
Question
Solve the following rational inequality for x:
4x−3
x+ 1 >2
Solution
Step 1: Begin by multiplying both sides of the inequality by (x+ 1) to get rid
of the denominator:
(x+ 1) ·4x−3
x+ 1 >2·(x+ 1)
4x−3>2x+ 2
Step 2: Simplify the inequality by combining like terms:
4x−3>2x+ 2
2x > 5
Step 3: Divide by 2 to solve for x:
x > 5
2
Therefore, the solution to the rational inequality is x > 5
2.
3
Question 5
Question
Find the horizontal asymptotes of the rational function:
f(x) = 4x3−2x2+ 3
2x3+ 5x+ 1
Solution
To find the horizontal asymptotes of the rational function f(x), we need to
consider the behavior of the function as xapproaches infinity and negative
infinity.
Step 1: Find the degrees of the numerator and denominator
Let nbe the degree of the numerator and dbe the degree of the denominator.
In this case, n= 3 and d= 3.
Step 2: Check if the degrees are equal
If n=d, there is a horizontal asymptote at the line y=an
bnwhere anis
the leading coefficient of the numerator and bnis the leading coefficient of the
denominator.
Step 3: Find the leading coefficients
The leading coefficient of the numerator (an) is 4 and the leading coefficient
of the denominator (bn) is 2.
Step 4: Determine the horizontal asymptote
Since n=dand the leading coefficients are 4 and 2 respectively, the hori-
zontal asymptote of the function is given by:
y=4
2= 2
Therefore, the horizontal asymptote of the rational function is y= 2.
Question 6
Question
Solve the inequality: 2x−1
x+ 3 >1
Solution
Step 1: Subtract 1 from both sides of the inequality.
2x−1
x+ 3 −1>0
4
2x−1
x+ 3 −x+ 3
x+ 3 >0
x−4
x+ 3 >0
Step 2: Find the critical points by setting the numerator and denominator
equal to zero and solving for x.
x−4 = 0 =⇒x= 4
x+ 3 = 0 =⇒x=−3
Step 3: Create a sign chart with the critical points and test points to deter-
mine where the expression is positive.
x < −3−3< x < 4x > 4
x−4− − +
x+ 3 −+ +
x−4
x+3 +−+
Step 4: Analyze the sign chart to see where x−4
x+3 >0.
The solution is x∈(−3,4)
Question 7
Question
Given the rational function f(x) = 4x2−7x−12
x2−3x−28 , find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph.
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero
and solving for x.
x2−3x−28 = 0
(x−7)(x+ 4) = 0
Setting each factor to zero gives x= 7 and x=−4. Therefore, the vertical
asymptotes are x= 7 and x=−4.
Step 2: Find the horizontal asymptote by comparing the degrees of the
numerator and denominator. Since the degree of the numerator is 2 and the de-
gree of the denominator is 2, divide the leading coefficients to find the horizontal
asymptote.
y=4
1= 4
Therefore, the horizontal asymptote is y= 4.
5
Step 3: To find any holes in the graph, factor the numerator and denominator
and simplify the function. Factor the numerator 4x2−7x−12 to get (4x+3)(x−
4). Therefore, the function can be simplified as f(x) = (4x+3)(x−4)
(x−7)(x+4) . There are
no common factors that cancel out, so there are no holes in the graph.
In conclusion, the rational function f(x) = 4x2−7x−12
x2−3x−28 has vertical asymp-
totes at x= 7 and x=−4, a horizontal asymptote at y= 4, and no holes in
the graph.
Question 8
Question
Let f(x) = x2+ 3x−10
x2−x−20 be a rational function. Find the vertical asymptotes,
horizontal asymptotes, and x-intercepts (if any) of f(x).
Solution
Step 1: To find the vertical asymptotes of f(x), we need to determine the values
of xthat make the denominator equal to zero. This will give us the vertical
lines that the graph approaches but never crosses. Thus, solve the equation
x2−x−20 = 0 for x.
Step 2: Factor the quadratic equation x2−x−20 = 0 to find its roots
(zeros). We find that (x+ 4)(x−5) = 0. Therefore, x=−4or x= 5. These
are the vertical asymptotes of f(x).
Step 3: To find the horizontal asymptote of f(x), we need to compare the
degrees of the numerator and the denominator. Since both the numerator and
denominator have the same degree (2), the horizontal asymptote is given by the
ratio of the leading coefficients. Thus, the horizontal asymptote is y=1
1= 1.
Step 4: To find the x-intercepts of f(x), we set the numerator equal to zero
and solve for x. Thus, solve the equation x2+ 3x−10 = 0 for x.
Step 5: Factor the quadratic equation x2+ 3x−10 = 0 to find its roots
(zeros). We find that (x+ 5)(x−2) = 0. Therefore, x=−5or x= 2. These
are the x-intercepts of f(x)(where the graph intersects the x-axis).
Therefore, the vertical asymptotes are x=−4and x= 5, the horizontal
asymptote is y= 1, and the x-intercepts are x=−5and x= 2.
Question 9
Question
Solve the rational equation:
1
x−3+3
x+ 2 =2
x2−x−6
6
Solution
Step 1: Find a common denominator for all three fractions.
To add the fractions, the common denominator is the least common multiple
of the denominators x−3,x+ 2, and x2−x−6 = (x−3)(x+ 2).
Therefore, the common denominator is (x−3)(x+ 2).
Step 2: Rewrite each fraction with the common denominator (x−3)(x+ 2).
(x+ 2)
(x+ 2)(x−3) +3(x−3)
(x+ 2)(x−3) =2
(x−3)(x+ 2)
Step 3: Combine the fractions on the left side of the equation.
x+2+3(x−3)
(x+ 2)(x−3) =2
(x−3)(x+ 2)
Simplify the expression by adding the terms in the numerator.
x+2+3x−9
(x+ 2)(x−3) =2
(x−3)(x+ 2)
4x−7
(x+ 2)(x−3) =2
(x−3)(x+ 2)
Step 4: Solve the equation.
Since the denominators are the same, we can equate the numerators.
4x−7 = 2
Solve for x.
4x= 9
x=9
4
Therefore, the solution to the given rational equation is x=9
4.
Question 10
Question
Solve the following rational equation for x:
3x
x+ 4 +2
x−1=5
x2+ 3x−4
7
Solution
Step 1: Simplify the fractions in the equation by finding a common denominator.
3x(x−1)
(x+ 4)(x−1) +2(x+ 4)
(x−1)(x+ 4) =5
(x−1)(x+ 4)
Step 2: Combine the fractions on the left side of the equation.
3x(x−1) + 2(x+ 4)
(x+ 4)(x−1) =5
(x−1)(x+ 4)
Step 3: Expand and simplify the numerators.
3x2−3x+ 2x+ 8
(x+ 4)(x−1) =5
(x−1)(x+ 4)
3x2−x+ 8
(x+ 4)(x−1) =5
(x−1)(x+ 4)
Step 4: Multiply both sides by the common denominator to clear the frac-
tions.
(3x2−x+ 8)(x−1) = 5
Step 5: Expand and simplify the left side of the equation.
3x3−3x2+ 8x−x2+x−8 = 5
3x3−4x2+ 9x−8 = 5
3x3−4x2+ 9x−13 = 0
Step 6: Factor the cubic equation if possible, or use numerical methods to
find the roots.
Question 11
Question
Find the domain of the rational function:
f(x) = 2x+ 1
x2−4
Solution
Step 1: To find the domain of the rational function, we need to identify the values
of xfor which the function is undefined. The function is undefined when the
denominator is equal to zero. Therefore, we must solve the equation x2−4 = 0
to find the values of xthat make the function undefined.
8
Step 2: Let’s solve the equation x2−4 = 0 to find the values that make the
function undefined.
Step 3: Add 4 to both sides to isolate x2:
x2= 4
Step 4: Take the square root of both sides:
x=±2
Step 5: The values of xthat make the function undefined are x= 2 and
x=−2. Therefore, the domain of the rational function is all real numbers
except x= 2 and x=−2, which can be expressed as:
(−∞,−2) ∪(−2,2) ∪(2,∞)
Question 12
Question
Solve the following rational equation for x:
4
x−2+3
x+ 1 =5
x2−x−2
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation. Step 2: Rewrite the equation with a common denominator. Step 3:
Simplify the equation by combining the fractions. Step 4: Factor the denomi-
nator on the right side of the equation. Step 5: Rewrite the equation with the
factored denominator. Step 6: Multiply both sides of the equation by x2−x−2
to eliminate the fractions. Step 7: Solve the resulting polynomial equation.
Step 8: Check for extraneous solutions.
Step 1: Find a common denominator for the fractions on the left side of
the equation.
The common denominator for x−2and x+ 1 is (x−2)(x+ 1) = x2−x−2.
Step 2: Rewrite the equation with a common denominator.
4(x+ 1)
(x−2)(x+ 1) +3(x−2)
(x−2)(x+ 1) =5
x2−x−2
Step 3: Simplify the equation by combining the fractions.
4x+4+3x−6
x2−x−2=5
x2−x−2
7x−2
x2−x−2=5
x2−x−2
9
Step 4: Factor the denominator on the right side of the equation.
The denominator x2−x−2factors as (x−2)(x+ 1).
Step 5: Rewrite the equation with the factored denominator.
7x−2
(x−2)(x+ 1) =5
(x−2)(x+ 1)
Step 6: Multiply both sides of the equation by x2−x−2to eliminate the
fractions.
(7x−2) = 5
Step 7: Solve the resulting polynomial equation.
7x−2 = 5
7x= 7
x= 1
Step 8: Check for extraneous solutions.
Check that x= 1 does not make any of the original denominators zero.
The values x= 1,x=−1, and x= 2 do not make any denominator zero, so
the solution x= 1 is valid.
Question 13
Question
Simplify the following rational expression:
3x2−x−4
x2−5x+ 6∇ · x2−4x−5
x2−3x−10
Solution
Step 1: Factorize the numerators and denominators of both fractions. Step
2: Rewrite the division as multiplication by the reciprocal of the second frac-
tion. Step 3: Simplify by canceling out common factors in the numerator and
denominator. Step 4: Factor the resulting expression if possible.
Step 1: Factorize the expressions:
3x2−x−4
x2−5x+ 6 =(3x+ 4)(x−1)
(x−3)(x−2)
x2−4x−5
x2−3x−10 =(x−5)(x+ 1)
(x−5)(x+ 2)
10
Step 2: Now rewrite the division as multiplication by the reciprocal:
(3x+ 4)(x−1)
(x−3)(x−2) ×(x+ 2)
(x+ 1)
Step 3: Now simplify by canceling common factors:
3x+ 4
x−3×x+ 2
x−2
Step 4: Factorize the resulting expression if possible:
(3x+ 4)(x+ 2)
(x−3)(x−2)
Question 14
Question
Solve the rational equation:
3
x−4−5
x+ 3 =4
x2−x−12
Solution
Step 1: Find a common denominator for all fractions in the equation. Since the
denominators are (x−4),(x+3), and (x2−x−12), we can rewrite the equation
with the common denominator as (x−4)(x+ 3).
Step 2: Rewrite the equation with the common denominator and simplify:
3(x+ 3)
(x−4)(x+ 3) −5(x−4)
(x−4)(x+ 3) =4
x2−x−12
Step 3: Combine the fractions on the left side of the equation:
3x+ 9 −5x+ 20
(x−4)(x+ 3) =4
x2−x−12
Step 4: Simplify the left side:
−2x+ 29
(x−4)(x+ 3) =4
x2−x−12
Step 5: Cross multiply to get rid of the fractions:
(−2x+ 29)(x2−x−12) = 4(x−4)(x+ 3)
Step 6: Expand both sides of the equation:
−2x3+ 2x2+ 24x−29x2+ 29x+ 348 = 4(x2−x−12)
11
Step 7: Simplify the equation further:
−2x3−27x2+ 53x+ 348 = 4x2−4x−48
Step 8: Rearrange the terms to set the equation to zero:
−2x3−27x2+ 53x+ 348 −4x2+ 4x+ 48 = 0
Step 9: Simplify the equation:
−2x3−31x2+ 57x+ 396 = 0
This is the final equation after simplifying.
Question 15
Question
Find the domain of the rational function:
f(x) = x2−5x+ 6
x2−4x−5.
Solution
Step 1: To find the domain of a rational function, we must identify any values
of xthat would make the denominator equal to zero, since division by zero is
undefined.
Step 2: Set the denominator of the rational function equal to zero and solve
for x:
x2−4x−5 = 0.
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0.
Step 4: Solve for x:
x−5 = 0 or x+ 1 = 0.
Step 5: So, the values that make the denominator zero are x= 5 and x=−1.
These values must be excluded from the domain.
Step 6: Therefore, the domain of the rational function f(x) = x2−5x+6
x2−4x−5is all
real numbers except x= 5 and x=−1.
Step 7: Thus, the domain is (−∞,−1) ∪(−1,5) ∪(5,∞).
12
Question 16
Question
Simplify the following rational expression:
3x2+ 7x−6
x2−4x−5∇ · 4x2−x−5
3x2−13x+ 12
Solution
Step 1: Factor the numerator and denominator of both fractions. Step 2:
Rewrite the division as multiplication by the reciprocal. Step 3: Multiply the
two fractions together and simplify.
Step 1: Factor the numerator and denominator of both fractions. The first
fraction: 3x2+ 7x−6
x2−4x−5=(3x−2)(x+ 3)
(x−5)(x+ 1)
The second fraction:
4x2−x−5
3x2−13x+ 12 =(4x+ 5)(x−1)
(3x−4)(x−3)
Step 2: Rewrite the division as multiplication by the reciprocal. This is
equivalent to multiplying by the reciprocal of the second fraction:
(3x−2)(x+ 3)
(x−5)(x+ 1) ×(3x−4)(x−3)
(4x+ 5)(x−1)
Step 3: Multiply the two fractions together and simplify. Multiplying the
fractions, we get:
(3x−2)(x+ 3)(3x−4)(x−3)
(x−5)(x+ 1)(4x+ 5)(x−1)
Expanding and simplifying, we get:
9x2−9x−8x+ 8
(x2−4x−5)(4x2−x−5)
=9x2−17x+ 8
(x−5)(x+ 1)(4x+ 5)(x−1)
Therefore, the simplified form of the rational expression is 9x2−17x+8
(x−5)(x+1)(4x+5)(x−1) .
Question 17
Question
Determine the domain of the rational function given by
f(x) = 3x−2
2x2+x−3.
13
Solution
Step 1: To find the domain of the rational function, we need to identify any
values of xthat would make the denominator equal to zero, as division by zero
is undefined.
Step 2: Set the denominator equal to zero and solve for x:
2x2+x−3 = 0.
Step 3: Factor the quadratic equation:
2x2+ 3x−2x−3 = 0
2x(x+ 1) −3(x+ 1) = 0
(2x−3)(x+ 1) = 0.
Step 4: Set each factor equal to zero and solve for x:
2x−3 = 0 or x+ 1 = 0
2x= 3 or x=−1
x=3
2or x=−1.
Step 5: The values x=3
2and x=−1make the denominator equal to zero.
Therefore, the domain of the rational function is all real numbers except x=3
2
and x=−1, or in interval notation, (−∞,−1) ∪(−1,3
2)∪(3
2,∞).
Question 18
Question
Simplify the following rational expression:
3x2−9x
x2−4
Solution
Step 1: Factor out common terms in the numerator and simplify where possible.
3x2−9x
x2−4=3x(x−3)
(x+ 2)(x−2)
Step 2: Simplify the expression by canceling out common factors.
3x(x−3)
(x+ 2)(x−2) =3(x−3)
x+ 2
Therefore, the simplified form of the given rational expression is 3(x−3)
x+2 .
14
Question 19
Question
Simplify the following rational expression:
6x2−3x−15
x2−4x−12
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify the ratio-
nal expression by canceling out any common factors. Step 3: Check for any
extraneous solutions.
Step 1: Factor the numerator and denominator. The numerator 6x2−3x−
15 can be factored as 3(2x2−x−5). The denominator x2−4x−12 can be
factored as (x−6)(x+ 2).
Step 2: Simplify the rational expression by canceling out any common
factors. 6x2−3x−15
x2−4x−12 =3(2x2−x−5)
(x−6)(x+ 2) =3(2x+ 5)(x−1)
(x−6)(x+ 2)
Step 3: Check for any extraneous solutions. Since the simplified expression
does not have any common factors that can be canceled further, we have the
final simplified form:
3(2x+ 5)(x−1)
(x−6)(x+ 2)
Question 20
Question
Solve the following rational inequality:
3x−1
x−2≥5
Solution
Step 1: Begin by rewriting the inequality with a zero on one side:
3x−1
x−2−5≥0
Step 2: Find a common denominator for the left side of the inequality:
3x−1
x−2−5(x−2)
x−2≥0
15
Step 3: Combine the fractions on the left side:
3x−1−5x+ 10
x−2≥0
Step 4: Simplify the numerator:
−2x+ 9
x−2≥0
Step 5: Now, find the critical points by setting the numerator and denomi-
nator equal to zero:
−2x+ 9 = 0 ⇒x=9
2
x−2 = 0 ⇒x= 2
Step 6: Create a sign chart using the critical points to test the intervals
between the critical points and beyond:
x < 9
2
9
2< x < 2x= 2 x > 2
−2x+ 9 −+ + +
x−2− − 0 +
−2x+9
x−2+−undefined +
Step 7: Determine the solution from the sign chart:
x∈(−∞,9
2)∪(2,∞)
Question 21
Question
Given the rational function f(x) = 3x2+4x−1
x2−5x+6 , find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph of f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to zero and
solve for x.
x2−5x+ 6 = 0
(x−2)(x−3) = 0
Therefore, x= 2 and x= 3 are the vertical asymptotes.
Step 2: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator. Since the degree of the numerator is equal to the degree
of the denominator, we divide the leading coefficients to find the horizontal
asymptote.
lim
x→∞
3x2
x2= lim
x→∞ 3 = 3
16
Therefore, the horizontal asymptote is y= 3.
Step 3: To find any holes in the graph, factor the numerator and see if there
are any common factors with the denominator.
3x2+ 4x−1 = (3x−1)(x+ 1)
There are no common factors between the numerator and denominator, so
there are no holes in the graph.
In conclusion, the rational function f(x) = 3x2+4x−1
x2−5x+6 has vertical asymptotes
at x= 2 and x= 3, a horizontal asymptote at y= 3, and no holes in its graph.
Question 22
Question
Let f(x) = x2−4
x2−5x+6 be a rational function. Determine the vertical asymptotes,
horizontal asymptotes, and holes, if any, of the graph of f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to zero and
solve for x.
x2−5x+ 6 = 0
(x−2)(x−3) = 0
x= 2 or x= 3
Thus, the vertical asymptotes are x= 2 and x= 3.
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and the denominator. Since the degrees are the same (2), we check
the leading coefficients. The horizontal asymptote is the ratio of the leading
coefficients:
lim
x→∞
x2−4
x2−5x+ 6 = 1
Therefore, the horizontal asymptote is y= 1.
Step 3: To find any holes, factor and cancel out common factors in the
numerator and denominator:
f(x) = x2−4
x2−5x+ 6 =(x+ 2)(x−2)
(x−2)(x−3)
We see that there is a hole at x= 2, where the factor (x−2) cancels out.
So, the graph of f(x)has vertical asymptotes at x= 2 and x= 3, a horizontal
asymptote at y= 1, and a hole at x= 2.
17
Question 23
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes for the ra-
tional function:
f(x) = 2x3−5x2−3x
x2−4
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to zero and
solve:
x2−4 = 0
(x+ 2)(x−2) = 0
x=−2,2
So, the vertical asymptotes are x=−2and x= 2.
Step 2: To find the horizontal asymptotes, compare the degrees of the nu-
merator and denominator: - If the degree of the numerator is less than the
degree of the denominator, there is a horizontal asymptote at y= 0. - If the
degree of the numerator is equal to the degree of the denominator, divide the
leading coefficients to find the horizontal asymptote. - If the degree of the nu-
merator is greater than the degree of the denominator, there is no horizontal
asymptote.
Here, the degree of the numerator is 3 and the degree of the denominator is
2, so there is no horizontal asymptote.
Step 3: To find any holes in the graph, factor the numerator and denominator
and simplify: Factor the numerator: 2x3−5x2−3x=x(2x2−5x−3) =
x(2x+ 1)(x−3)
So, the simplified form of the function is:
f(x) = x(2x+ 1)(x−3)
(x+ 2)(x−2)
Since there is a common factor of (x−2) in both the numerator and denom-
inator, there is a hole at x= 2.
Therefore, the vertical asymptotes are x=−2and x= 2, there is no hori-
zontal asymptote, and there is a hole at x= 2.
Question 24
Question
Solve the following rational inequality for x:
x−2
x+ 3 ≤1
2
18
Solution
To solve this rational inequality, we will first find the critical points by setting
the numerator of the left-hand side equal to the right-hand side. We will then
test each interval created by the critical points to determine where the inequality
holds true.
Step 1: Find the critical points
x−2
x+ 3 =1
2
2(x−2) = (x+ 3)
2x−4 = x+ 3
x= 7
So, the critical point is x= 7.
Step 2: Test the intervals We have three intervals: (−∞,−3),(−3,7),
and (7,∞).
Testing the interval (−∞,−3):Pick x=−4, then −4−2
−4+3 =−6
−1= 6 >1
2
So, this interval is not a solution.
Testing the interval (−3,7):Pick x= 0, then 0−2
0+3 =−2
3<1
2
So, this interval is a solution.
Testing the interval (7,∞):Pick x= 8, then 8−2
8+3 =6
11 <1
2
So, this interval is not a solution.
Step 3: Conclusion The solution to the inequality x−2
x+3 ≤1
2is x∈(−3,7].
Question 25
Question
Given the rational function f(x) = 3x3+2x2−5x+1
x2−4x+3 , determine the vertical asymp-
totes, horizontal asymptotes, and the x-intercepts, if any.
Solution
To find the vertical asymptotes of the rational function f(x), we need to find
the values of xwhich make the denominator equal to zero. These values will
give us the equations of the vertical asymptotes.
Step 1: Find the vertical asymptotes:
Set the denominator equal to zero and solve for x:
x2−4x+ 3 = 0
Factoring the quadratic equation:
(x−3)(x−1) = 0
So, the vertical asymptotes are at x= 1 and x= 3.
19
Step 2: Find the horizontal asymptote:
To find the horizontal asymptote, we compare the degrees of the numerator
and the denominator of the rational function f(x).
Since the degree of the numerator is equal to the degree of the denominator,
we look at the ratio of the leading coefficients:
3
1= 3
So, the horizontal asymptote is y= 3.
Step 3: Find the x-intercepts:
To find the x-intercepts, we set the numerator equal to zero and solve for x:
3x3+ 2x2−5x+ 1 = 0
Unfortunately, there is no straightforward way to solve this cubic equation
by factoring or using the quadratic formula. Hence, we need to use numerical
methods or graphical methods to approximate the x-intercepts of the function.
Therefore, the vertical asymptotes are x= 1 and x= 3, the horizontal
asymptote is y= 3, and the x-intercepts need to be approximated.
Question 26
Question
Solve the rational inequality: 2x−5
x−3≤1.
Solution
Step 1: Begin by solving the related equation 2x−5
x−3= 1.
2x−5
x−3= 1
2x−5 = x−3
2x−x= 3 −5
x=−2
Step 2: Next, determine the critical points by finding where the denominator
is equal to zero.
x−3 = 0
x= 3
Step 3: Now, create a number line and mark the critical points x=−2and
x= 3.
−2 3
20
Step 4: Test the intervals determined by the critical points x=−2and x= 3
by selecting test values in each interval and determining the sign of 2x−5
x−3−1. -
For x < −2, choose x=−3:
2(−3) −5
−3−3−1 = −11
−6−1 = 11
6−1>0
- For −2< x < 3, choose x= 0:
2(0) −5
0−3−1 = −5
−3−1 = 5
3−1>0
- For x > 3, choose x= 4:
2(4) −5
4−3−1 = 3
1−1 = 3 −1>0
Step 5: Therefore, the solution to the inequality 2x−5
x−3≤1is x∈(−∞,−2] ∪
(3,∞).
Question 27
Question
Simplify the following rational expression:
3x2−2x−5
2x2+ 7x−15
Solution
Step 1: Factor both the numerator and the denominator.
Step 2: Factor 3x2−2x−5.
3x2−2x−5 = (3x+ 1)(x−5)
Step 3: Factor 2x2+ 7x−15.
2x2+ 7x−15 = (2x−3)(x+ 5)
Step 4: Rewrite the expression with factored numerator and denominator.
(3x+ 1)(x−5)
(2x−3)(x+ 5)
Therefore, the simplified form of the given rational expression is (3x+1)(x−5)
(2x−3)(x+5) .
21
Question 28
Question
Find the domain of the rational function:
f(x) = x2−3x−4
x2+ 2x−8
Solution
Step 1: We start by finding the values of xfor which the denominator is equal
to zero, since division by zero is undefined.
x2+ 2x−8 = 0
Factoring the quadratic equation, we get:
(x+ 4)(x−2) = 0
So, the values of xthat make the denominator zero are x=−4and x= 2.
Step 2: Next, we need to consider any restrictions on the domain due to
simplifying the expression. Since the numerator and denominator are both
polynomials, the only restrictions on the domain occur when the denominator
is zero.
Step 3: Therefore, the domain of the rational function is all real numbers
except for the values of xthat make the denominator equal to zero. Hence, the
domain of f(x)is (−∞,−4) ∪(−4,2) ∪(2,∞).
Question 29
Question
Let f(x) = 2x2−5x−3
x2−4x−5. Find the domain of f(x).
Solution
Step 1: To find the domain of f(x), we need to identify any values of xthat
would make the denominator x2−4x−5equal to zero, since division by zero is
undefined.
Step 2: To find the values of xthat make the denominator zero, we solve
the equation x2−4x−5 = 0.
Step 3: Factoring the quadratic equation x2−4x−5 = 0, we get (x−5)(x+
1) = 0.
Step 4: Setting each factor to zero and solving for x, we find x= 5 and
x=−1.
Step 5: Therefore, the domain of the function f(x)is all real numbers except
x= 5 and x=−1.
Step 6: In interval notation, the domain of f(x)is (−∞,−1)∪(−1,5)∪(5,∞).
22
Question 30
Question
Solve the inequality: 2x+1
x−2<3.
Solution
Step 1: Find the critical points by setting the inequality equal to zero:
2x+ 1
x−2= 3.
Step 2: Multiply both sides by x−2to get rid of the denominator:
2x+ 1 = 3(x−2).
Step 3: Expand the right side and simplify the equation:
2x+ 1 = 3x−6.
Step 4: Move all terms to one side to get the equation in standard form:
2x+ 1 −3x+ 6 = 0.
Step 5: Simplify the equation further:
−x+ 7 = 0.
Step 6: Solve for xby isolating x:
x= 7.
So, x= 7 is the critical point.
Step 7: Test the intervals defined by the critical point x= 7 on the original
inequality. Choose a test point on the intervals x < 7and x > 7: - Test ˜x= 6
for x < 7:2(6) + 1
6−2=13
4<3.
- Test ˜x= 8 for x > 7:2(8) + 1
8−2=17
6≮3.
Step 8: Write the final solution as an inequality: Since x < 7satisfies the
inequality, the solution to the inequality 2x+1
x−2<3is:
x < 7.
23
Question 31
Question
Find the vertical asymptotes, horizontal asymptotes, and holes (if any) of the
rational function:
f(x) = 4x3−16x
x2−2x−3
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes, we need to determine where the denominator
is equal to zero:
x2−2x−3 = 0
Factoring the quadratic gives:
x2−2x−3 = (x−3)(x+ 1) = 0
Setting each factor equal to zero, we find the potential vertical asymptotes
at x= 3 and x=−1.
Step 2: Horizontal Asymptotes
To find the horizontal asymptotes, we compare the degrees of the numerator
and denominator: - If the degree of the numerator is greater, there is no hor-
izontal asymptote. - If the degrees are equal, the horizontal asymptote is the
ratio of the leading coefficients. - If the degree of the denominator is greater,
the horizontal asymptote is the x-axis, y= 0.
In our case, the degree of the numerator is 3, and the degree of the denomi-
nator is 2. So, there is no horizontal asymptote.
Step 3: Holes
To find any potential holes in the graph, we look for common factors between
the numerator and the denominator. Canceling common factors gives:
f(x) = 4x(x2−4)
(x−3)(x+ 1) =4x(x+ 2)(x−2)
(x−3)(x+ 1)
This simplification shows a hole at x= 2, as (x−2) cancels out in the
expression.
Therefore, the vertical asymptotes are x= 3 and x=−1, there are no
horizontal asymptotes, and there is a hole at x= 2.
Question 32
Question
Solve the rational inequality:
x2−4x−5
x+ 4 ≥0
24
Solution
Step 1: Find the critical points by setting the numerator and denominator equal
to zero.
x2−4x−5 = 0 =⇒(x−5)(x+ 1) = 0
x+ 4 = 0 =⇒x=−4
So the critical points are x= 5,x=−1, and x=−4.
Step 2: Create a sign chart using the critical points.
x < −4−4< x < −1−1< x < 5x > 5
x+ 4 −+ + +
x2−4x−5 + + −+
x2−4x−5
x+4 + + −+
Step 3: Determine the solution based on the sign chart. The inequality is
satisfied when x2−4x−5
x+4 ≥0, which means the function is positive or zero. So
the solution is x∈(−∞,−4] ∪(−1,5].
Question 33
Question
Find the domain of the rational function:
f(x) = 2x2−5x
x2−4x−12.
Solution
Step 1: We need to find the values of xfor which the denominator of the rational
function is not equal to zero, since division by zero is undefined. Therefore, we
set the denominator x2−4x−12 not equal to zero and solve for x:
x2−4x−12 = 0.
Step 2: Factor the quadratic equation x2−4x−12:
x2−4x−12 = 0.
(x−6)(x+ 2) = 0.
Step 3: Set each factor to zero and solve for x:
x−6 = 0 =⇒x= 6,
x+ 2 = 0 =⇒x=−2.
Step 4: The domain of the rational function f(x)is all real numbers except
x= 6 and x=−2. Therefore, the domain of f(x)is:
(−∞,−2) ∪(−2,6) ∪(6,∞).
25
Question 34
Question
Simplify the rational function:
f(x) = x2−2x−8
x2−5x+ 6
Solution
Step 1: Factor both the numerator and the denominator:
f(x) = (x−4)(x+ 2)
(x−3)(x−2)
Step 2: Simplify the expression by canceling out common factors:
f(x) = x−4
x−3
Therefore, the simplified form of the rational function is f(x) = x−4
x−3.
Question 35
Question
Find the domain of the rational function: f(x) = 1
x2−4x−5.
Solution
Step 1: We first need to find where the denominator is equal to zero since the
function is undefined at those points.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0
Step 4: Find the roots:
x= 5 or x=−1
Step 5: Therefore, the function f(x)is undefined at x= 5 and x=−1, so
the domain is all real numbers except x= 5 and x=−1. In interval notation,
the domain of f(x)is (−∞,−1) ∪(−1,5) ∪(5,∞).
26
Step 7: Solve for xin each equation:
3x=−4⇒x=−4
3
x= 5
Therefore, the solutions to the rational function are x=−4
3and x= 5.
Question 2
Question
Let f(x) = x2−5x−6
x2−3x−10 . Find the domain of the function f(x).
Solution
Step 1: The domain of a rational function is all real numbers except for the
values that make the denominator equal to zero, since division by zero is un-
defined. Therefore, we need to find the values of xthat make the denominator
x2−3x−10 equal to zero.
Step 2: Let’s solve the equation x2−3x−10 = 0 to find the values that
make the denominator zero. This is a quadratic equation and we can factor it
as (x−5)(x+ 2) = 0.
Step 3: Setting each factor to zero gives us x−5 = 0 and x+ 2 = 0. Solving
these equations gives us x= 5 and x=−2.
Step 4: Since x= 5 and x=−2make the denominator zero, the domain of
f(x)is all real numbers except x= 5 and x=−2.
Step 5: Therefore, the domain of the function f(x) = x2−5x−6
x2−3x−10 is (−∞,−2) ∪(−2,5) ∪(5,∞).
Question 3
Question
Solve the rational inequality: x+ 1
x−4≤2.
Solution
Step 1: Find the critical points by setting the numerator equal to zero.
x+ 1 = 0
x=−1
Step 2: Find the critical points by setting the denominator equal to zero and
checking the validities of each.
x−4 = 0
x= 4
2
Because x= 4 is excluded from the domain, it is also a critical point.
Step 3: Rewrite the inequality with the critical points. We have x+ 1
x−4≤2.
Now, we rewrite the inequality as two separate inequalities: x+ 1 ≤2(x−4)
and x−4>0.
Step 4: Solve the first inequality x+ 1 ≤2(x−4).
x+ 1 ≤2x−8
1+8≤2x−x
9≤x
Step 5: Solve the second inequality x−4>0.
x > 4
Step 6: Combine the solutions from Step 4 and Step 5. The solution to the
original inequality x+ 1
x−4≤2is 9≤x < 4.
Question 4
Question
Solve the following rational inequality for x:
4x−3
x+ 1 >2
Solution
Step 1: Begin by multiplying both sides of the inequality by (x+ 1) to get rid
of the denominator:
(x+ 1) ·4x−3
x+ 1 >2·(x+ 1)
4x−3>2x+ 2
Step 2: Simplify the inequality by combining like terms:
4x−3>2x+ 2
2x > 5
Step 3: Divide by 2 to solve for x:
x > 5
2
Therefore, the solution to the rational inequality is x > 5
2.
3
Question 5
Question
Find the horizontal asymptotes of the rational function:
f(x) = 4x3−2x2+ 3
2x3+ 5x+ 1
Solution
To find the horizontal asymptotes of the rational function f(x), we need to
consider the behavior of the function as xapproaches infinity and negative
infinity.
Step 1: Find the degrees of the numerator and denominator
Let nbe the degree of the numerator and dbe the degree of the denominator.
In this case, n= 3 and d= 3.
Step 2: Check if the degrees are equal
If n=d, there is a horizontal asymptote at the line y=an
bnwhere anis
the leading coefficient of the numerator and bnis the leading coefficient of the
denominator.
Step 3: Find the leading coefficients
The leading coefficient of the numerator (an) is 4 and the leading coefficient
of the denominator (bn) is 2.
Step 4: Determine the horizontal asymptote
Since n=dand the leading coefficients are 4 and 2 respectively, the hori-
zontal asymptote of the function is given by:
y=4
2= 2
Therefore, the horizontal asymptote of the rational function is y= 2.
Question 6
Question
Solve the inequality: 2x−1
x+ 3 >1
Solution
Step 1: Subtract 1 from both sides of the inequality.
2x−1
x+ 3 −1>0
4
2x−1
x+ 3 −x+ 3
x+ 3 >0
x−4
x+ 3 >0
Step 2: Find the critical points by setting the numerator and denominator
equal to zero and solving for x.
x−4 = 0 =⇒x= 4
x+ 3 = 0 =⇒x=−3
Step 3: Create a sign chart with the critical points and test points to deter-
mine where the expression is positive.
x < −3−3< x < 4x > 4
x−4− − +
x+ 3 −+ +
x−4
x+3 +−+
Step 4: Analyze the sign chart to see where x−4
x+3 >0.
The solution is x∈(−3,4)
Question 7
Question
Given the rational function f(x) = 4x2−7x−12
x2−3x−28 , find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph.
Solution
Step 1: Find the vertical asymptotes by setting the denominator equal to zero
and solving for x.
x2−3x−28 = 0
(x−7)(x+ 4) = 0
Setting each factor to zero gives x= 7 and x=−4. Therefore, the vertical
asymptotes are x= 7 and x=−4.
Step 2: Find the horizontal asymptote by comparing the degrees of the
numerator and denominator. Since the degree of the numerator is 2 and the de-
gree of the denominator is 2, divide the leading coefficients to find the horizontal
asymptote.
y=4
1= 4
Therefore, the horizontal asymptote is y= 4.
5
Step 3: To find any holes in the graph, factor the numerator and denominator
and simplify the function. Factor the numerator 4x2−7x−12 to get (4x+3)(x−
4). Therefore, the function can be simplified as f(x) = (4x+3)(x−4)
(x−7)(x+4) . There are
no common factors that cancel out, so there are no holes in the graph.
In conclusion, the rational function f(x) = 4x2−7x−12
x2−3x−28 has vertical asymp-
totes at x= 7 and x=−4, a horizontal asymptote at y= 4, and no holes in
the graph.
Question 8
Question
Let f(x) = x2+ 3x−10
x2−x−20 be a rational function. Find the vertical asymptotes,
horizontal asymptotes, and x-intercepts (if any) of f(x).
Solution
Step 1: To find the vertical asymptotes of f(x), we need to determine the values
of xthat make the denominator equal to zero. This will give us the vertical
lines that the graph approaches but never crosses. Thus, solve the equation
x2−x−20 = 0 for x.
Step 2: Factor the quadratic equation x2−x−20 = 0 to find its roots
(zeros). We find that (x+ 4)(x−5) = 0. Therefore, x=−4or x= 5. These
are the vertical asymptotes of f(x).
Step 3: To find the horizontal asymptote of f(x), we need to compare the
degrees of the numerator and the denominator. Since both the numerator and
denominator have the same degree (2), the horizontal asymptote is given by the
ratio of the leading coefficients. Thus, the horizontal asymptote is y=1
1= 1.
Step 4: To find the x-intercepts of f(x), we set the numerator equal to zero
and solve for x. Thus, solve the equation x2+ 3x−10 = 0 for x.
Step 5: Factor the quadratic equation x2+ 3x−10 = 0 to find its roots
(zeros). We find that (x+ 5)(x−2) = 0. Therefore, x=−5or x= 2. These
are the x-intercepts of f(x)(where the graph intersects the x-axis).
Therefore, the vertical asymptotes are x=−4and x= 5, the horizontal
asymptote is y= 1, and the x-intercepts are x=−5and x= 2.
Question 9
Question
Solve the rational equation:
1
x−3+3
x+ 2 =2
x2−x−6
6
Solution
Step 1: Find a common denominator for all three fractions.
To add the fractions, the common denominator is the least common multiple
of the denominators x−3,x+ 2, and x2−x−6 = (x−3)(x+ 2).
Therefore, the common denominator is (x−3)(x+ 2).
Step 2: Rewrite each fraction with the common denominator (x−3)(x+ 2).
(x+ 2)
(x+ 2)(x−3) +3(x−3)
(x+ 2)(x−3) =2
(x−3)(x+ 2)
Step 3: Combine the fractions on the left side of the equation.
x+2+3(x−3)
(x+ 2)(x−3) =2
(x−3)(x+ 2)
Simplify the expression by adding the terms in the numerator.
x+2+3x−9
(x+ 2)(x−3) =2
(x−3)(x+ 2)
4x−7
(x+ 2)(x−3) =2
(x−3)(x+ 2)
Step 4: Solve the equation.
Since the denominators are the same, we can equate the numerators.
4x−7 = 2
Solve for x.
4x= 9
x=9
4
Therefore, the solution to the given rational equation is x=9
4.
Question 10
Question
Solve the following rational equation for x:
3x
x+ 4 +2
x−1=5
x2+ 3x−4
7
Solution
Step 1: Simplify the fractions in the equation by finding a common denominator.
3x(x−1)
(x+ 4)(x−1) +2(x+ 4)
(x−1)(x+ 4) =5
(x−1)(x+ 4)
Step 2: Combine the fractions on the left side of the equation.
3x(x−1) + 2(x+ 4)
(x+ 4)(x−1) =5
(x−1)(x+ 4)
Step 3: Expand and simplify the numerators.
3x2−3x+ 2x+ 8
(x+ 4)(x−1) =5
(x−1)(x+ 4)
3x2−x+ 8
(x+ 4)(x−1) =5
(x−1)(x+ 4)
Step 4: Multiply both sides by the common denominator to clear the frac-
tions.
(3x2−x+ 8)(x−1) = 5
Step 5: Expand and simplify the left side of the equation.
3x3−3x2+ 8x−x2+x−8 = 5
3x3−4x2+ 9x−8 = 5
3x3−4x2+ 9x−13 = 0
Step 6: Factor the cubic equation if possible, or use numerical methods to
find the roots.
Question 11
Question
Find the domain of the rational function:
f(x) = 2x+ 1
x2−4
Solution
Step 1: To find the domain of the rational function, we need to identify the values
of xfor which the function is undefined. The function is undefined when the
denominator is equal to zero. Therefore, we must solve the equation x2−4 = 0
to find the values of xthat make the function undefined.
8
Step 2: Let’s solve the equation x2−4 = 0 to find the values that make the
function undefined.
Step 3: Add 4 to both sides to isolate x2:
x2= 4
Step 4: Take the square root of both sides:
x=±2
Step 5: The values of xthat make the function undefined are x= 2 and
x=−2. Therefore, the domain of the rational function is all real numbers
except x= 2 and x=−2, which can be expressed as:
(−∞,−2) ∪(−2,2) ∪(2,∞)
Question 12
Question
Solve the following rational equation for x:
4
x−2+3
x+ 1 =5
x2−x−2
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation. Step 2: Rewrite the equation with a common denominator. Step 3:
Simplify the equation by combining the fractions. Step 4: Factor the denomi-
nator on the right side of the equation. Step 5: Rewrite the equation with the
factored denominator. Step 6: Multiply both sides of the equation by x2−x−2
to eliminate the fractions. Step 7: Solve the resulting polynomial equation.
Step 8: Check for extraneous solutions.
Step 1: Find a common denominator for the fractions on the left side of
the equation.
The common denominator for x−2and x+ 1 is (x−2)(x+ 1) = x2−x−2.
Step 2: Rewrite the equation with a common denominator.
4(x+ 1)
(x−2)(x+ 1) +3(x−2)
(x−2)(x+ 1) =5
x2−x−2
Step 3: Simplify the equation by combining the fractions.
4x+4+3x−6
x2−x−2=5
x2−x−2
7x−2
x2−x−2=5
x2−x−2
9
Step 4: Factor the denominator on the right side of the equation.
The denominator x2−x−2factors as (x−2)(x+ 1).
Step 5: Rewrite the equation with the factored denominator.
7x−2
(x−2)(x+ 1) =5
(x−2)(x+ 1)
Step 6: Multiply both sides of the equation by x2−x−2to eliminate the
fractions.
(7x−2) = 5
Step 7: Solve the resulting polynomial equation.
7x−2 = 5
7x= 7
x= 1
Step 8: Check for extraneous solutions.
Check that x= 1 does not make any of the original denominators zero.
The values x= 1,x=−1, and x= 2 do not make any denominator zero, so
the solution x= 1 is valid.
Question 13
Question
Simplify the following rational expression:
3x2−x−4
x2−5x+ 6∇ · x2−4x−5
x2−3x−10
Solution
Step 1: Factorize the numerators and denominators of both fractions. Step
2: Rewrite the division as multiplication by the reciprocal of the second frac-
tion. Step 3: Simplify by canceling out common factors in the numerator and
denominator. Step 4: Factor the resulting expression if possible.
Step 1: Factorize the expressions:
3x2−x−4
x2−5x+ 6 =(3x+ 4)(x−1)
(x−3)(x−2)
x2−4x−5
x2−3x−10 =(x−5)(x+ 1)
(x−5)(x+ 2)
10
Step 2: Now rewrite the division as multiplication by the reciprocal:
(3x+ 4)(x−1)
(x−3)(x−2) ×(x+ 2)
(x+ 1)
Step 3: Now simplify by canceling common factors:
3x+ 4
x−3×x+ 2
x−2
Step 4: Factorize the resulting expression if possible:
(3x+ 4)(x+ 2)
(x−3)(x−2)
Question 14
Question
Solve the rational equation:
3
x−4−5
x+ 3 =4
x2−x−12
Solution
Step 1: Find a common denominator for all fractions in the equation. Since the
denominators are (x−4),(x+3), and (x2−x−12), we can rewrite the equation
with the common denominator as (x−4)(x+ 3).
Step 2: Rewrite the equation with the common denominator and simplify:
3(x+ 3)
(x−4)(x+ 3) −5(x−4)
(x−4)(x+ 3) =4
x2−x−12
Step 3: Combine the fractions on the left side of the equation:
3x+ 9 −5x+ 20
(x−4)(x+ 3) =4
x2−x−12
Step 4: Simplify the left side:
−2x+ 29
(x−4)(x+ 3) =4
x2−x−12
Step 5: Cross multiply to get rid of the fractions:
(−2x+ 29)(x2−x−12) = 4(x−4)(x+ 3)
Step 6: Expand both sides of the equation:
−2x3+ 2x2+ 24x−29x2+ 29x+ 348 = 4(x2−x−12)
11
Step 7: Simplify the equation further:
−2x3−27x2+ 53x+ 348 = 4x2−4x−48
Step 8: Rearrange the terms to set the equation to zero:
−2x3−27x2+ 53x+ 348 −4x2+ 4x+ 48 = 0
Step 9: Simplify the equation:
−2x3−31x2+ 57x+ 396 = 0
This is the final equation after simplifying.
Question 15
Question
Find the domain of the rational function:
f(x) = x2−5x+ 6
x2−4x−5.
Solution
Step 1: To find the domain of a rational function, we must identify any values
of xthat would make the denominator equal to zero, since division by zero is
undefined.
Step 2: Set the denominator of the rational function equal to zero and solve
for x:
x2−4x−5 = 0.
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0.
Step 4: Solve for x:
x−5 = 0 or x+ 1 = 0.
Step 5: So, the values that make the denominator zero are x= 5 and x=−1.
These values must be excluded from the domain.
Step 6: Therefore, the domain of the rational function f(x) = x2−5x+6
x2−4x−5is all
real numbers except x= 5 and x=−1.
Step 7: Thus, the domain is (−∞,−1) ∪(−1,5) ∪(5,∞).
12
Question 16
Question
Simplify the following rational expression:
3x2+ 7x−6
x2−4x−5∇ · 4x2−x−5
3x2−13x+ 12
Solution
Step 1: Factor the numerator and denominator of both fractions. Step 2:
Rewrite the division as multiplication by the reciprocal. Step 3: Multiply the
two fractions together and simplify.
Step 1: Factor the numerator and denominator of both fractions. The first
fraction: 3x2+ 7x−6
x2−4x−5=(3x−2)(x+ 3)
(x−5)(x+ 1)
The second fraction:
4x2−x−5
3x2−13x+ 12 =(4x+ 5)(x−1)
(3x−4)(x−3)
Step 2: Rewrite the division as multiplication by the reciprocal. This is
equivalent to multiplying by the reciprocal of the second fraction:
(3x−2)(x+ 3)
(x−5)(x+ 1) ×(3x−4)(x−3)
(4x+ 5)(x−1)
Step 3: Multiply the two fractions together and simplify. Multiplying the
fractions, we get:
(3x−2)(x+ 3)(3x−4)(x−3)
(x−5)(x+ 1)(4x+ 5)(x−1)
Expanding and simplifying, we get:
9x2−9x−8x+ 8
(x2−4x−5)(4x2−x−5)
=9x2−17x+ 8
(x−5)(x+ 1)(4x+ 5)(x−1)
Therefore, the simplified form of the rational expression is 9x2−17x+8
(x−5)(x+1)(4x+5)(x−1) .
Question 17
Question
Determine the domain of the rational function given by
f(x) = 3x−2
2x2+x−3.
13
Solution
Step 1: To find the domain of the rational function, we need to identify any
values of xthat would make the denominator equal to zero, as division by zero
is undefined.
Step 2: Set the denominator equal to zero and solve for x:
2x2+x−3 = 0.
Step 3: Factor the quadratic equation:
2x2+ 3x−2x−3 = 0
2x(x+ 1) −3(x+ 1) = 0
(2x−3)(x+ 1) = 0.
Step 4: Set each factor equal to zero and solve for x:
2x−3 = 0 or x+ 1 = 0
2x= 3 or x=−1
x=3
2or x=−1.
Step 5: The values x=3
2and x=−1make the denominator equal to zero.
Therefore, the domain of the rational function is all real numbers except x=3
2
and x=−1, or in interval notation, (−∞,−1) ∪(−1,3
2)∪(3
2,∞).
Question 18
Question
Simplify the following rational expression:
3x2−9x
x2−4
Solution
Step 1: Factor out common terms in the numerator and simplify where possible.
3x2−9x
x2−4=3x(x−3)
(x+ 2)(x−2)
Step 2: Simplify the expression by canceling out common factors.
3x(x−3)
(x+ 2)(x−2) =3(x−3)
x+ 2
Therefore, the simplified form of the given rational expression is 3(x−3)
x+2 .
14
Question 19
Question
Simplify the following rational expression:
6x2−3x−15
x2−4x−12
Solution
Step 1: Factor the numerator and denominator. Step 2: Simplify the ratio-
nal expression by canceling out any common factors. Step 3: Check for any
extraneous solutions.
Step 1: Factor the numerator and denominator. The numerator 6x2−3x−
15 can be factored as 3(2x2−x−5). The denominator x2−4x−12 can be
factored as (x−6)(x+ 2).
Step 2: Simplify the rational expression by canceling out any common
factors. 6x2−3x−15
x2−4x−12 =3(2x2−x−5)
(x−6)(x+ 2) =3(2x+ 5)(x−1)
(x−6)(x+ 2)
Step 3: Check for any extraneous solutions. Since the simplified expression
does not have any common factors that can be canceled further, we have the
final simplified form:
3(2x+ 5)(x−1)
(x−6)(x+ 2)
Question 20
Question
Solve the following rational inequality:
3x−1
x−2≥5
Solution
Step 1: Begin by rewriting the inequality with a zero on one side:
3x−1
x−2−5≥0
Step 2: Find a common denominator for the left side of the inequality:
3x−1
x−2−5(x−2)
x−2≥0
15
Step 3: Combine the fractions on the left side:
3x−1−5x+ 10
x−2≥0
Step 4: Simplify the numerator:
−2x+ 9
x−2≥0
Step 5: Now, find the critical points by setting the numerator and denomi-
nator equal to zero:
−2x+ 9 = 0 ⇒x=9
2
x−2 = 0 ⇒x= 2
Step 6: Create a sign chart using the critical points to test the intervals
between the critical points and beyond:
x < 9
2
9
2< x < 2x= 2 x > 2
−2x+ 9 −+ + +
x−2− − 0 +
−2x+9
x−2+−undefined +
Step 7: Determine the solution from the sign chart:
x∈(−∞,9
2)∪(2,∞)
Question 21
Question
Given the rational function f(x) = 3x2+4x−1
x2−5x+6 , find the vertical asymptotes,
horizontal asymptotes, and any holes in the graph of f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to zero and
solve for x.
x2−5x+ 6 = 0
(x−2)(x−3) = 0
Therefore, x= 2 and x= 3 are the vertical asymptotes.
Step 2: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator. Since the degree of the numerator is equal to the degree
of the denominator, we divide the leading coefficients to find the horizontal
asymptote.
lim
x→∞
3x2
x2= lim
x→∞ 3 = 3
16
Therefore, the horizontal asymptote is y= 3.
Step 3: To find any holes in the graph, factor the numerator and see if there
are any common factors with the denominator.
3x2+ 4x−1 = (3x−1)(x+ 1)
There are no common factors between the numerator and denominator, so
there are no holes in the graph.
In conclusion, the rational function f(x) = 3x2+4x−1
x2−5x+6 has vertical asymptotes
at x= 2 and x= 3, a horizontal asymptote at y= 3, and no holes in its graph.
Question 22
Question
Let f(x) = x2−4
x2−5x+6 be a rational function. Determine the vertical asymptotes,
horizontal asymptotes, and holes, if any, of the graph of f(x).
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to zero and
solve for x.
x2−5x+ 6 = 0
(x−2)(x−3) = 0
x= 2 or x= 3
Thus, the vertical asymptotes are x= 2 and x= 3.
Step 2: To find the horizontal asymptote, compare the degrees of the nu-
merator and the denominator. Since the degrees are the same (2), we check
the leading coefficients. The horizontal asymptote is the ratio of the leading
coefficients:
lim
x→∞
x2−4
x2−5x+ 6 = 1
Therefore, the horizontal asymptote is y= 1.
Step 3: To find any holes, factor and cancel out common factors in the
numerator and denominator:
f(x) = x2−4
x2−5x+ 6 =(x+ 2)(x−2)
(x−2)(x−3)
We see that there is a hole at x= 2, where the factor (x−2) cancels out.
So, the graph of f(x)has vertical asymptotes at x= 2 and x= 3, a horizontal
asymptote at y= 1, and a hole at x= 2.
17
Question 23
Question
Find the vertical asymptotes, horizontal asymptotes, and any holes for the ra-
tional function:
f(x) = 2x3−5x2−3x
x2−4
Solution
Step 1: To find the vertical asymptotes, set the denominator equal to zero and
solve:
x2−4 = 0
(x+ 2)(x−2) = 0
x=−2,2
So, the vertical asymptotes are x=−2and x= 2.
Step 2: To find the horizontal asymptotes, compare the degrees of the nu-
merator and denominator: - If the degree of the numerator is less than the
degree of the denominator, there is a horizontal asymptote at y= 0. - If the
degree of the numerator is equal to the degree of the denominator, divide the
leading coefficients to find the horizontal asymptote. - If the degree of the nu-
merator is greater than the degree of the denominator, there is no horizontal
asymptote.
Here, the degree of the numerator is 3 and the degree of the denominator is
2, so there is no horizontal asymptote.
Step 3: To find any holes in the graph, factor the numerator and denominator
and simplify: Factor the numerator: 2x3−5x2−3x=x(2x2−5x−3) =
x(2x+ 1)(x−3)
So, the simplified form of the function is:
f(x) = x(2x+ 1)(x−3)
(x+ 2)(x−2)
Since there is a common factor of (x−2) in both the numerator and denom-
inator, there is a hole at x= 2.
Therefore, the vertical asymptotes are x=−2and x= 2, there is no hori-
zontal asymptote, and there is a hole at x= 2.
Question 24
Question
Solve the following rational inequality for x:
x−2
x+ 3 ≤1
2
18
Solution
To solve this rational inequality, we will first find the critical points by setting
the numerator of the left-hand side equal to the right-hand side. We will then
test each interval created by the critical points to determine where the inequality
holds true.
Step 1: Find the critical points
x−2
x+ 3 =1
2
2(x−2) = (x+ 3)
2x−4 = x+ 3
x= 7
So, the critical point is x= 7.
Step 2: Test the intervals We have three intervals: (−∞,−3),(−3,7),
and (7,∞).
Testing the interval (−∞,−3):Pick x=−4, then −4−2
−4+3 =−6
−1= 6 >1
2
So, this interval is not a solution.
Testing the interval (−3,7):Pick x= 0, then 0−2
0+3 =−2
3<1
2
So, this interval is a solution.
Testing the interval (7,∞):Pick x= 8, then 8−2
8+3 =6
11 <1
2
So, this interval is not a solution.
Step 3: Conclusion The solution to the inequality x−2
x+3 ≤1
2is x∈(−3,7].
Question 25
Question
Given the rational function f(x) = 3x3+2x2−5x+1
x2−4x+3 , determine the vertical asymp-
totes, horizontal asymptotes, and the x-intercepts, if any.
Solution
To find the vertical asymptotes of the rational function f(x), we need to find
the values of xwhich make the denominator equal to zero. These values will
give us the equations of the vertical asymptotes.
Step 1: Find the vertical asymptotes:
Set the denominator equal to zero and solve for x:
x2−4x+ 3 = 0
Factoring the quadratic equation:
(x−3)(x−1) = 0
So, the vertical asymptotes are at x= 1 and x= 3.
19
Step 2: Find the horizontal asymptote:
To find the horizontal asymptote, we compare the degrees of the numerator
and the denominator of the rational function f(x).
Since the degree of the numerator is equal to the degree of the denominator,
we look at the ratio of the leading coefficients:
3
1= 3
So, the horizontal asymptote is y= 3.
Step 3: Find the x-intercepts:
To find the x-intercepts, we set the numerator equal to zero and solve for x:
3x3+ 2x2−5x+ 1 = 0
Unfortunately, there is no straightforward way to solve this cubic equation
by factoring or using the quadratic formula. Hence, we need to use numerical
methods or graphical methods to approximate the x-intercepts of the function.
Therefore, the vertical asymptotes are x= 1 and x= 3, the horizontal
asymptote is y= 3, and the x-intercepts need to be approximated.
Question 26
Question
Solve the rational inequality: 2x−5
x−3≤1.
Solution
Step 1: Begin by solving the related equation 2x−5
x−3= 1.
2x−5
x−3= 1
2x−5 = x−3
2x−x= 3 −5
x=−2
Step 2: Next, determine the critical points by finding where the denominator
is equal to zero.
x−3 = 0
x= 3
Step 3: Now, create a number line and mark the critical points x=−2and
x= 3.
−2 3
20
Step 4: Test the intervals determined by the critical points x=−2and x= 3
by selecting test values in each interval and determining the sign of 2x−5
x−3−1. -
For x < −2, choose x=−3:
2(−3) −5
−3−3−1 = −11
−6−1 = 11
6−1>0
- For −2< x < 3, choose x= 0:
2(0) −5
0−3−1 = −5
−3−1 = 5
3−1>0
- For x > 3, choose x= 4:
2(4) −5
4−3−1 = 3
1−1 = 3 −1>0
Step 5: Therefore, the solution to the inequality 2x−5
x−3≤1is x∈(−∞,−2] ∪
(3,∞).
Question 27
Question
Simplify the following rational expression:
3x2−2x−5
2x2+ 7x−15
Solution
Step 1: Factor both the numerator and the denominator.
Step 2: Factor 3x2−2x−5.
3x2−2x−5 = (3x+ 1)(x−5)
Step 3: Factor 2x2+ 7x−15.
2x2+ 7x−15 = (2x−3)(x+ 5)
Step 4: Rewrite the expression with factored numerator and denominator.
(3x+ 1)(x−5)
(2x−3)(x+ 5)
Therefore, the simplified form of the given rational expression is (3x+1)(x−5)
(2x−3)(x+5) .
21
Question 28
Question
Find the domain of the rational function:
f(x) = x2−3x−4
x2+ 2x−8
Solution
Step 1: We start by finding the values of xfor which the denominator is equal
to zero, since division by zero is undefined.
x2+ 2x−8 = 0
Factoring the quadratic equation, we get:
(x+ 4)(x−2) = 0
So, the values of xthat make the denominator zero are x=−4and x= 2.
Step 2: Next, we need to consider any restrictions on the domain due to
simplifying the expression. Since the numerator and denominator are both
polynomials, the only restrictions on the domain occur when the denominator
is zero.
Step 3: Therefore, the domain of the rational function is all real numbers
except for the values of xthat make the denominator equal to zero. Hence, the
domain of f(x)is (−∞,−4) ∪(−4,2) ∪(2,∞).
Question 29
Question
Let f(x) = 2x2−5x−3
x2−4x−5. Find the domain of f(x).
Solution
Step 1: To find the domain of f(x), we need to identify any values of xthat
would make the denominator x2−4x−5equal to zero, since division by zero is
undefined.
Step 2: To find the values of xthat make the denominator zero, we solve
the equation x2−4x−5 = 0.
Step 3: Factoring the quadratic equation x2−4x−5 = 0, we get (x−5)(x+
1) = 0.
Step 4: Setting each factor to zero and solving for x, we find x= 5 and
x=−1.
Step 5: Therefore, the domain of the function f(x)is all real numbers except
x= 5 and x=−1.
Step 6: In interval notation, the domain of f(x)is (−∞,−1)∪(−1,5)∪(5,∞).
22
Question 30
Question
Solve the inequality: 2x+1
x−2<3.
Solution
Step 1: Find the critical points by setting the inequality equal to zero:
2x+ 1
x−2= 3.
Step 2: Multiply both sides by x−2to get rid of the denominator:
2x+ 1 = 3(x−2).
Step 3: Expand the right side and simplify the equation:
2x+ 1 = 3x−6.
Step 4: Move all terms to one side to get the equation in standard form:
2x+ 1 −3x+ 6 = 0.
Step 5: Simplify the equation further:
−x+ 7 = 0.
Step 6: Solve for xby isolating x:
x= 7.
So, x= 7 is the critical point.
Step 7: Test the intervals defined by the critical point x= 7 on the original
inequality. Choose a test point on the intervals x < 7and x > 7: - Test ˜x= 6
for x < 7:2(6) + 1
6−2=13
4<3.
- Test ˜x= 8 for x > 7:2(8) + 1
8−2=17
6≮3.
Step 8: Write the final solution as an inequality: Since x < 7satisfies the
inequality, the solution to the inequality 2x+1
x−2<3is:
x < 7.
23
Question 31
Question
Find the vertical asymptotes, horizontal asymptotes, and holes (if any) of the
rational function:
f(x) = 4x3−16x
x2−2x−3
Solution
Step 1: Vertical Asymptotes
To find the vertical asymptotes, we need to determine where the denominator
is equal to zero:
x2−2x−3 = 0
Factoring the quadratic gives:
x2−2x−3 = (x−3)(x+ 1) = 0
Setting each factor equal to zero, we find the potential vertical asymptotes
at x= 3 and x=−1.
Step 2: Horizontal Asymptotes
To find the horizontal asymptotes, we compare the degrees of the numerator
and denominator: - If the degree of the numerator is greater, there is no hor-
izontal asymptote. - If the degrees are equal, the horizontal asymptote is the
ratio of the leading coefficients. - If the degree of the denominator is greater,
the horizontal asymptote is the x-axis, y= 0.
In our case, the degree of the numerator is 3, and the degree of the denomi-
nator is 2. So, there is no horizontal asymptote.
Step 3: Holes
To find any potential holes in the graph, we look for common factors between
the numerator and the denominator. Canceling common factors gives:
f(x) = 4x(x2−4)
(x−3)(x+ 1) =4x(x+ 2)(x−2)
(x−3)(x+ 1)
This simplification shows a hole at x= 2, as (x−2) cancels out in the
expression.
Therefore, the vertical asymptotes are x= 3 and x=−1, there are no
horizontal asymptotes, and there is a hole at x= 2.
Question 32
Question
Solve the rational inequality:
x2−4x−5
x+ 4 ≥0
24
Solution
Step 1: Find the critical points by setting the numerator and denominator equal
to zero.
x2−4x−5 = 0 =⇒(x−5)(x+ 1) = 0
x+ 4 = 0 =⇒x=−4
So the critical points are x= 5,x=−1, and x=−4.
Step 2: Create a sign chart using the critical points.
x < −4−4< x < −1−1< x < 5x > 5
x+ 4 −+ + +
x2−4x−5 + + −+
x2−4x−5
x+4 + + −+
Step 3: Determine the solution based on the sign chart. The inequality is
satisfied when x2−4x−5
x+4 ≥0, which means the function is positive or zero. So
the solution is x∈(−∞,−4] ∪(−1,5].
Question 33
Question
Find the domain of the rational function:
f(x) = 2x2−5x
x2−4x−12.
Solution
Step 1: We need to find the values of xfor which the denominator of the rational
function is not equal to zero, since division by zero is undefined. Therefore, we
set the denominator x2−4x−12 not equal to zero and solve for x:
x2−4x−12 = 0.
Step 2: Factor the quadratic equation x2−4x−12:
x2−4x−12 = 0.
(x−6)(x+ 2) = 0.
Step 3: Set each factor to zero and solve for x:
x−6 = 0 =⇒x= 6,
x+ 2 = 0 =⇒x=−2.
Step 4: The domain of the rational function f(x)is all real numbers except
x= 6 and x=−2. Therefore, the domain of f(x)is:
(−∞,−2) ∪(−2,6) ∪(6,∞).
25
Question 34
Question
Simplify the rational function:
f(x) = x2−2x−8
x2−5x+ 6
Solution
Step 1: Factor both the numerator and the denominator:
f(x) = (x−4)(x+ 2)
(x−3)(x−2)
Step 2: Simplify the expression by canceling out common factors:
f(x) = x−4
x−3
Therefore, the simplified form of the rational function is f(x) = x−4
x−3.
Question 35
Question
Find the domain of the rational function: f(x) = 1
x2−4x−5.
Solution
Step 1: We first need to find where the denominator is equal to zero since the
function is undefined at those points.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0
Step 4: Find the roots:
x= 5 or x=−1
Step 5: Therefore, the function f(x)is undefined at x= 5 and x=−1, so
the domain is all real numbers except x= 5 and x=−1. In interval notation,
the domain of f(x)is (−∞,−1) ∪(−1,5) ∪(5,∞).
26
