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# Solved Examples on Rational Functions

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Review:
Given a Rational Function:
To Find the Vertical Asymptote (VA):
(1.) Simplify the function
(2.) Set the denominator to zero (if applicable after simplifying the function)
(3.) Solve for the value of $x$
(4.) $VA: x = value$

To Find the Horizontal Asymptote (HA):
(1.) Arrange the numerator in standard form
(2.) Arrange the denominator in standard form
(3.) If the degree of the numerator is less than the degree of the denominator, $HA: y = 0$
(4.) If the degree of the numerator is the same as the degree of the denominator,
$HA: y = \dfrac{leading\;\;coefficient\;\;of\;\;the\;\;numerator}{leading\;\;coefficient\;\;of\;\;the\;\;denominator}$

If the degree of the numerator is greater than the degree of the denominator, then onto the Slant Asymptote (or Oblique Asymptote)

To Find the Slant Asymptote (SA):
If the degree of the numerator is greater than the degree of the denominator,
$SA: y = quotient\;\;of\;\;\dfrac{numerator}{denominator}$

To Find the $x-intercept$:
(1.) Set $y = 0$
(2.) Solve for the value of $x$
(3.) $x-intercept = (value, 0)$

To Find the $y-intercept$:
(1.) Set $x = 0$
(2.) Solve for the value of $y$
(3.) $y-intercept = (0, value)$

Solve all questions.
Show all work.

(1.) Determine the $x-intercept$ and the $y-intercept$ for the function.

$f(x) = \dfrac{x + 7}{x^2 + 36} \\[5ex]$

$f(x) = \dfrac{x + 7}{x^2 + 36} \\[5ex] y = \dfrac{x + 7}{x^2 + 36} \\[5ex] \underline{y-intercept} \\[3ex] Set\;\;x = 0 \;\;and\;\;solve\;\;for\;\;y \\[3ex] y = \dfrac{0 + 7}{0^2 + 36} \\[5ex] y = \dfrac{7}{36} \\[5ex] y-intercept = \left(0, \dfrac{7}{36}\right) \\[5ex] \underline{x-intercept} \\[3ex] Set\;\;y = 0 \;\;and\;\;solve\;\;for\;\;x \\[3ex] 0 = \dfrac{x + 7}{x^2 + 36} \\[5ex] Multiply\;\;both\;\;sides\;\;by\;\; x^2 + 36 \\[3ex] 0 = x + 7 \\[3ex] x + 7 = 0 \\[3ex] x = 0 - 7 \\[3ex] x = -7 \\[3ex] x-intercept = (-7, 0)$
(2.) ACT Consider the graph of the equation $y = \dfrac{3x - 12}{2x - 6}$ in the standard $(x, y)$ coordinate plane.
Which of the following equations represents the vertical asymptote of the graph?

$F.\:\: x = 2 \\[3ex] G.\:\: x = 3 \\[3ex] H.\:\: x = 4 \\[3ex] J.\:\: x = 6 \\[3ex] K.\:\: x = 12 \\[3ex]$

$\underline{Vertical\;\;Asymptote:\;VA} \\[3ex] y = \dfrac{3x - 12}{2x - 6} \\[5ex] y = \dfrac{3(x - 4)}{2(x - 3)} \\[5ex] Nothing\;\;cancels \\[3ex] Set\;\;the\;\;denominator\;\;to\;\;0; \;\;\;\;solve\;\;for\;\;x \\[3ex] 2x - 6 = 0 \\[3ex] 2x = 0 + 6 \\[3ex] 2x = 6 \\[3ex] x = \dfrac{6}{2} \\[5ex] x = 3 \\[3ex] VA:\;\;x = 3$
(3.) Determine the intercepts and the vertical asymptote of

$f(x) = \dfrac{x^2 - 3x - 4}{x - 5} \\[5ex]$

$f(x) = \dfrac{x^2 - 3x - 4}{x - 5} \\[5ex] y = \dfrac{x^2 - 3x - 4}{x - 5} \\[5ex] x-intercept:\;\; \\[3ex] set\;\;y = 0; \;\;\;\;solve\;\;for\;\;x \\[3ex] y = \dfrac{x^2 - 3x - 4}{x - 5} \\[5ex] 0 = \dfrac{x^2 - 3x - 4}{x - 5} \\[5ex] 0(x - 5) = x^2 - 3x - 4 \\[3ex] x^2 - 3x - 4 = 0 \\[3ex] (x + 1)(x - 4) = 0 \\[3ex] x + 1 = 0 \;\;OR\;\;x - 4 = 0 \\[3ex] x = -1 \;\;OR\;\; x = 4 \\[3ex] x-intercepts\;\;are:\;\; (-1, 0) \;\;and\;\; (4, 0) \\[3ex] y-intercept:\;\; \\[3ex] set\;\;x = 0; \;\;\;\;solve\;\;for\;\;y \\[3ex] y = \dfrac{x^2 - 3x - 4}{x - 5} \\[5ex] y = \dfrac{0^2 - 3(0) - 4}{0 - 5} \\[5ex] y = \dfrac{0 - 0 - 4}{-5} \\[5ex] y = \dfrac{-4}{-5} \\[5ex] y = \dfrac{4}{5} \\[5ex] y-intercept\;\;is:\;\; \left(0, \dfrac{4}{5}\right) \\[5ex] \underline{Vertical\;\;Asymptote:\;VA} \\[3ex] y = \dfrac{x^2 - 3x - 4}{x - 5} \\[5ex] y = \dfrac{(x + 1)(x - 4)}{x - 5} \\[5ex] Nothing\;\;cancels \\[3ex] set\;\;denominator\;\;to\;\;0; \;\;\;\;solve\;\;for\;\;x \\[3ex] x - 5 = 0 \\[3ex] x = 5 \\[3ex] VA:\;\;x = 5$
(4.) ACT Which of the following linear equations gives the vertical asymptote for the graph of $y = \dfrac{201x + 202}{203x + 204}$ in the standard $(x, y)$ coordinate plane?

$A.\:\: x = -\dfrac{201}{203} \\[5ex] B.\:\: x = -\dfrac{202}{201} \\[5ex] C.\:\: x = -\dfrac{202}{204} \\[5ex] D.\:\: x = -\dfrac{204}{203} \\[5ex] E.\:\: x = -\dfrac{403}{407} \\[5ex]$

$\underline{Vertical\;\;Asymptote:\;VA} \\[3ex] y = \dfrac{201x + 202}{203x + 204} \\[5ex] Nothing\;\;cancels \\[3ex] Set\;\;the\;\;denominator\;\;to\;\;0; \;\;\;\;solve\;\;for\;\;x \\[3ex] 203x + 204 = 0 \\[3ex] 203x = 0 - 204 \\[3ex] 203x = -204 \\[3ex] x = -\dfrac{204}{203} \\[5ex] VA:\;\;x = -\dfrac{204}{203}$
(5.) ACT At what point in the standard $(x, y)$ coordinate plane asymptotes of the function $y = \dfrac{2x(x + 2)}{x - 3}$, graphed below, intersect?

$F.\;\; \left(-\dfrac{7}{3}, 3\right) \\[5ex] G.\;\; \left(\dfrac{7}{3}, 10\right) \\[5ex] H.\;\; (3, 10) \\[3ex] J.\;\; (3, 16) \\[3ex] K.\;\; (3, 31) \\[3ex]$

Only two asymptotes apply to the function: the Vertical Asymptote (VA) and the Slant Asymptote (SA)
So, we shall find both asymptotes.
Then, we shall answer the question (the intersection of both asymptotes)

$\underline{Vertical\;\;Asymptote:\;VA} \\[3ex] y = \dfrac{2x(x + 2)}{x - 3} \\[5ex] Nothing\;\;cancels \\[3ex] Set\;\;the\;\;denominator\;\;to\;\;0; \;\;\;\;solve\;\;for\;\;x \\[3ex] x - 3 = 0 \\[3ex] x = 0 + 3 \\[3ex] x = 3 \\[3ex] VA:\;\;x = 3 \\[3ex] \underline{Slant\;\;Asymptote:\;SA} \\[3ex] y = \dfrac{2x(x + 2)}{x - 3} \\[5ex] y = \dfrac{2x^2 + 4x}{x - 3} \\[5ex] y = \dfrac{2x^2 + 4x + 0}{x - 3} \\[5ex] \require{enclose} \begin{array}{rl6} 2x + 10 \\[-3pt] x - 3 \enclose{longdiv}{2x^2 + 4x + 0}\kern-.5ex \\[-3pt] \underline{2x^2 - 6x\phantom{000}} \\[-3pt] 10x + 0\phantom{0} \\[-3pt] \underline{\phantom{0}10x - 30} \\[-3pt] \phantom{0}30 \\[-3pt] \end{array} \\[3ex] SA:\;\; y = 2x + 10 \\[3ex] \underline{Intersection\;\;of\;\;VA\;\;and\;\;SA} \\[3ex] VA:\;\;x = 3 \\[3ex] SA:\;\; y = 2x + 10 \\[3ex] y = 2(3) + 10 \\[3ex] y = 6 + 10 \\[3ex] y = 16 \\[3ex] Intersection:\;\;(x, y) = (3, 16)$
(6.) ACT In the standard $(x, y)$ coordinate plane, the graph of which of the following equations has the line $x = 2$ as a vertical asymptote?

$A.\;\; y = \dfrac{4x + 3}{2x} \\[5ex] B.\;\; y = \dfrac{4x + 3}{x + 2} \\[5ex] C.\;\; y = \dfrac{4x + 3}{x - 2} \\[5ex] D.\;\; y = \dfrac{x + 2}{4x + 3} \\[5ex] E.\;\; y = \dfrac{x - 2}{4x + 3} \\[5ex]$

To determine the vertical asymptote:
(1.) Simplify the function
(2.) Set the denominator to zero
(3.) Solve for $x$

$All\;\;the\;\;functions\;\;are\;\;already\;\;simplified \\[3ex] Looking\;\;at\;\;the\;\;denominators: \\[3ex] x - 2 = 0 \\[3ex] x = 2 \\[3ex] Correct\;\;Option:\;\; C$
(7.)

(8.) ACT The graph of the function $f(x) = \dfrac{x^2 + 3x + 2}{x}$ is shown in the standard $(x, y)$ coordinate plane below.
Which of the following, if any, is a list of each of the vertical asymptotes of $f(x)$?

$F.\;\; x = 0 \\[3ex] G.\;\; x = -1 \;\;and\;\; x = -2 \\[3ex] H.\;\; y = x + 3 \\[3ex] J.\;\; y = 3x + 2 \\[3ex] K.\;\; This\;\;function\;\;has\;\;no\;\;vertical\;\;asymptote \\[3ex]$

To determine the vertical asymptote:
(1.) Simplify the function
(2.) Set the denominator to zero
(3.) Solve for $x$

$\underline{Vertical\;\;Asymptote:\;VA} \\[3ex] f(x) = \dfrac{x^2 + 3x + 2}{x} \\[5ex] f(x) = \dfrac{(x + 2)(x + 1)}{x} \\[5ex] Nothing\;\;cancels \\[3ex] Set\;\;the\;\;denominator\;\;to\;\;0; \;\;\;\;solve\;\;for\;\;x \\[3ex] x = 0 \\[3ex] VA:\;\;x = 0$
(9.)

(10.) ACT At what value(s) of x is $\dfrac{(x - 3)^2}{x^2}$ undefined?

F. 0 only
G. 0 and 3 only
H. -3 only
J. -3 and 0 only
K. -3, 0, and 3 only

The function is undefined when the denominator is zero.
In other words, find the vertical asymptote of the function.

$\underline{Vertical\;\;Asymptote:\;VA} \\[3ex] f(x) = \dfrac{(x - 3)^2}{x^2} \\[5ex] Nothing\;\;cancels \\[3ex] Set\;\;the\;\;denominator\;\;to\;\;0; \;\;\;\;solve\;\;for\;\;x \\[3ex] x^2 = 0 \\[3ex] x = \pm \sqrt{0} \\[3ex] x = \pm 0 \\[3ex] x = 0 \\[3ex] VA:\;\;x = 0$
(11.)

(12.) ACT The equation $y = \dfrac{2x^2 - 18}{x^2 - 5x}$ has 2 vertical asymptotes and 1 horizontal asymptote.
What is the horizontal asymptote?

$F.\;\; x = 0 \\[3ex] G.\;\; x = 3 \\[3ex] H.\;\; x = 9 \\[3ex] J.\;\; y = 0 \\[3ex] K.\;\; y = 2 \\[3ex]$

$\underline{Horizontal\;\;Asymptote:\;HA} \\[3ex] numerator\;\;in\;\;standard\;\;form ...yes \\[3ex] denominator\;\;in\;\;standard\;\;form ... yes \\[3ex] Degree\;\;of\;\;Numerator = Degree\;\;of\;\;Denominator = 2 \\[3ex] HA: y = \dfrac{leading\;\;coefficient\;\;of\;\;the\;\;numerator}{leading\;\;coefficient\;\;of\;\;the\;\;denominator} \\[5ex] HA: y = \dfrac{2}{1} \\[5ex] HA: y = 2$
(13.)

(14.) ACT In the standard $(x, y)$ coordinate plane, when $a \ne 0$ and $b \ne 0$, the graph of $f(x) = \dfrac{2x + b}{x + a}$ has a horizontal asymptote at:

$F.\;\; y = 2 \\[3ex] G.\;\; y = a \\[3ex] H.\;\; y = -a \\[3ex] J.\;\; y = -\dfrac{b}{2} \\[5ex] K.\;\; y = \dfrac{b}{a} \\[5ex]$

$\underline{Horizontal\;\;Asymptote:\;HA} \\[3ex] numerator\;\;in\;\;standard\;\;form ...yes \\[3ex] denominator\;\;in\;\;standard\;\;form ... yes \\[3ex] Degree\;\;of\;\;Numerator = Degree\;\;of\;\;Denominator = 1 \\[3ex] HA: y = \dfrac{leading\;\;coefficient\;\;of\;\;the\;\;numerator}{leading\;\;coefficient\;\;of\;\;the\;\;denominator} \\[5ex] HA: y = \dfrac{2}{1} \\[5ex] HA: y = 2$

Use the following information to answer questions 15 - 17

ACT Consider the rational function $f(x) = \dfrac{x^2 - 9}{x - 7}$, whose graph is shown in the standard $(x, y)$ coordinate plane below.

(15.) ACT What is the value of $f(x)$ at $x = 4$?

$F.\;\; 4 \\[3ex] G.\;\; 2 \\[3ex] H.\;\; \dfrac{5}{3} \\[5ex] J.\;\; -1 \\[3ex] K.\;\; -\dfrac{7}{3} \\[5ex]$

$f(x) = \dfrac{x^2 - 9}{x - 7} \\[5ex] When\;\; x = 4 \\[3ex] f(4) \\[3ex] = \dfrac{x^2 - 9}{x - 7} \\[5ex] = \dfrac{4^2 - 9}{4 - 7} \\[5ex] = \dfrac{16 - 9}{-3} \\[5ex] = \dfrac{7}{-3} \\[5ex] = -\dfrac{7}{3}$
(16.) ACT What is the domain of $f(x)$?
(Note: The domain of a function is all the $x-values$ for which the function is defined.)

A. All real values of $x$ except $\pm 3$

B. All real values if $x$ except $\dfrac{9}{7}$

C. All real values of $x$ except $7$

D. All real values of $x$ except $\pm 3$ and $7$

E. All real values of $x$ where $x \le -6$ or $x \ge 8$

In this case: (based on this question):
Th denominator cannot be zero
Why?
Because division by zero is undefined
Therefore, to find the domain, set the denominator to zero; and exclude the value of $x$ that makes the denominator to be zero

$x - 7 = 0 \\[3ex] x = 0 + 7 \\[3ex] x = 7 \\[3ex]$ Therefore, the domain is the set of all the real values of $x$ except $7$
(17.) ACT How many horizontal and/or vertical asymptotes are there for the graph of $f(x)$?

$F.\;\; 4 \\[3ex] G.\;\; 3 \\[3ex] H.\;\; 2 \\[3ex] J.\;\; 1 \\[3ex] K.\;\; 0 \\[3ex]$

To determine the vertical asymptote:
(1.) Simplify the function
(2.) Set the denominator to zero
(3.) Solve for $x$

$\underline{Vertical\;\;Asymptote:\;VA} \\[3ex] f(x) = \dfrac{x^2 - 9}{x - 7} \\[5ex] f(x) = \dfrac{x^2 - 3^2}{x - 7} \\[5ex] f(x) = \dfrac{(x + 3)(x - 3)}{x - 7} \\[5ex] Nothing\;\;cancels \\[3ex] Set\;\;the\;\;denominator\;\;to\;\;0; \;\;\;\;solve\;\;for\;\;x \\[3ex] x - 7 = 0 \\[3ex] x = 7 \\[3ex] VA:\;\;x = 7...only\;\;1\;\;vertical\;\;asymptote \\[5ex] \underline{Horizontal\;\;Asymptote:\;HA} \\[3ex] numerator\;\;in\;\;standard\;\;form ...yes \\[3ex] denominator\;\;in\;\;standard\;\;form ... yes \\[3ex] Degree\;\;of\;\;Numerator = 2 \\[3ex] Degree\;\;of\;\;Denominator = 1 \\[3ex]$ Because the Degree of the Numerator is greater than the Degree of the Denominator, there is no horizontal asymptote.
In this case, we have a slant asymptote.
Therefore, we have only 1 vertical asymptote.
The correct option is J.
(18.)

$numerator\;\;in\;\;standard\;\;form ...yes \\[3ex] denominator\;\;in\;\;standard\;\;form ... yes \\[3ex] Degree\;\;of\;\;Numerator = Degree\;\;of\;\;Denominator = 1 \\[3ex] HA: y = \dfrac{leading\;\;coefficient\;\;of\;\;the\;\;numerator}{leading\;\;coefficient\;\;of\;\;the\;\;denominator} \\[5ex] HA: y = \dfrac{2}{1} \\[5ex] HA: y = 2$