#12Consider theX+ Icational function plx) = X-4A Find the doin and all the interepts of P(x)B)End all vertical and horizontal asymptotesc) Sketch the graph of p(x)

Answered: #12 Consider Hhe cahional function plx)…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.5: Rational Functions

Problem 54E

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Question

#12
Consider the
X+ I
cational function plx) = X-4
A Find the doin and all the interepts of P(x)
B)End all vertical and horizontal asymptotes
c) Sketch the graph of p(x)

Expert Solution

Given,

$p (x) = \frac{x + 1}{x^{2} - 4}$

A).

Domain is the set of all values of x for which denominator is non-zero. That is,

$x^{2} - 4 \neq 0 \Rightarrow x^{2} \neq 4 \Rightarrow x \neq \pm 2$

Therefore domain is,

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$ .

Now to find x-intercepts, put $p (x) = 0$ , we get

$\frac{x + 1}{x^{2} - 4} = 0 \Rightarrow x + 1 = 0 \Rightarrow x = - 1$

Therefore x-intercept is $(- 1, 0)$ .

Now to find y-intercept, put $x = 0$ , we get

$p (0) = \frac{0 + 1}{0^{2} - 4} \Rightarrow p (0) = - 0.25$

Therefore y-intercept is $(0, - 0.25)$ .

Step 2

B).

Vertical asymptotes are the values of x for which denominator is zero. That is,

$x^{2} - 4 = 0 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$

Therefore vertical asymptotes are $x = - 2 & x = 2$ .

Now since the degree of numerator (1) is less than the degree of the denominator (2), therefore the horizontal asymptote is x-axis. That is,

The horizontal asymptote is $y = 0$ .

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#12 Consider Hhe cahional function plx) = x²-4 A) Find the domin and all the intereepts of P(x) B) Find all vertical and horizontal asymptotes c) Sketch the graph of plx) X+1