Question

the function $f(x)=\frac{x + 2}{x^{2}+2x}$ is a rational function.
a. determine the coordinates of any removable discontinuities.
b. sketch the graph.
c. find all x - intercepts or state that the function has no x - intercepts.
d. find the y - intercept or state that the function does not have a y - intercept.
e. find the equation(s) of all vertical asymptotes.
f. find the equation(s) of all horizontal asymptotes.

Explanation:

Step1: Simplify the function

First, factor the denominator \(x^{2}+2x=x(x + 2)\). Then \(f(x)=\frac{x + 2}{x(x + 2)}\). For \(x
eq-2\), \(f(x)=\frac{1}{x}\).

Step2: Find removable discontinuity

A removable discontinuity occurs when a factor cancels out. Set \(x+2 = 0\), \(x=-2\). Substitute \(x=-2\) into the simplified function (but note the original function is undefined here). The limit as \(x
ightarrow-2\) of \(f(x)\) is \(\lim_{x
ightarrow-2}\frac{1}{x}=-\frac{1}{2}\). So the removable discontinuity is at the point \((-2,-\frac{1}{2})\).

Step3: Find \(x\) - intercept

Set \(y = 0\), \(\frac{1}{x}=0\) (no solution for \(x\in R\)). But if we consider the original function \(f(x)=\frac{x + 2}{x^{2}+2x}\), set numerator \(x + 2=0\) (excluding \(x=-2\) which is a removable discontinuity). So no \(x\) - intercept.

Step4: Find \(y\) - intercept

Set \(x = 0\), the function \(f(x)=\frac{x + 2}{x^{2}+2x}\) is undefined. So no \(y\) - intercept.

Step5: Find vertical asymptote

For \(y=\frac{1}{x}\) (after removing the removable discontinuity), set denominator \(x=0\). So the vertical asymptote is \(x = 0\).

Step6: Find horizontal asymptote

For \(y=\frac{1}{x}\), since the degree of the numerator is less than the degree of the denominator. \(\lim_{x
ightarrow\pm\infty}\frac{1}{x}=0\). So the horizontal asymptote is \(y = 0\).

Answer:

a. \((-2,-\frac{1}{2})\)
c. No \(x\) - intercept
d. No \(y\) - intercept
e. \(x = 0\)
f. \(y = 0\)