Question

graph the function $y = x^{2}-2x - 8$ by identifying the domain and any symmetries, finding the derivatives $y$ and $y$, finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any
the domain is $(-\infty,\infty)$
(type your answer in interval notation.)
identify any symmetries. choose the correct answer below.
a. the function is an even function that is symmetric about the y - axis
b. the function is an even function that is symmetric about the origin
c. the function is an odd function that is symmetric about the y - axis
d. the function is an odd function that is symmetric about the origin
e. the function is neither even nor odd

Explanation:

Step1: Recall the definitions of even and odd functions

An even function satisfies \(f(-x)=f(x)\) for all \(x\) in the domain. An odd function satisfies \(f(-x)=-f(x)\) for all \(x\) in the domain.
For the function \(y = f(x)=x^{2}-2x - 8\), find \(f(-x)\):

$$
LATEXBLOCK0
$$

Step2: Check if \(f(-x)=f(x)\) or \(f(-x)=-f(x)\)

Check \(f(-x)=f(x)\):
\(x^{2}+2x - 8
eq x^{2}-2x - 8\) (since \(2x
eq - 2x\) for \(x
eq0\))
Check \(f(-x)=-f(x)\):
\(-f(x)=-(x^{2}-2x - 8)=-x^{2}+2x + 8\)
And \(x^{2}+2x - 8
eq -x^{2}+2x + 8\) (since \(x^{2}
eq -x^{2}\) for \(x
eq0\))

Answer:

E. The function is neither even nor odd