Question

Check please
Identify the vertical asymptote(s) of the function f(x) =x^2-1/2x^2-8

Answer 1

would it be x=0?

Answer 2

@tcarroll010 can you check my answer

Answer 3

The vertical asymptote is where you have unallowable values in the denominator. So, if your full denominator is (x^2 - 8). Is that your denominator? If it is, then see where that expression equals "0".

Answer 4

it would be 2

Answer 5

Otherwise, if the expression is:

x^2 - 1/(2x^2) - 8

then you are correct.

Answer 6

The reason I am presenting you with 2 choices is that I can't tell if that expression is really supposed to be the one in my last post or if you mean:

(x^2 - 1) / (2x^2 - 8)

Answer 7

without parenthesis
ill put a pic so i can know where to put the parenthesis next time

Answer 8

Okay, so your denominator is :

(2x^2 - 8)

and that will be 0 at 2 different values of x. Just solve:

2x^2 - 8 = 0

Answer 9

2?

Answer 10

That's one of the values. Don't forget that you are dealing with a square root. Do you remember something about positive and negative roots?

Answer 11

ahh

Answer 12

Hint:

sqrt(x^2) = +-x

Answer 13

x=2 x=-2

Answer 14

horizontal is y=1/2

Answer 15

Yes! You got It!