Question

State the vertical asymptote of the rational function.
f(x) = quantity x minus six times quantity x plus six divided by quantity x squared minus nine.

Answer 1

\(\bf \cfrac{x-6x+6}{x^2-9} \ \ ?\)

Answer 2

(x-6)(x+6)
---------
x2 - 9

Answer 3

well, keep in mind vertical asymptotes for a rational expression occurs at the zeros of the denominator, so long they don't make the numerator 0
so if we set the denominator to 0
\(\bf x^2-9 =0\\
\text{solve for "x", keeping in mind that}\\
(a-b)(a+b) = (a^2-b^2)\)

Answer 4

what would you get on that for "x"?

Answer 5

3?

Answer 6

is that right??

Answer 7

\(\bf x^2-9 =0\\
x^2-3^2 \implies (x-3)(x+3)\)
so the values for "x" are +3 and -3
at that value for "x", the denominator is 0

now let's check those values against the numerator, replace "x" in the numerator for +3 and then for -3, does it give 0?

Answer 8

so, (3-6)(3+6) = -18
(-3-6)(-3+6) = -18

so zeros of the denominator, do not make the numerator 0, thus those are your vertical asymptotes :)

x = 3, x = -3

Answer 9

omg thats so helpful! thank you so much! <33333

Answer 10

yw

Answer 11

would you mind helping me with a few more? this chapter is new and nobody else can help really well

Answer 12

is easier to ask in the channel, that way we can all see it and help and revise each other :)

Answer 13

the channel? :o im new here idk what you mean :(

Answer 14

ohh, where it says, "ask a question"

Answer 15

if we can see it, we can jump in and revise each other too :)

Answer 16

ok thanks!!