Identify the vertical asymptotes of f(x) = x-4/x^2+13x+36

Question

Identify the vertical asymptotes of                          f(x) = x-4/x^2+13x+36

anonymous · Answer

Parenthesis please...  I'll rewrite it for what I think you meant

anonymous · Answer

$$(x-4)/(x^2+13x+36)=\frac{x-4}{x^2+13x+36}$$
This?

anonymous · Answer

yes thats it

anonymous · Answer

What you wrote initially was:
$$x-\frac{4}{x^2}+13x+36$$

So yeah, definitely NOT the same thing :-D  It's kinda important, just for future reference.

Here is your function factored:
$$ \frac{(x-4)}{(x+4)(x+9)} $$

Check:
x*x=x^2
4x+9x=13x
4*9=36

Vertical asymptotes are x = some constant, numerical #

You want to solve the denominator for when it's equal to zero
(x+4)(x+9)=0


See if you can find the answer now :-)

anonymous · Answer

If you graph it on a TI-84 or something like that, when you trace along the function to those x values you'll get errors or blanks for the "y=" display.