Question

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

Answer 1

assumin ya meant
$f(x)=\frac{x-4}{x^2+13x+36}$

to solve, fimplify then set the denomenator equal to 0

$f(x)=\frac{x-4}{x^2+13x+36}=\frac{x-4}{(x+4)(x+9)}$
hum, no simplificatoin today

so set denom equal to 0
(x+4)(x+9)=0
x+4=0
x=-4

x+9=0
x=-9

so vertical assemtotes at x=-9 and x=-4

Answer 2

Answer:

Vertical asymptote at $x= -9 \ and \ x=-4$

Step-by-step explanation:

Identify the vertical asymptotes of $\frac{x-4}{x^2+13x+36}$

To find vertical asymptote we set the denominator =0 and solve for x

Set $x^2+13x+36=0$ and solve for x

now we factor x^2 +13x+36

Product is 36 and sum is 13

$9 \cdot  4=36$

$9+4=13$

so the equation becomes

$(x+9)(x+4)=0$

Now set each factor=0 and solve for x

$x+9=0, x=-9$

$x+4=0, x=-4$

Vertical asymptote at $x= -9 \ and \ x=-4$