Question

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

Answer 1

assumin ya meant

`f(x)=(x-4)/(x^2+13x+36)`

to solve, fimplify then set the denomenator equal to 0


`f(x)=(x-4)/(x^2+13x+36)=(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

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

Explanation:

Identify the vertical asymptotes of
`(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`