Question

Identify the vertical asymptotes of f(x) = quantity x minus 4 over quantity x squared plus 13 x plus 36

Answer 1

vertical asymptotes, occur when the rational is undefined, and that happens when the denominator turns to 0, for this specific case, is when  $\bf \cfrac{x-4}{x^2+13x+36}\implies \cfrac{x-4}{0}\impliedby un de fined$

when does that  happen?  well, it happens at 

$\bf x^2+13x+36=0\implies (x+4)(x+9)=0\implies x= \begin{cases} -4\\ -9 \end{cases}$

Answer 2

Answer:

vertical asymptotes of the given function $f(x)=\frac{x-4}{x^2+13x+36}$ is $x=-4$ and $x=-9$

Explanation:

We have been given with the function $f(x)=\frac{x-4}{x^2+13x+36}$  and we need to find the vertical asymptotes of the function

So, to find the  vertical asymptotes we equate the denominator of given function to zero

we will get  ${x^2+13x+36}$=0

So, as to find the value of x we will factorise the above equation we will get

$(x+9)$  and $(x+4)$

Hence, vertical asymptotes are at $x=-9$ and $x=-4$