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 of the given function
`f(x)=(x-4)/(x^2+13x+36)` is
`x=-4` and
`x=-9`

Explanation:

We have been given with the function
`f(x)=(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`

Answer 2

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}`