What are the vertical asymptotes of the function f(x) = the quantity of 2 x plus 8, all over x squared plus 5 x plus 6?
You'd find the vertical asymptotes by seeing where the denominator equals zero; you can do so by factoring the denominator.
In this case, you can factor the denominator into (x+3)(x+2), so if you set each of those equal to zero you can find the equations of the vertical asymptotes (x=-3 and x=-2).
Answer:
Step-by-step explanation:
Alright, lets get started.
The given rational function is :
\(f(x)=\frac{2x+8}{x^2+5x+6}\)
For finding the vertical asymptotes of a rational function, we must set the denominator equal to zero.
So, equaling denominator to zero :
\(x^2 +5x+6 = 0\)
factoring
\((x+3)(x+2)=0\)
This will give two values of x
\(x=-3, x=-2\)
So, these two are vertical asymptotes : Answer
Hope it will help :)