How do you identify the vertical asymptotes of #f(x) =2/(x^2+3x-10)#?

Question

Elias Ackerman · Accepted Answer

Find the zeros of the denominator and make sure they are not also zeros of the numerator. (If any of them are: factor, reduce the quotient and try again.)

#f(x) =2/(x^2+3x-10)# The denominator is #0# when #x^2+3x-10=0# #(x+5)(x-2)=0# The zeros are #-5# and #2# (For a quadratic equation, I always try factoring first, because it's fast. But don't spend a lot of time on factoring, because you can use the quadratic formula to get zeros.) Obviously, neither of these numbers is a zero of the numerator so: The vertical asymptotes are the lines whose equations are: #x=-5# and #x=2#

Anna Allen · Answer

undefined To identify the vertical asymptotes of the function f(x) = 2/(x^2 + 3x - 10), we need to find the values of x that make the denominator equal to zero, since vertical asymptotes occur where the function is undefined.

To do this, we set the denominator equal to zero and solve for x:
x^2 + 3x - 10 = 0

Next, we factor the quadratic equation:
(x + 5)(x - 2) = 0

Then, we set each factor equal to zero and solve for x:
x + 5 = 0  =>  x = -5
x - 2 = 0  =>  x = 2

So, the vertical asymptotes of the function f(x) = 2/(x^2 + 3x - 10) occur at x = -5 and x = 2.