Question

How do you find the zeros of the function `f(x)=(x^2-x-12)/(x^2+2x-35)`?

Answer 1

`x=-3" and "x=4`

Explanation:

`"to find the zeros equate the numerator to zero and solve"`

`"solve "x^2-x-12=0`

`"The factors of - 12 which sum to - 1 are - 4 and + 3"`

`rArr(x-4)(x+3)=0`

`"equate each factor to zero and solve for x"`

`x-4=0rArrx=4`

`x+3=0rArrx=-3`
graph{(x^2-x-12)/(x^2+2x-35 [-10, 10, -5, 5]}