How do you find the zeros of #f(x)=(x^2-3x-4)/(x^2-x-12)#?
1 Answer
Jan 20, 2017
Explanation:
Factorising and simplifying f(x) as follows.
#f(x)=(x^2-3x-4)/(x^2-x-12)=(cancel((x-4))(x+1))/(cancel((x-4))(x-3))=(x+1)/(x-3)# with exclusion x ≠ 4
The zeros of f(x) are the values of x which make f(x) equal zero. That is f(x)=0.
For
#f(x)=(x+1)/(x+3)# The denominator of f(x) cannot be zero as this would make f(x) undefined. The numerator is the only part of the rational function that can equal zero.
#rArrx+1=0rArrx=-1" is the zero of f(x)"#
