Question

Consider the function given below.

Answer 1

Explanation:

`f(x)=x^3+2x^2-x-2=x^2(x+2)-1(x+2)=(x+2)(x^2-1)\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\f(x)=(x+2)(x-1)(x+1)\\\\\text{The zeros (x-intercepts):}\\\\f(x)=0\iff(x+2)(x-1)(x+1)=0\iff \\x+2=0\ \vee\ x-1=0\ \vee\ x+1=0\\\\x=-2\ \vee\ x=1\ \vee\ x=-1\\\\\text{y-intercept}:\\\text{Put x = 0:}\\\\f(0)=0^3+2(0^2)-0+2=2\to(0,\ 2)`