Chapter 8 - Section 8.6 - Further Graphing of Quadratic Functions - Exercise Set - Page 526: 23

Vertex: $(-2,-9)$ Opens upward Y-intercept: -5 X-intercepts: 1, -5

$f(x)=x^2+4x-5$ $f(x)=(x^2+4x)-5$ $f(x)=(x^2+4x+(4/2)^2)-5-(4/2)^2$ $f(x)=(x^2+4x+(2)^2)-5-(2)^2$ $f(x)=(x^2+4x+4)-5-4$ $f(x)=(x+2)^2-9$ Vertex: $(-2,-9)$ $x=0$ $f(x)=x^2+4x-5$ $f(0)=0^2+4*0-5$ $f(0)=0+0-5$ $f(0)=-5$ $0=x^2+4x-5$ $0=(x+5)(x-1)$ $x+5=0$ $x+5-5=0-5$ $x=-5$ $x-1=0$ $x-1+1=0+1$ $x=1$