Chapter 11 - Section 11.6 - Further Graphing of Quadratic Functions - Exercise Set - Page 820: 17

Vertex: $(0, -4)$ Opens upward x-intercept:-2, 2 y-intercept: -4

$f(x)=x^2-4$ $f(x)=(x-2)(x+2)$ $a=1$, $b=0$, $c=-4$ Vertex: $x=-b/2a$ $x=-0/2*1$ $x=0$ $x=0$ $f(x)=x^2-4$ $f(0)=0^2-4$ $f(0)=0-4$ $f(0)=-4$ Vertex: $(0, -4)$ Opens upward $y=0$ $f(x)=(x-2)(x+2)$ $0=(x-2)(x+2)$ $0=x-2$ $0+2=x-2+2$ $2=x$ $0=x+2$ $0-2=x+2-2$ $-2=x$ $x=0$ $f(x)=x^2-4$ $f(0)=0^2-4$ $f(0)=0-4$ $f(0)=-4$ x-intercept:-2, 2 y-intercept: -4