Chapter 11 - Section 11.6 - Further Graphing of Quadratic Functions - Exercise Set - Page 820: 13

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) $x=0$ $f(x)=x^2+4x−5$ $f(0)=0^2+4∗0−5$ $f(0)=0+0−5$ $f(0)=−5$ $0=x^2+4x−5$ $0=(x+5)(x−1)$ $x+5=0$ $x+5−5=0−5$ $x=−5$ $x−1=0$ $x−1+1=0+1$ $x=1$