Answer
Opens upward
Vertex: (−1,−4)
Y-intercept: -3
X-intercepts: -3, 1
Work Step by Step
$f(x)=x^2+2x−3$
$f(x)=(x^2+2x)−3$
$f(x)=(x^2+2x+(2/2)^2)−3−(2/2)^2$
$f(x)=(x^2+2x+1^2)−3−1^2$
$f(x)=(x^2+2x+1)−3−1$
$f(x)=(x+1)^2−4$
Vertex: (−1,−4)
$x=0$
$f(x)=x^2+2x−3$
$f(0)=0^2+2∗0−3$
$f(0)=0+0−3$
$f(0)=−3$
x-intercepts:
$f(x)=(x+1)^2−4$
$0=(x+1)^2−4$
$0+4=(x+1)^2−4+4$
$4=(x+1)^2$
$\sqrt4=\sqrt {(x+1)^2}$
$±2=x+1$
$2=x+1$
$2−1=x+1−1$
$1=x$
$−2=x+1$
$−2−1=x+1−1$
$−3=x$
