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:
$f(x)=(x+1)^2-4$
$0=(x+1)^2-4$
$0+4=(x+1)^2-4+4$
$4=(x+1)^2$
$\sqrt 4 = \sqrt {(x+1)^2}$
$±2 = x+1$
$2=x+1$
$2-1=x+1-1$
$1=x$
$-2=x+1$
$-2-1=x+1-1$
$-3=x$
