Answer
Vertex: $(-1,-3)$
Opens upward
x-intercepts: -2.2, .2
y-intercept: -1
Work Step by Step
$f(x)=2x^2+4x-1$
$x=0$
$f(x)=2x^2+4x-1$
$f(0)=2*0^2+4*0-1$
$f(0)=2*0+0-1$
$f(0)=0-1$
$f(0)=-1$
$f(x)=2x^2+4x-1$
$y=2x^2+4x-1$
$y=2(x^2+2x)-1$
$y=2(x^2+2x+(2/2)^2)-1-2*(2/2)^2$
$y=2(x^2+2x+1^2)-1-2*1^2$
$y=2(x^2+2x+1)-1-2*1$
$y=2(x+1)^2-1-2$
$y=2(x+1)^2-3$
Vertex: $(-1,-3)$
Coefficient of $x^2$ is positive, so graph opens upward
$y=0$
$y=2(x+1)^2-3$
$0=2(x+1)^2-3$
$0+3=2(x+1)^2-3+3$
$3=2(x+1)^2$
$3/2=2(x+1)^2/2$
$3/2 = (x+1)^2$
$\sqrt {3/2} = \sqrt {(x+1)^2}$
$±\sqrt {1.5} = x+1$
$\sqrt {1/5}=x+1$
$\sqrt {1.5}-1 = x+1-1$
$\sqrt {1.5}-1 = x$
$1.22-1 =x$
$.22 =x$
$x=.2$
$-\sqrt {1/5}=x+1$
$-\sqrt {1.5}-1 = x+1-1$
$-\sqrt {1.5}-1 = x$
$-1.22-1=x$
$-2.22=x$
$x=-2.2$
