Answer
Vertex: $(1/4, -25/8)$
Opens upward
x-intercepts: -1, 3/2
y-intercept: -3
Work Step by Step
$f(x)=2x^2-x-3$
$y=2x^2-x-3$
$y+3=2x^2-x-3+3$
$y+3=2x^2-x$
$y+3=2(x^2-1/2*x)$
$y+3+2*(1/2*2)^2=2(x^2-1/2*x+(1/2*2)^2)$
$y+3+2*(1/4)^2=2(x^2-1/2*x+(1/4)^2)$
$y+3+2*(1/16) =2(x^2-1/2*x+(1/16))$
$y+3+1/8 =2(x^2-1/2*x+1/16)$
$y+25/8 =2(x-1/4)^2$
$y+25/8-25/8 =2(x-1/4)^2-25/8$
$y=2(x-1/4)^2-25/8$
Vertex: $(1/4, -25/8)$
Opens upward
$x=0$
$f(x)=2x^2-x-3$
$f(0)=2*0^2-0-3$
$f(0)=2*0-3$
$f(0)=0-3$
$f(0)=-3$
$y=0$
$y+25/8 =2(x-1/4)^2$
$0+25/8 =2(x-1/4)^2$
$ 25/8 =2(x-1/4)^2$
$ 25/8*1/2 =1/2*2(x-1/4)^2$
$ 25/16 = (x-1/4)^2$
$\sqrt {25/16} =\sqrt {(x-1/4)^2}$
$±5/4 = x-1/4$
$-5/4 =x-1/4$
$-5/4 +1/4 =x-1/4 +1/4$
$-1 = x$
$5/4=x-1/4$
$5/4+1/4=x-1/4+1/4$
$3/2 = x$
x-intercepts: -1, 3/2
y-intercept: -3
