Answer
Vertex: $(-3/4, 289/8)$
Opens downward
x-intercepts: -5, 7/2
y-intercept: 35
Work Step by Step
$f(x)=-2x^2-3x+35$
$f(x)=-2(x^2+3/2x)+35$
$f(x)=-2(x^2+3/2x)+35+(3/2*1/2)^2-(3/2*1/2)^2$
$f(x)=-2(x^2+3/2x)+35+(-2)(3/4)^2-(-2)(3/4)^2$
$f(x)=-2(x^2+3/2x)+35+(-2)(9/16)-(-2)(9/16) $
$f(x)=-2(x^2+3/2x+9/16)+35-(-2)(9/16) $
$f(x)=-2(x+3/4)^2+35+(9/8) $
$f(x)=-2(x+3/4)^2+289/8$
Vertex: $(-3/4, 289/8)$
Opens downward
$x=0$
$f(x)=-2x^2-3x+35$
$f(0)=-2*0^2-3*0+35$
$f(0)=-2*0-0+35$
$f(0)=0-0+35$
$f(0)=35$
$y=0$
$f(x)=-2(x+3/4)^2+289/8$
$0=-2(x+3/4)^2+289/8$
$0-289/8=-2(x+3/4)^2+289/8-289/8$
$-289/8=-2(x+3/4)^2$
$-289/8*-1/2=-2(x+3/4)^2*-1/2$
$289/16=(x+3/4)^2$
$\sqrt {289/16}=\sqrt {(x+3/4)^2}$
$±17/4 = x+3/4$
$17/4=x+3/4$
$17/4-3/4 = x+3/4-3/4$
$14/4=x$
$7/2=x$
$-17/4=x+3/4$
$-17/4-3/4 = x+3/4-3/4$
$-20/4=x$
$-5=x$
x-intercepts: -5, 7/2
y-intercept: 35
