## Algebra 1

$y=2x^2+5$ axis of symmetry: $x=-b/2a$ $x=0/2*2$ $x=0$ vertex: $y=2x^2+5$ $y=2*0^2+5$ $y=2*0+5$ $y=5$ One other point on the curve: $x=1$ $y=2x^2+5$ $y=2*1^2+5$ $y=2*1+5$ $y=7$ $1-0=1$, so we also know $(-1,7)$ is a point on the curve.