## Algebra 1

$y=5x^2+8$ vertex: $x=-b/2a$ $x=-0/2*5$ $x=0$ $y=5x^2+8$ $y=5*0^2+8$ $y=5*0+8$ $y=8$ $(0,8)$ One other point on the line $x=1$ $y=5(1)^2+8$ $y=5*1+8$ $y=5+8=13$ $(1,13)$ The blue line is the axis of symmetry, and the vertex is labeled.