## Algebra 1

$y=3x^2+x-5$ axis of symmetry: $x=-b/2a$ $x=-b/2a$ $x=-1/2*3$ $x = -1/6$ $y=3x^2+x-5$ $y=3(-1/6)^2+(-1/6)-5$ $y= 3*1/36 -1/6-5$ $y=3/36-1/6-5$ $y=1/12-31/6$ $y=1/12-62/12$ $y=-61/12$ $(-1/6, -61/12)$ is the vertex Since $0$ is 1/6 from the vertex, we also know $-1/3$ is 1/6 from the vertex (and both values have the same y-value). $y=3x^2+x-5$ $y=3*0^2+0-5$ $y=3*0+0-5$ $y=0+0-5$ $y=-5$