Answer
Please see the graph.
Work Step by Step
$y=3x^2+x-5$
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).
$y=3x^2+x-5$
$y=3*0^2+0-5$
$y=3*0+0-5$
$y=0+0-5$
$y=-5$
