## Algebra 1: Common Core (15th Edition)

We are given: $y=3x^2+x-5$ Let's list some values: $x=-2 \rightarrow y=1$ $x=-1 \rightarrow y=\frac{-1}{2}$ $x=0 \rightarrow y=-3$ $x=1 \rightarrow y=\frac{-1}{2}$ $x=2 \rightarrow y=1$ The x-coordinate of the vertex is given by $x=\frac{-b}{2a}=\frac{-1}{2.3}=-\frac{1}{6}$ Find the y-coordinate of the vertex $y=3(-\frac{1}{6})^2+(-\frac{1}{6})-5=\frac{-61}{12}$ Hence, the vertex is $(-\frac{1}{6},\frac{-61}{12})$.