91Ó°ÊÓ

Consider the given function,

$y=6x^2+2x-1$

Compare localid="1645194963571" $y=a{x}^2+bx+c$with the given function,

localid="1645194979505" $$\begin{gathered} a=6, \\ b=2, \\ c=-1 \end{gathered}$$

Since $a=6$, the parabola opens to the right.

The axis of symmetry, localid="1645196732423" $x=-\frac{b}{2a}$. Then,

localid="1645196736005" $$\begin{gathered} x=-\frac{2}{2\left(6\right)} \\ =-\frac{2}{12} \\ =-\frac{1}{6} \end{gathered}$$

The vertex is on the line$x=-\frac{1}{6}$.

Substitute $x=-\frac{1}{6}$ in the original equation,

$$\begin{gathered} y=6{\left(-\frac{1}{6}\right)}^2+2\left(-\frac{1}{6}\right)-1 \\ =\frac{1}{6}-\frac{1}{3}-1 \\ =\frac{-7}{6} \end{gathered}$$

Therefore, the vertex is $\left(-\frac{1}{6},-\frac{7}{6}\right)$.

Substitute $y=0$in the original equation,

$$\begin{gathered} 0=6x^2+2x-1 \\ x≈\frac{-1-\sqrt{7}}{6},\frac{-1+\sqrt{7}}{6} \end{gathered}$$

Therefore, the x-intercepts are $\left(\frac{-1-\sqrt{7}}{6},0\right),\left(\frac{-1+\sqrt{7}}{6},0\right)$.

Substitute $x=0$in the original equation,

$$\begin{gathered} y=6{\left(0\right)}^2+2\left(0\right)-1 \\ =-1 \end{gathered}$$

Therefore, the y-intercept is $\left(0,-1\right)$.

The point symmetric tolocalid="1645197343108" $\left(0,1\right)$across the axis of symmetry is localid="1645197346856" $\left(-\frac{1}{3},-1\right)$.

On graphing the parabola we get,