To determine the horizontal intercepts, we'll solve \(0=5x^2-2x-\frac{1}{3}\) using the quadratic formula. One option is to directly use \(a=5\text{,}\) \(b=-2\text{,}\) and \(c=-\frac{1}{3}\text{.}\) Another option is to first multiply each side of that equation by \(3\text{,}\) which will yield the equivalent equation \(0=15x^2-6x-1\text{.}\) From there, we can use the quadratic formula with \(a=10\text{,}\) \(b=-6\) and \(c=-1\text{:}\)

\begin{align*} 0=\amp=5x^2-2x-\frac{1}{3}\\ \multiplyleft{3}(0)\amp=\multiplyleft{3}\left(5x^2-2x-\frac{1}{3}\right)\\ 0\amp=15x^2-6x-1\\ x\amp=\frac{-(-6)\pm \sqrt{(-6)^2-4(15)(-1)}}{2(15)}\\ x\amp=\frac{6\pm \sqrt{96}}{30}\\ \end{align*}

Simplifying, we get:

\begin{align*} x\amp=\frac{6\pm 4\sqrt{6}}{30}\\ x\amp=\frac{3\pm 2\sqrt{6}}{15}\\ \end{align*}

Rouding, we have:

\begin{align*} x\amp\approx -0.1266,\ x\approx 0.5266 \end{align*}

The horizontal intercepts occur at the points \(\left(\frac{3+ 2\sqrt{6}}{15},0\right)\) or \(\left(\frac{3-2\sqrt{6}}{15},0\right)\) , or at the approximate points \((-0.1266,0)\) and \((0.5266,0)\text{.}\)

To determine the vertical intercept, we'll set \(x=0\text{:}\)

\begin{align*} y\amp=5(\substitute{0})^2-2(\substitute{0})-\frac{1}{3}\\ \amp=-\frac{1}{3} \end{align*}

Thus the \(y\)-intercept is \(\left(0,-\frac{1}{3}\right)\text{.}\)

Using \(a=5\) and \(b=-2\text{,}\) we see that the vertex occurs at:

\begin{align*} x\amp=-\frac{\substitute{-2}}{2(\substitute{5})}\\ \amp=\frac{1}{5} \end{align*}

Now using this to determine the \(y\)-value, we have:

\begin{align*} y\amp=5\left(\substitute{\frac{1}{5}}\right)^2-2\left(\substitute{\frac{1}{5}}\right)-\frac{1}{3} \end{align*}