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*}