This equation can be represented as follows(quadratic equation). The vertex of the quadratic equation is where the slope is 0.
[tex]ax^2+bx+c=0[/tex][tex]\begin{gathered} y=\frac{1}{5}x^2+2x-8 \\ \frac{dy}{dx}=\frac{x}{5}\text{ }\times2+2 \\ The\text{ slope = }\frac{dy}{dx} \\ \frac{dy}{dx}\text{ = 0} \\ \frac{2}{5}x+2=0 \\ \frac{2}{5}x=-2 \\ 2x=-10 \\ x=-5 \end{gathered}[/tex]The corresponding y coordinate can be find by substituting x = -5 in the equation
[tex]\begin{gathered} y=\frac{1}{5}x^2+2x-8 \\ y=\text{ }\frac{-5^2}{5}+2(-5)-8 \\ y=\frac{25}{5}-10-8 \\ y=\text{ 5-10-8} \\ y=-13 \end{gathered}[/tex]