Question

What is the vertex of y= 1/5x^2+ 2x-8

Answer 1

`y=(1)/(5)x^2+2x-8`

This equation can be represented as follows(quadratic equation). The vertex of the quadratic equation is where the slope is 0.

`ax^2+bx+c=0`
`\begin{gathered} y=(1)/(5)x^2+2x-8 \\ (dy)/(dx)=(x)/(5)\text{ }*2+2 \\ The\text{ slope = }(dy)/(dx) \\ (dy)/(dx)\text{ = 0} \\ (2)/(5)x+2=0 \\ (2)/(5)x=-2 \\ 2x=-10 \\ x=-5 \end{gathered}`

The corresponding y coordinate can be find by substituting x = -5 in the equation

`\begin{gathered} y=(1)/(5)x^2+2x-8 \\ y=\text{ }(-5^2)/(5)+2(-5)-8 \\ y=(25)/(5)-10-8 \\ y=\text{ 5-10-8} \\ y=-13 \end{gathered}`