Question

How do you find the vertex of `f(x)=-5x^2+2x+7`?

Answer 1

`(1/5,36/5)`

Explanation:

`"for the standard quadratic function"`

`• y=ax^2+bx+c; a!=0`

`"the x-coordinate of the vertex is "`

`x_(color(red)"vertex")=-b/(2a)`

`"for " f(x)=-5x^2+2x+7`

`a=-5,b=2" and " c=7`

`rArrx_(color(red)"vertex")=-2/(-10)=1/5`

`"substitute this value into f(x) for y-coordinate"`

`rArry_(color(red)"vertex")=-5(1/5)^2+2(1/5)+7`

`color(white)(rArryverte)=-1/5+2/5+35/5=36/5`

`rArrcolor(magenta)"vertex " =(1/5,36/5)`