Question

How do you find the vertex and the intercepts for `f(x)=-x^2+6x+6`?

Answer 1

`"see explanation"`

Explanation:

`"for the quadratic equation in standard form"`

`•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0`

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

`f(x)=-x^2+6x+6" is in standard form"`

`"with "a=-1,b=6,c=6`

`rArrx_(color(red)"vertex")=-6/(-2)=3`

`"substitute this value into the equation for y"`

`y=f(3)=-(3)^2+6(3)+6=15`

`rArrcolor(magenta)"vertex "=(3,15)`

`color(blue)"for intercepts"`

`• " let x = 0, in equation for y-intercept"`

`• " let y = 0, in equation for x-intercepts"`

`x=0toy=6larrcolor(red)" y-intercept"`

`y=0to-x^2+6x+6=0`

`"solve for x using the "color(blue)"quadratic formula"`

`x=(-6+-sqrt(36+24))/(-2)`

`color(white)(x)=(-6+-sqrt60)/(-2)`

`color(white)(x)=(-6+-2sqrt15)/(-2)=3+-sqrt15`

`rArrx~~ -0.87,x~~ 6.87larrcolor(red)" x-intercepts"`