How do you find the vertex and the intercepts for f(x)=-x^2+6x+6 f(x)=x2+6x+6?

1 Answer
Sep 2, 2017

"see explanation"see explanation

Explanation:

"for the quadratic equation in standard form"for the quadratic equation in standard form

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0xy=ax2+bx+cx;a0

x_(color(red)"vertex")=-b/(2a)xvertex=b2a

f(x)=-x^2+6x+6" is in standard form"f(x)=x2+6x+6 is in standard form

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

rArrx_(color(red)"vertex")=-6/(-2)=3xvertex=62=3

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

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

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

color(blue)"for intercepts"for intercepts

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

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

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

y=0to-x^2+6x+6=0y=0x2+6x+6=0

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

x=(-6+-sqrt(36+24))/(-2)x=6±36+242

color(white)(x)=(-6+-sqrt60)/(-2)x=6±602

color(white)(x)=(-6+-2sqrt15)/(-2)=3+-sqrt15x=6±2152=3±15

rArrx~~ -0.87,x~~ 6.87larrcolor(red)" x-intercepts"x0.87,x6.87 x-intercepts