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

1 Answer
Sep 2, 2017

#"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"#