How do you find the vertex and the intercepts for #f(x)=-x^2+6x+6 #?
1 Answer
Sep 2, 2017
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"#