What is the axis of symmetry and vertex for the graph #f(x)=-3x^2+6x+12#?

2 Answers
Jun 16, 2017

Axis of symmetry is #x=1#, vertex is at #(1,15)#.

Explanation:

#f(x)= -3x^2+6x+12= -3(x^2-2x)+12 = -3(x^2-2x+1)+3+12#

#= -3(x-1)^2+15# . Comparing with standard vertex form of equation #f(x)= a(x-h)^2+k ; (h,k)# being vertex.

Here #h=1,k=15# . So vertex is at #(1,15)#.

Axis of symmetry is #x=1#

graph{-3x^2+6x+12 [-40, 40, -20, 20]} [Ans]

Jun 16, 2017

#x=1, "vertex "=(1,15)#

Explanation:

#"for a parabola in standard form " y=ax^2+bx+c#

#"the x-coordinate of the vertex is " x_(color(red)"vertex")=-b/(2a)#

#y=-3x^2+6x+12" is in standard form"#

#"with " a=-3, b=6" and " c=12#

#rArrx_(color(red)"vertex")=-6/(-6)=1#

#"substitute this value into function for y-coordinate"#

#y_(color(red)"vertex")=-3+6+12=15#

#rArrcolor(magenta)"vertex "=(1,15)#

#"since " a<0" then graph has a maximum " nnn#

#"the axis of symmetry passes through the vertex "#

#rArrx=1" is equation of axis of symmetry"#
graph{(y+3x^2-6x-12)(y-1000x+1000)=0 [-40, 40, -20, 20]}