Question

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

Answer 1

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]

Answer 2

`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]}