Question

How do I use the vertex formula to determine the vertex of the graph for `-3x^2+12x-9`?

Answer 1

Vertex form: `f(x) = a(x+b/(2a))^2 + f(-b/(2a))`

x of vertex:`x =( -b/(2a) )= -4/-2 = 2`

y of vertex: f(-2) = -12 + 24 - 9 = 3

Vertex form: `f(x) = -3(x - 2)^2 + 3`

Check by developing:

`f(x) = -3(x^2 - 4x + 4) + 3 = -3x^2 + 12x - 9`. OK

Answer 2

For `y=ax^2+bx+c`, the vertex formula tells us that the `x`-coordinate of the vertex is `-b/(2a)`

For `y=-3x^2+12x-9`, we have: `a=-3` and `b=12`

Therefore, the vertex formula gives us `x` coordinate is

`-12/(2(-3)) = -12/(-6) = 1(-2) = 2`

We still need the `y` coordinate of the vertex, so use the equation to find it:

`y = -3(2)^2+12(2)-9 = -12+24-9 = 3`

The vertex is `(2,3)`

graph{y=-3x^2+12x-9 [-10, 10, -5, 5]}