How do you find the vertex and the intercepts for `-3x^2+12x-9`?

The given equation is `-3x^2 +12x - 9 = y`.
Given that the coefficient(constant number in front) of `x^2` is negative 3, for any NEGATIVE coefficient, the graph is a SAD smiley (has a maximum point).
Note for a POSITIVE coefficient, the graph is a HAPPY smiley (has a minimum point).

Complete the square to find the vertex:

`-3x^2 +12x -9 =0`
`-3( x^2 - 4x) - 9 = 0`
`-3(x - 2)^2 - (-3)(2)^2 - 9 = 0`
`-3(x - 2)^2 +12 - 9 = 0`
`-3(x - 2)^2 + 3 =0`

In this form `-3(x-2)^2 + 3 = 0`,
the constant 3 is the y coordinate of the vertex.
And (x-2) = 0 is the x coordinate of the vertex, equate for x,
x=2 is the x coordinate. Hence the vertex is (2, 3).

To find the intercepts:

For y-intercept, equate x=0,
`-3(0)^2 +12(0) -9 = y`
`0+0 -9 = y`
y= -9 is the y coordinate of the y-intercept.

For x-intercept, equate y=0, then factorise to find the 2 x intercepts,
`-3x^2 +12x - 9 = 0`
`(-3x+3)(x -3 ) = 0`
Hence, the factors of the equation is (-3x+3) and (x-3), solve for x,
`(-3 x +3) =0`
`x = (-3)/-3`
`x = 1`

(x-3) =0,
x=3
Hence the x intercepts are x=1 and x =3.