Question

How do you find the vertex of `y= -3x^2+ 12x-10`?

Answer 1

Vertex: `(2, 2)`

Explanation:

`y = -3x^2 + 12x -10`

Use completing of the squares to put the equation in standard form: `y = a(x-h)^2 + k`, where vertex: `(h, k)`, axis of symmetry: `x = h`

1. Factor the `x` terms: `y = -3(x^2 - 4x) -10`

2. Take `1/2` of the `x-`term coefficient: `1/2 * -4 = -2`:
   `(x-2)^2` is the completed square.

3. Square the value from step 2: `(-2)^2 = 4`

4. Multiply the value from step 3 by the factored value `-3`: `4*-3 = -12`. This means we need to add `12` to the equation because when we completed the square we subtracted `-12`:`-3(x -2)^2 = -3(x^2 - 4x +4) = -3x^2 +12x -12`

5. `y = -3(x -2)^2 - 10 +12`

6. `y = -3(x -2)^2 + 2`

7. vertex: `(2, 2)`

Answer 2

There is a sort of cheat (not really) way of doing this

Vertex`->(x,y)=(2,2)`

Explanation:

Write as `y=-3(x^2-4x)-10`

This part way to completing the square.

`x_("vertex")=(-1/2)xx(-4) = +2`

Determine y by substitution

`y=-3(2)^2+12(2)-10`

`y_("vertex")= +2`

Vertex`->(x,y)=(2,2)`