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

Mar 3, 2017

Vertex: $\left(2 , 2\right)$

#### Explanation:

$y = - 3 {x}^{2} + 12 x - 10$

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

1. Factor the $x$ terms: $y = - 3 \left({x}^{2} - 4 x\right) - 10$

2. Take $\frac{1}{2}$ of the $x -$term coefficient: $\frac{1}{2} \cdot - 4 = - 2$:
${\left(x - 2\right)}^{2}$ is the completed square.

3. Square the value from step 2: ${\left(- 2\right)}^{2} = 4$

4. Multiply the value from step 3 by the factored value $- 3$: $4 \cdot - 3 = - 12$. This means we need to add $12$ to the equation because when we completed the square we subtracted $- 12$:$- 3 {\left(x - 2\right)}^{2} = - 3 \left({x}^{2} - 4 x + 4\right) = - 3 {x}^{2} + 12 x - 12$

5. $y = - 3 {\left(x - 2\right)}^{2} - 10 + 12$

6. $y = - 3 {\left(x - 2\right)}^{2} + 2$

7. vertex: $\left(2 , 2\right)$

Mar 3, 2017

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

Vertex$\to \left(x , y\right) = \left(2 , 2\right)$

#### Explanation:

Write as $y = - 3 \left({x}^{2} - 4 x\right) - 10$

This part way to completing the square.

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(- 4\right) = + 2$

Determine y by substitution

$y = - 3 {\left(2\right)}^{2} + 12 \left(2\right) - 10$

${y}_{\text{vertex}} = + 2$

Vertex$\to \left(x , y\right) = \left(2 , 2\right)$