# What is the vertex of y= 2x^2 -4x - 12?

Mar 18, 2017

Vertex$\text{ "->" } \left(x , y\right) = \left(1 , - 14\right)$

#### Explanation:

I am going to use part of the process of completing the square.

Write as:$\text{ } y = 2 \left({x}^{2} - \frac{4}{2} x\right) - 12$

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

So by substitution:

${y}_{\text{vertex}} = 2 {\left(1\right)}^{2} - 4 \left(1\right) - 12 = - 14$

Vertex$\text{ "->" } \left(x , y\right) = \left(1 , - 14\right)$

Mar 18, 2017

The vertex is $= \left(1 , - 14\right)$

#### Explanation:

We need

${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$

Let complete the squares and factorise

$y = 2 {x}^{2} - 4 x - 12$

$y = 2 \left({x}^{2} - 2 x\right) - 12$

$y = 2 \left({x}^{2} - 2 x + 1\right) - 12 - 2$

$y = 2 {\left(x - 1\right)}^{2} - 14$

Therefore,

the vertex is $= \left(1 , - 14\right)$

graph{(y-(2x^2-4x-12))((x-1)^2+(x+14)^2-0.01)=0 [-7.8, 6.25, -14.32, -7.295]}