What is the vertex of $y = 2 x^{2} - 4 x - 12$ $y= 2x^2 -4x - 12$ ?

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Answer 1 · 2017-03-18T13:00:57

Vertex $\to (x, y) = (1, - 14)$ $" "->" "(x,y)=(1,-14)$

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

Write as: $y = 2 (x^{2} - \frac{4}{2} x) - 12$ $" "y=2(x^2-4/2x)-12$

$x_{vertex} = (- \frac{1}{2}) \times (- \frac{4}{2}) = + 1$ $x_("vertex")=(-1/2)xx(-4/2) = + 1$

So by substitution:

$y_{vertex} = 2 {(1)}^{2} - 4 (1) - 12 = - 14$ $y_("vertex")=2(1)^2-4(1)-12 = -14$

Vertex $\to (x, y) = (1, - 14)$ $" "->" "(x,y)=(1,-14)$

Tony B

Answer 2 · 2017-03-18T13:06:52

The vertex is $= (1, - 14)$ $=(1,-14)$

We need

$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2=(a-b)^2$

Let complete the squares and factorise

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

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

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

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

Therefore,

the vertex is $= (1, - 14)$ $=(1,-14)$

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

What is the vertex of y=2x2−4x−12y= 2x^2 -4x - 12?