# How do you find the vertex of y=2x^2-6x+1?

Feb 28, 2016

By a process called completing the square, we find that the vertex is at (3/2, -7/2).

#### Explanation:

Completing the square goes like this. Take the quadratic and linear terms, 2(x^2)-6x. We can imagine this as 2*((x^2)-3x). Now think of the identity

(a+b)^2 = (a^2)+2ab+(b^2).

We can think of a as x, and if we take b equal to -3/2 then the -3x fits with 2ab. So the (b^2) part that completes the square is 9/4. We then have:

(x-3/2)^2 = (x^2)-3x+9/4

Multiplying by 2 because we started with 2(x^2):

2*(x-3/2)^2 = 2(x^2)-6x+9/2

That is "almost right". But we have

y = 2(x^2)-6x+1.

To make up our difference we subtract 7/2 from our completed-square expression and we get:

y = 2*(x-3/2)^2 - 7/2

Now we can read off the vertex. The x value is where the squared quantity is zero, thus x = 3/2. And the y value at the vertex is -7/2 because that's what's left when the completed square is zero.

Completing the square solves quadratic equations too. Look for that phrase on Google or whatever search engine you use.

Feb 28, 2016

$\left(\frac{3}{2} , - \frac{7}{2}\right)$

#### Explanation:

The given equation
$y = 2 {x}^{2} - 6 x + 1$
Dividing both sides by 2 we have
$\frac{y}{2} = {x}^{2} - 3 x + \frac{1}{2}$
$\implies \frac{y}{2} = {x}^{2} - 2 x \frac{3}{2} + {\left(\frac{3}{2}\right)}^{2} - \frac{9}{4} + \frac{1}{2}$
$\implies \frac{y}{2} = {\left(x - \frac{3}{2}\right)}^{2} - \frac{7}{4}$
$\implies \frac{y}{2} + \frac{7}{4} = {\left(x - \frac{3}{2}\right)}^{2}$
$\implies \frac{1}{2} \left(y + \frac{7}{2}\right) = {\left(x - \frac{3}{2}\right)}^{2}$
putting
y+7/2=Y & (x-3/2)=X we have
new equation of parabola as
$\frac{Y}{2} = {X}^{2}$ having vertex at (0,0)
So putting X=0 and Y =0 we will get
$x = \frac{3}{2}$ and $y = - \frac{7}{2}$

$\therefore$Coordinate of vertex $= \left(\frac{3}{2} , - \frac{7}{2}\right)$