Question

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

Answer 1

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.

Answer 2

`(3/2,-7/2)`

Explanation:

The given equation
`y=2x^2-6x+1`
Dividing both sides by 2 we have
`y/2=x^2-3x+1/2`
`=>y/2=x^2-2x3/2+(3/2)^2-9/4+1/2`
`=>y/2=(x-3/2)^2-7/4`
`=>y/2+7/4=(x-3/2)^2`
`=>1/2(y+7/2)=(x-3/2)^2`
putting
`y+7/2=Y & (x-3/2)=X` we have
new equation of parabola as
`Y/2=X^2` having vertex at (0,0)
So putting X=0 and Y =0 we will get
`x= 3/2` and `y =-7/2`

`:.`Coordinate of vertex `= (3/2,-7/2)`