Question

Question #2cf05

Answer 1

`y=2(x-(3/2))^2-(3/2)`
`vertex (3/2,-(3/2))`

Explanation:

`y=2x^2-6x+3`
`y-3=2x^2-3x`
`y-3=2(x^2-(3/2)x)`
`y-3=2(x^2-(3/2)x++9/4)-(9/2)`
`y=2(x-(3/2))^2-(9/2)+3`
`y=2(x-(3/2))^2-(3/2)`

`h=3/2, k=-(3/2)`

Answer 2

The vertex of `y=2x^2-6x+3` is `y=2(x-3/2)^2+3/2`

Explanation:

To solve this problem, we need to convert this quadratic equation from standard form to vertex form.

Let's start with the original equation:

`y=2x^2-6x+3`

Now let's turn the coefficient of the `x^2` term into `1`. This way, we're setting the stage for the process of completing the square.

`1/2y=x^2-3x+3/2`

Now we can start completing the square:

`1/2y+(-3/2)^2=x^2-3x+(-3/2)^2+3/2`

Then we simplify the left side and write `x^2-3x+(-3/2)^2` as the square of a binomial (a two-term polynomial):

`1/2y+9/4=(x-3/2)^2+3/2`

We then subtract `9/4` from both sides and simplify:

`1/2y=(x-3/2)^2+3/2-9/4=(x-3/2)^2+6/4-9/4=(x-3/2)^2-3/4`

All we need to do now is multiply by `2` to isolate `y` and obtain our vertex form:

`y=2((x-3/2)^2-3/4)=2(x-3/2)^2-2(-3/4)=2(x-3/2)^2+3/2`

And so our vertex form is:

`y=2(x-3/2)^2+3/2`