Question

What is the vertex form of `2y=5x^2-3x+11`?

Answer 1

see explanation

Explanation:

...I can never remember it, so I always have to look it up.

The vertex form of a quadratic equation is:

`f(x) = a(x - h)^2 + k`

So, for your original equation `2y = 5x^2 - 3x + 11`, you have to do some algebraic manipulation.

Firstly, you need the `x^2` term to have a multiple of 1, not 5.
So divide both sides by 5:

`2/5y = x^2 - 3/5x + 11/5`

...now you have to perform the infamous "complete the square" maneuver. Here's how I go about it:

Say that your `-3/5` coefficient is `2a`. Then `a = -3/5 * 1/2 = -3/10`

And `a^2` would be `9/100`.

So, if we add and subract this from the quadratic equation, we'd have:

`2/5y = x^2 - 3/5x + 9/100 - 9/100 + 11/5`

...and now the 1st 3 terms of the right side are a perfect square in form `(x - a)^2 = x^2 - 2ax + a^2`

...so you can write:

`2/5y = (x - 3/10)^2 + (11/5 - 9/100)`

`2/5y = (x - 3/10)^2 + (220 - 9)/100`

`2/5y = (x - 3/10)^2 + 211/100`

So now, all you gotta do is multiply through by `5/2`, giving:

`y = 5/2(x-3/10)^2 + 5/2*211/100`

`y = 5/2(x-3/10)^2 + 211/40`

which is vertex form, `y = a(x-h)^2 + k`

where `a = 5/2`, `h = 3/10`, and `k = 211/40`