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

1 Answer
Sep 12, 2017

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