What is the vertex form of y=4x^2-17x-16?

1 Answer
Feb 1, 2016

y=4(x-17/8)^2-545/16

Explanation:

We start with 4x^2-17x-16=y
4x^2-17x-16 cannot be factored, so we will have to complete the square. To do that, we first have to make the coefficient of x^2 1. That makes the equation now 4(x^2-17/4x-4).

The way completing the square works is, because x^2-17/4x isn't factorable, we find a value that makes it factorable. We do that by taking the middle value, -17/4x, dividing it by two and then squaring the answer. In this case it would look this: (-17/4)/2, which equals -17/8. If we square it, that becomes 289/64.

We can rewrite the equation as 4(x^2-17/4x+289/64-4), but we can't just stick a number into an equation and not add it on both sides. We could add 289/64 to both sides, but I would prefer to just add 289/64 and then immedietly subtract it.

Now, we can rewrite this equation as 4(x^2-17/4x+289/64-289/64-4). Because x^2-17/4x+289/64 is factorable, I can rewrite it as (x-17/8)^2. Putting it together we have 4(x-17/8)^2-289/64-4 or 4(x-17/8)^2-545/64. The last step is to multiply -545/64 by 4.

The final form is 4(x-17/8)^2-545/16