How change to Vertex Form?

Question

D=3t^2-78t+500
Can someone explain this please?

Answer 1

To change that into vertex form, you have to use a method known as "Complete the Square". Essentially, what completing the square would do is put your equation in the following form:

D = A(t - h)^2 + k

And the coordinates of your vertex would be (h,k).

Back to your problem.

D = 3t^2 - 78t + 500.

Your first step would be to factor the coefficient of t^2 from the first two terms. In this case, the coefficient is 3.

D = 3(t^2 - 26t) + 500

Your next step would be to add a magic number that will make the bracketed terms square. This magic number is "half squared" of the coefficient of t.

By half-squared, this is what I mean: Take half of the number, and then square it. In our case, the number in question is -26. Half of -26 is 13, and then 13 squared is 169. The magic number we have to add is 169, so we add this within the brackets.

D = 3(t^2 - 26t) + 500
D = 3(t^2 - 26t + 169) + 500 + ?

Note the question mark I added in there. I put that there because we can't just ADD a number into an equation for no reason; we have to offset the number we added so that it stays an equation. We added a 169 out of nowhere, but since we added it INSIDE the brackets with a 3 on the outside, what we *really* added was 3(169), or 507. The only way to compensate for this would be to SUBTRACT 507. That's what goes in our question mark.

D = 3(t^2 - 26t + 169) + 500 + 507

Now, we can complete the square.

D = 3(t - 13)^2 + 500 + 507

See what I did there? What goes in the brackets is half of -26.

Now, we can add 500 and 507 together.

D = 3(t - 13)^2 + 1007

This is the vertex form. Our coordinates of the vertex would be the NEGATIVE of what's inside the brackets (the negative of -13, which is 13), and the number outside of the brackets, 1007.

The coordinates of the vertex is therefore (13, 1007).

Answer 2

I got this from the link below. It's a good link with a longer explanation.

The vertex form of a quadratic is given by y = a(x – h)2 + k, where (h, k) is the vertex. The "a" in the vertex form is the same "a" as in y = ax2 + bx + c (that is, both a's have exactly the same value). The sign on "a" tells you whether the quadratic opens up or opens down. Think of it this way: A positive "a" draws a smiley, and a negative "a" draws a frowny. (Yeah, it's stupid-- but you'll remember it now!)

For your problem, you want to express d = 3(t-h)^2 + k

to do this, move the loose term to the left
d-500 = 3(t^2 - 78/3t)
d - 500 + (78/2)^2 = 3(t^2 - 78/3 t + (78/3/2)^2)
Now we have a perfect square on the right:
t^2 - 78/3t + (78/3/2)*(78/3/2) = (t-78/3/2)^2
We're almost there:

d = 3(t-78/3/2)^2 + 500 - (78/2)^2

simplifying:
d = 3*(t-13)^2 -1021
You can simplify