Question

How do you turn a quadratic equation from general form into vertex form?
y=-16t²+960t+0

Answer 1

Hint: Complete the square.

Answer 2

yea I know that so would the first step be forcing the lead coefficent to = 1 so, would i divide y=-16t²+960t+0 by -16 and get t²-60t+0 then complete the square?

Answer 3

Uh, be careful. Do not divide specifically unless you have a reason for it. Here's the guide through this one:
\[\begin{align}
y&=-16t^2+960t+0\implies\\
y&=-16(t^2-60t)+0
\end{align}
\]We have that, using the method for completing the square:
\[
\left(\frac{-60}{2}\right)^2=(-30)^2=900
\]So:
\[
y=-16(t^2-60t+900)-(-16)\cdot900\implies\\
\](Since we added 900 inside of the expression)
\[
y=-16(t^2-60t+900)+14400\implies\\
y=-16(t-30)^2+14400
\]Thus, we are done.

Answer 4

oh ok i got that but using finding the x value by using this formula:
\[\frac{ -b }{ 2a }\] meaning using y=t²-60t+0 as i told you by dividing the original fraction by -16. So i go30 for the x value and plugged it in the original equation and got 14400. So why did that also worked?

Answer 5

The quadratic formula (of which -b/2a is the first term) is derived from completing the square. They are really the same operation, but you use when when you want to solve and the other for getting vertex form.

Answer 6

oh ok so in this case would it be better to use it or not?

Answer 7

The -b/2a term is the x-coordinate of the vertex, so subbing it back into the equation will give you the y-coordinate of the vertex, but if you want the vertex form of the equation, you want to keep that -16 as it affects the shape of the graph.

Answer 8

I recommend getting very good at completing the square and using the QF to check your work. If you just wan to solve the equation for the x-intercepts then QF is typically the weapon of choice (if it's not easily factorable).
For vertex form and later if you want to find equations of ellipses and hyperbolas, you'll need to complete the square.

Answer 9

Oh ok well thanks.