Question

HELP!!!!!!!!!!! Write formula h=-5t^2 + 50t + 40 to y= a(x-h)^2 + k.

Answer 1

\[\large h=-5t^2+50t+40\]Let's start by factoring a -5 out of each term,\[\large h=-5(t^2-10t-8)\]Err actually, let's ignore the 40 for right now, let's just factor -5 from the t terms.\[\large h=-5(t^2-10t)+40\]
So we'll try to complete the square on the part inside the brackets.
Understand what we've done so far?

Answer 2

Yes

Answer 3

So let's look at this part for now,
\[\large t^2-10t\]

This is of the form, \(\large x^2+bx\)
To complete the square, we'll need to add some value.
The value that we want is found this way:
we'll take the b term, divide it by 2, and then square it.

In this case, our b term is -10.
So the value we want to add is: \(\large \left(\dfrac{-10}{2}\right)^2 \qquad = \qquad 25\)
That value will complete the square for us.

Answer 4

Adding this value gives us,\[\large \large t^2-10t+25\]Which will factor down to this form: \(\large \left(x+\dfrac{b}{2}\right)^2 \qquad \rightarrow \qquad \left(t-5\right)^2\)

Answer 5

Completely the square can be kinda tricky, let me know D:
Ok but we have a problem, we can't just add 25 to an equation, that will unbalance it.

We're going to implement a little math trick.
We're going to `add` 25 and `subtract` 25, that is the same as adding 0, so it doesn't unbalance our equation.

\[\large \large \color{orangered}{t^2-10t+25}-25 \qquad = \qquad \color{orangered}{(t-5)^2}-25\]

Answer 6

Oh okay :)

Answer 7

So this is what we're doing,\[\large h=-5(\color{orangered}{t^2-10t})+40\]We're adding 25 to complete the square, and we're subtracting 25 to keep things balanced.\[\large h=-5(\color{orangered}{t^2-10t+25}-25)+40\]Simplifying our perfect square gives us,\[\large h=-5\left(\color{orangered}{(t-5)^2}-25\right)+40\]

Answer 8

It's ALMOST in the form that we want, but that pesky -25 is in the way.

Answer 9

To take it OUTSIDE of the black brackets, we have to multiply it by the -5.
\[\large h=-5\left(\color{orangered}{(t-5)^2}\right)-25(-5)+40\]Which gives us (Oh I'll drop the black brackets and color now also, since we don't need them anymore),\[\large h=-5(t-5)^2+125+40\]

Answer 10

Combine the constants and Tada! It's of the form that we wanted!\[\large h=-5(t-5)^2+165\]

Answer 11

Any part of that really confusing? It's quite a few steps to get through :)

Answer 12

But when I put it in the graphing calculator they don't have the same parabola :( I got the same answer as you did

Answer 13

Hmm Ok lemme see if I made a mistake somewhere one sec :o

Answer 14

OMG im so sorry! I put in a ( + ) after the t instead a ( - )!!!

Answer 15

Oh dats silly :3

Answer 16

I'm so sorry! Thank you for your help!

Answer 17

Oh did you already have an answer?
Was that too slow the way I approached that? XD lol

Answer 18

Just wanted to check your work? :D

Answer 19

LOL no it wasn't slow :P & yes, I put in a ( + ) in my answer instead of a ( - )