How do you complete the square and find the vertex of:
$$f(t)=−16t^2−32t+128$$

Question

How do you complete the square and find the vertex of:
$$f(t)=−16t^2−32t+128$$

wilder.monday · Answer

Can you factor out a GCF ( greatest common factor)

camerondoherty · Answer

No, I've already factored it, I need to complete the square...

camerondoherty · Answer

Don't you have to first set the equation to zero and get rid of any constants?

wilder.monday · Answer

May I see all the work you did?

wilder.monday · Answer

First set it equal to y. Then, subtract 128 from both sides the left side should look like y - 128

displayerror · Answer

Divide the entire equation by the coefficient in front of the t^2 term:
$$f(t) = t^2 + 2t - 8$$
Move the 8 to the other side:
$$t^2 + 2t = 8$$
Divide the coefficient in front of the t term by 2, then square that result
$$(\frac{2}{2})^2 = 1$$
Add the squared result (1) to both sides of the equation
$$t^2 + 2t + 1= 8+1$$
Factor
$$(t+1)^2= 9$$
Solve for t
$$t = \pm \sqrt{9} - 1$$
$$t = -4, 2$$
Find the corresponding y value for t=-4 and t=2

wilder.monday · Answer

Is this multiple choice? I just want to make sure what I'm doing goes along its the choices.

acxbox22 · Answer

for the vertex you can find the the x value of the vertex by using x=-b/2a
from the original equation
then sub that x value into the equation to get the y value of the vertex

camerondoherty · Answer

To find the y value, do you sub the value for y and solve?

camerondoherty · Answer

*t

displayerror · Answer

Yeah, we found two values of t, so plug into f(t) to find your y value(s).

camerondoherty · Answer

Ok, so:
$$f(t)=−16t^2−32t+128 $$$$f(t)=−16(-4)^2−32t+128 $$and$$f(t)=−16(2)^2−32t+128? $$

camerondoherty · Answer

Sheesh, Openstudy has a real character issue
If you see diamonds, refresh so it goes back to normal cx

acxbox22 · Answer

@DisplayError found the solutions now we have to find the vertex
for vertex we can use x=-b/2a
x= 32/(-16)(2)
x=32/-32
x=-1
now sub -1 for x into the equation to find y or f(t)

displayerror · Answer

Oh wait, my bad. It was already assumed in completing the square that the equation was equal to 0 (completed the square to find the zeroes of the function), so we don't have to plug the x-values back into f(t)--they're 0 already.

camerondoherty · Answer

$$f(t)=−16t^2−32t+128 $$
$$f(t)=−16(-1)^2−32t+128 $$
$$f(t)=−16(1)−32t+128 $$
$$f(t)=−16−32t+128 $$
$$f(t)=−32t+112 $$

acxbox22 · Answer

sorry @camerondoherty 
in this case x is t but it is the same thing
plug in -1 for t

camerondoherty · Answer

I did

acxbox22 · Answer

you forgot to plug it in for the term -32t

camerondoherty · Answer

Oh, lol, sorry cx
Just noticed

camerondoherty · Answer

$$f(t) = -16 - 32(-1) + 128$$
$$f(t) = -16+32 + 128$$
$$f(t) = 16+ 128$$
$$f(t)=144 $$

hartnn · Answer

let me give the steps to complete the square, see if you guys did the same thing,

$\Large x^2+bx+c \ = \Large (x^2 + bx \color{blue}{+b^2/4}) + (c\color{blue}{-b^2/4}) \ \Large = (x+(b/2))^2 +(c-b^2/4)$

have you guys done the similar thing ?

acxbox22 · Answer

yes so we have the x and y values of the vertex now @camerondoherty 
where x=-1 and y/f(t) = 144
so the vertex is (-1,144)

camerondoherty · Answer

Ohhhh ok, Thank You c:

acxbox22 · Answer

no problem :)