Question

PLEASE HELP!!!!!!!!!!!!!!!!

A sandbag was thrown downward from a building. The function f(t) = -16t^2 - 64t + 512 shows the height f(t), in feet, of the sandbag after t seconds.

Complete the square of the expression for f(x) to determine the vertex of the graph of f(x). Would this be a maximum or minimum on the graph?

Answer 1

@Hero @ganeshie8 @tHe_FiZiCx99 @zzr0ck3r

Answer 2

where are you stuck ?

Answer 3

f(t) = -16t^2 - 64t + 512

you need to complete the square, right ?

Answer 4

yes

Answer 5

im so bad at it

Answer 6

basically we need to change it to this form :
\[\large f(c) = a(x-h)^2 + k\]

Answer 7

start by factoring out "-16" from first two terms :
\[\large \begin{align} \\ f(t) &= -16t^2 - 64t + 512 \\ &= -16(t^2+4t) + 512\end{align}\]

Answer 8

next, recall the identity : \(\large \color{red}{(a+b)^2 = a^2+2ab+b^2}\)

Answer 9

ok

Answer 10

it will be a maximum, as the graph gets more negative as t increases

Answer 11

so it turns into\[f(t) = -16(t^2 + 4t + 4) + 512\]

Answer 12

right?

Answer 13

Perfect !

Answer 14

after factorising you get -16[(t+2)^2 - 36]

Answer 15

wait a sec, you have added 4, so u better subtract it also, right ?

Answer 16

right :)

Answer 17

the stuff inside parenthesis can be written into this form, if we're clever enough we will see it immediately :

start by factoring out "-16" from first two terms :
\[\large \begin{align} \\ f(t) &= -16t^2 - 64t + 512 \\ &= -16(t^2+4t) + 512 \\ &= -16(t^2+4t + \color{red}{2^2} ) + 512 - \color{red}{(-16*2^2)}\end{align}\]

Answer 18

see if that looks okay ^^

Answer 19

yup :)

Answer 20

so then u simplify?

Answer 21

yes

Answer 22

see if you get -16[(t+2)^2 - 36]

Answer 23

after that you can sub in t= 1 and t=2 to check if it is a maximum or minimum curve

Answer 24

yes before simplifying, complete the square so that we feel accomplished a bit :)

\[\large \begin{align} \\ f(t) &= -16t^2 - 64t + 512 \\ &= -16(t^2+4t) + 512 \\ &= -16(t^2+4t + \color{red}{2^2} ) + 512 - \color{red}{(-16*2^2)} \\
&= -16(t+\color{red}{2})^2 + 512 - \color{red}{(-16*2^2)} \\
\end{align}\]

Answer 25

i got\[f(x) = -16(t+2)^2 + 576\]

Answer 26

am i right @ganeshie8 ?

Answer 27

Looks good ^^

Answer 28

compare it with :
\[\large f(x) = a(x-\color{Red}{h})^2 + \color{Red}{k}\]

vertex = \((\color{Red}{h}, \color{Red}{k})\) = ?

Answer 29

(-2, 576) right?

Answer 30

Correct !!

Answer 31

so the value of function at vertex is 576
Is that a maximum or minimum value ?

Answer 32

maximum, right?

Answer 33

thats right ! but may i knw why do you think it is maximum ? :)

Answer 34

cuz the leading coefficient of the original equation is negative. :)

Answer 35

Excellent ! good job :)

Answer 36

THANK YOU SO MUCH!!!!!! :) <3 :) <3 ✱