Question

A sandbag was thrown downward from a building. The function f(t) = -16t2 - 32t + 128 shows the height f(t), in feet, of the sandbag after t seconds.

Part B: 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?

Part C: Use your answer in part B to determine the axis of symmetry for f(x)?

Answer 1

@mathmale @mathstudent55 @satellite73 @triciaal @ganeshie8 @sleepyjess

Answer 2

you do not (repeat do NOT) need to complete the square to do this

Answer 3

How do you complete a square?

Answer 4

the first coordinate of the vertex of \(y=ax^2+bx+c\) is always \[x=-\frac{b}{2a}\]

Answer 5

@satellite73 Perhaps, but the question is asking me to complete the square nonetheless.

Answer 6

the second coordinate is what you get when you replace \(x\) by the first coordinate

Answer 7

in your case \(a=-16,b=-32\)

Answer 8

And I already know how to calculate the vertex, thanks! But how do I complete a square @satellite73 ?

(This is unrelated, but how did you get a 100 smartscore, and are you the only person to ever get a 100 smartscore?)

Answer 9

ok first off lets note that the question makes no sense at all physically, since \(t\) represents time, and the first coordinate of the vertex is going to be negative
so unless you can go backwards in time, (which you cannot) the question is meaningless

Answer 10

Yeah, I pointed that out in Part A

Answer 11

But anyways, how do you complete a square?

Answer 12

Theoretically you could go "forward in time" by using a black hole's gravitational pull, but that's unrelated.

Answer 13

think about this: you throw a rock DOWN from an initial height, and you want to know the maximum height
pretty clearly the max is where you started from

Answer 14

Yeah I get it, the question doesn't make much sense. But how do you complete a square anyways?

Answer 15

if you still want to complete the square, we can do it

Answer 16

Yes please

Answer 17

first off, factor out the \(-16\) from the first two terms
what do you get ?

Answer 18

You can factor out -16 from all 3 terms, shouldn't I do that?

Answer 19

no

Answer 20

k

f(t) = -16(t2 - 2t) + 128

Answer 21

ok so we are at \[f(t)=-16(t^2-2t)+128\]now to complete the square, what is half of \(-2\)?

Answer 22

-1?

Answer 23

yes

Answer 24

and \((-1)^2=?\)

Answer 25

-1?

Answer 26

no

Answer 27

oh wait, sorry.

1

Answer 28

eys

Answer 29

yes

Answer 30

got it

Answer 31

ok that was wrong, let me back up

Answer 32

we got \((-1)^2=1\) right?

Answer 33

wait, it was 2t, where did the t go?

Answer 34

lets ignore the \(-16\) for the moment, and concentrate only on \(t^2-2t\)

Answer 35

right

Answer 36

but how did you make the t go away?

Answer 37

half of \(-2\) is \(-1\) so we are going to complete the square by writing \((t-1)^2\)

Answer 38

now the \(t\) did not go away, because \[(t-1)^2=t^2-2t+1\] right?

Answer 39

but something has changed
we started with \[t^2-2t\] and ended with \[t^2-2t+1\] so we completed the square by adding \(1\)

Answer 40

I dont really get it...

Answer 41

when you complete the square, you are making a change to write it as a perfect square

Answer 42

(t−1)2=t2−2t+1

Answer 43

we have to compensate for that change

Answer 44

oh, because you divided 2 in half, so you added 1 somewhere else?

Answer 45

right, add or subtract as the case may be

Answer 46

got it

Answer 47

now is where it gets a little complicated so lets go slow

Answer 48

k

Answer 49

we have \[-16(t^2-2t)+128\] and we write the first term as a perfect square as \[-16(t-1)^2+128\pm?\]

Answer 50

alright

Answer 51

now the first term is a perfect square, but we have to figure out what change we made so we can compensate for it

Answer 52

and compensate for it next to 128?

Answer 53

right

Answer 54

got it

Answer 55

i think we only took away 1. That's the only change i think, correct?

Answer 56

yes, but we actually have to worry about the \(-16\) in front

Answer 57

oh

Answer 58

we have also made a mistake i see, so lets correct it first

Answer 59

wheres the mistake?

Answer 60

\[-16t^2-32t+128=-16(t^2+2t)+128\]

Answer 61

i don't see any mistake

Answer 62

the process is still the same, half of 2 is 1, one squared is 1

Answer 63

we had \(-2t\) when it should have been \(+2t\)

Answer 64

oh yeah

Answer 65

so now we are at \[-16(t+1)^2+128\pm?\]

Answer 66

why is it 128 +-?

Answer 67

now we didn't really add 1, because of that \(-16\) out fron

Answer 68

that is what we are going to correct

Answer 69

we have to figure out what we added or subtracted so we can compensate for it

Answer 70

wait, how did we not really add 1. Okay, this is the most complicated math problem i've ever had

Answer 71

gaah

Answer 72

How does that -16 affect us not actually adding a 1

Answer 73

it looks like we added one (don't fret, i am going to show you a real snap way to do this )

Answer 74

I'm confused :(

Answer 75

\[-16(t+1)^2=-16(t^2+2t+1)=-16t^2-32t-16\]

Answer 76

is that part clear ?

Answer 77

It might help if you reversed the order, but yeah, I think i get that part

Answer 78

ok so we actually did not add one, the -16 out front means we SUBTRACTED 16

Answer 79

...

Answer 80

you see it right? the \(-16\) out at the end there

Answer 81

my expression isn't too far off from my profile pic right now

Answer 82

cause I really have no idea what you just said, or how you got to that answer @satellite73

Answer 83

(This is unrelated, but how did you get a 100 smartscore, and are you the only person to ever get a 100 smartscore?)

Answer 84

\[-16(t+1)^2=-16(t^2+2t+1)=-16t^2-32t\color{red}{-16}\]

Answer 85

you do see it right? it is red

Answer 86

yeah, i see the -16, but how does that mean we never added a 1?

Answer 87

we added 1 to complete the square, but because of the -16 out front we really subtracted 16

Answer 88

so we have to add it back

Answer 89

WAIT HOLD ON

Answer 90

\[-16t^2-32t+128=-16(t+1)^2+128+16\] that equal sign is for real

Answer 91

how does it mean you actually subtracted 16? I really don't get this

Answer 92

Ya know, I might be able to complete this question by saying you don't need to find the square to find the vertex...

Answer 93

we changed \(t^2+2t\) in to \((t+1)^2\) by adding 1 right, because \((t+1)^2=t^2+2t+1\)
is that clear or no?

Answer 94

yes

Answer 95

And then you would compensate by adding 1 somewhere else

Answer 96

yes, if that was all we had, that is what we would do (subtract 1 to compensate, not add)

Answer 97

we add one, we have to subtract one