Question

really need help :(
A sandbag was thrown downward from a building . The function f(t)=-16t^2-32+128 shows the height f(t), in feet of the sandbag after t seconds.
Part A. Factor the function f(t) and use the factors to interpret the meaning of the x-intercept of the function.
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 max or min

what is the axis of symm

Answer 1

I'll get you started on part A


-16t^2-32t+128


-16*t^2-16*2t+16*8


-16(t^2+2t-8) ... factor out the GCF 16


and let you finish up

Answer 2

is that how I would get a and b

Answer 3

What do you get when you factor t^2+2t-8

Answer 4

that's what im having trouble with I cant take out anything

Answer 5

find two numbers that multiply to -8 (last term) and add to 2 (middle coefficient)

Answer 6

I got (t-2) ( t+4)

Answer 7

good

Answer 8

therefore, -16t^2-32t+128 factors to -16(t-2)(t+4)

Answer 9

how can you use this to find the x-intercepts?

Answer 10

do I plug in o for t?

Answer 11

@jim_thompson5910

Answer 12

more like you set it equal to zero and solve for t

Answer 13

-16t^2-32t+128 = 0

-16(t-2)(t+4) = 0

keep going...

Answer 14

o ok thank you

Answer 15

you're welcome

Answer 16

is it 96t+32

Answer 17

-16(t-2)(t+4) = 0

(t-2)(t+4) = 0

t-2 = 0 or t+4=0

solve each for t

Answer 18

those are both the x intercepts>

Answer 19

which are what?

Answer 20

t= 2 and t=-4

Answer 21

yep, the x-intercepts are 2 and -4 (basically the graph crosses the x axis at 2 and -4)

Answer 22

your the best!!! can you help me with part b and c

Answer 23

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 max or min

Answer 24

To find the vertex, you first need to find the axis of symmetry

Answer 25

how do we do this?

Answer 26

I suck at math idk :(

Answer 27

@jim_thompson5910

Answer 28

hint: for ax^2 + bx + c, the axis of symmetry is

x = -b/(2a)

Answer 29

x=-1 ?

Answer 30

sorry for asking all these questions.. its just all this is due tonight @jim_thompson5910

Answer 31

think of -16t^2-32t+128 as -16x^2-32x+128

what are a, b, c?

Answer 32

a is 16 b is -32 and c is 128

Answer 33

a = -16
b = -32
c = 128

Answer 34

x = -b/(2a)

x = -(-32)/(2*(-16))

x = 32/(-32)

x = -1

So you are correct. The axis of symmetry is x = -1

Answer 35

Now plug this into

y = -16x^2-32x+128

to find y

Answer 36

is y =144

Answer 37

yep

Answer 38

for the vertex: the x coordinate is -1, the y coordinate is 144

so (h,k) = (-1,144)

h = -1
k = 144

Answer 39

a = -16
h = -1
k = 144

so...

y = a(x-h)^2 + k

y = -16(x-(-1))^2 + 144

y = -16(x+1)^2 + 144

Answer 40

which means -16x^2-32x+128 transforms into y = -16(x+1)^2 + 144 after completing the square

Answer 41

so what would I put for b

Answer 42

what's the vertex?

Answer 43

-1. 144

Answer 44

(-1,144)

Answer 45

The y coordinate of that vertex is 144

So that's the min or the max. Which one is it? Do you know how you can tell?

Answer 46

its that max because it positive right?

Answer 47

no, the y coordinate of the vertex doesn't determine min or max

look at the value of 'a'

Answer 48

its -16 so its a min?

Answer 49

it's actually the opposite (yeah confusing I know)

if a < 0, then we have a max
if a > 0, then we have a min

Answer 50

o so its a ma
x haha

Answer 51

therefore, y = 144 is the max

Answer 52

thank you so much you don't know how much your helping me man

Answer 53

I have one more set of questions if your not to busy

Answer 54

Sure I can do one more question

Answer 55

THE FUNCTION H(t)=-16t^2+vt+s shows the height H (t) in feet of a projectile launched vertically from s feet above the ground after y seconds. The intila speed of the projectile is v feet per second

Answer 56

Part A. the projectile was launched from a height of 100 feet with an intial velocity of 60 feet per second. Create an equation to find the time taken by the projectile to fall on the ground.

Answer 57

Part b. What is the maxium height that the projectile will reach?

Answer 58

@jim_thompson5910

Answer 59

@agent0smith

Answer 60

plz someone help @agent0smith @jim_thompson5910

Answer 61

For the equation H(t)=-16t^2+vt+s,

v = initial launch velocity
s = initial launch height

Answer 62

thanks @jim_thompson5910

Answer 63

so what is the equation?

Answer 64

i, trying to do it but i don't know how to get it

Answer 65

what are v and s?

Answer 66

60 and s is 100

Answer 67

so H(t)=-16t^2+vt+s turns into ????

Answer 68

16(60)^2+160+100

Answer 69

v = 60
s = 100

Answer 70

we haven't said what the value of t was yet

Answer 71

so 16t^2+60(t)+100

Answer 72

yes, H(t)=-16t^2+vt+s becomes H(t)=-16t^2+60t+100

Answer 73

now solve -16t^2+60t+100 = 0 to find when the projectile will hit the ground

Answer 74

how would i solve that

Answer 75

by using the quadratic formula

Answer 76

so its 60 + or - square root of -60^2 - 4(-16) (100)/2(-16)

Answer 77

@jim_thompson5910

Answer 78

that -60^2 should be (60)^2

Answer 79

but otherwise, so far, so good

Answer 80

How did you get 144 for -16*-1^2+32*-1+128??

Answer 81

I keep getting 112. @jim_thompson5910 @jayhawk31

Answer 82

The original function is f(t)=-16t^2-32t+128

Plug t = -1 into this function to get

f(t)=-16t^2-32t+128

f(-1)=-16(-1)^2-32(-1)+128

f(-1)=144

So you just forgot to place -1 in parenthesis and it's -32 (not +32)

Answer 83

oh okay thanks, i realized my mistake after :3

Answer 84

That looks correct

Answer 85

Thank you

Answer 86

np