f(x) = -16t^2 + 96t
what is the x and y intercept

- anonymous

f(x) = -16t^2 + 96t
what is the x and y intercept

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- schrodinger

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- freckles

where is the function of x?
is t suppose to be x?

- freckles

well I mean that could be a function of x
it would just be a constant function of x

- anonymous

http://assets.openstudy.com/updates/attachments/53ce7ea6e4b081ba963957d9-coltjoco-1406041829668-paulkicks.jpg
this is the graph

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## More answers

- Vocaloid

|dw:1436976311744:dw|

- Vocaloid

|dw:1436976350942:dw|

- Vocaloid

the x-intercepts are the points where the function is equal to 0
the y-intercept is the point where x = 0

- anonymous

is the x intercept 1,6?

- Vocaloid

yes, or more specifically (1,0) and (6,0)

- anonymous

also could you help me with a few more questions

- anonymous

Find the vertex algebraically. Write as an ordered pair.

- anonymous

What is the maximum height of the ball? Write your answer as an integer.

- freckles

(1,0) isn't on the graph though :p

- anonymous

so it would 6,0

- freckles

(0,0) and (6,0)
if we are talking about the function in the graph

- anonymous

okay

- freckles

\[f(x) = -16t^2 + 96t \]
so I guess that that one x was suppose to be a t :p

- anonymous

oh

- anonymous

Find the vertex algebraically. Write as an ordered pair.
would the vertex 6,0 ?

- freckles

wait and it looks like they called the function h(t) instead of f(t)

- freckles

but whatever

- freckles

the vertex would be that peak point

- freckles

if you wanted to find it using that graph you could observe that the vertex occurs midway between the x-intercepts

- anonymous

which is 3?

- freckles

I will also show you how to find it algebraically since you asked

- freckles

but first using the graph the vertex appears to be at (3,a little above 140)
I will let you approximate that y value

- freckles

but anyways algebraically ...

- freckles

\[y=at^2+bt+c \\ \text{ we want to put this in vertex form } \\\]
\[y-c =at^2+bt \text{ I subtract } c \text{ on both sides } \\ \frac{y-c}{a} =t^2+\frac{b}{a} t \text{ I divided both sides by } a \\ \frac{y-c}{a} +(\frac{b}{2a})^2=t^2+\frac{b}{a}t +(\frac{b}{2a})^2 \\ \text{ I threw in a number on both sides so I could } \\ \text{ complete the square on one side } \\ \frac{y-c}{a}+(\frac{b}{2a})^2= (t+\frac{b}{2a})^2 \\ \text{ now \let's re-isolate } y \\ \text{ subtract} (\frac{b}{2a})^2 \text{ on both sides } \\ \frac{y-c}{a}=(t+\frac{b}{2a})^2-(\frac{b}{2a})^2 \\ \text{ multiply } a \text{ on both sides } \\ y-c=a(t+\frac{b}{2a})^2-a (\frac{b}{2a})^2 \\ \text{ add } c \text{ on both sides } \\ y=a(t+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c\]
the vertex is given by:
\[(-\frac{b}{2a},-a(\frac{b}{2a})^2+c)\]

- anonymous

3o would the vertex be 3,14 or 3,140?

- freckles

neither

- freckles

look at the graph
you should see where that mountain's peak is just a hair above 140
the y value isn't exactly 140
it is just above 140

- anonymous

160?

- freckles

160 is too high

- freckles

it is closer to 140 than 160

- freckles

|dw:1436977378401:dw|

- freckles

|dw:1436977410220:dw|

- anonymous

120

- freckles

how is 120 just a little above 140 ?

- freckles

you can find the y-coordinate of the vertex exactly by using the steps I gave you above

- anonymous

sorry I was confused

- freckles

but you can approximate it using the visual it looks like it could be maybe less than 150 but greater than 140
and I say less than 150 because the peak doesn't reach halfway between 140 and 160
it is less than halfway
I would approximate saying 142
but that could be off a little because I did find this visually and not algebraically as the instructions|dw:1436977561263:dw| suggested

- freckles

|dw:1436977596987:dw|
so you know the vertex is gonna be somewhere close to (3,142)

- freckles

you can find the answer algebraically using the method above I suggested

- freckles

this one:
\[y-c =at^2+bt \text{ I subtract } c \text{ on both sides } \\ \frac{y-c}{a} =t^2+\frac{b}{a} t \text{ I divided both sides by } a \\ \frac{y-c}{a} +(\frac{b}{2a})^2=t^2+\frac{b}{a}t +(\frac{b}{2a})^2 \\ \text{ I threw in a number on both sides so I could } \\ \text{ complete the square on one side } \\ \frac{y-c}{a}+(\frac{b}{2a})^2= (t+\frac{b}{2a})^2 \\ \text{ now \let's re-isolate } y \\ \text{ subtract} (\frac{b}{2a})^2 \text{ on both sides } \\ \frac{y-c}{a}=(t+\frac{b}{2a})^2-(\frac{b}{2a})^2 \\ \text{ multiply } a \text{ on both sides } \\ y-c=a(t+\frac{b}{2a})^2-a (\frac{b}{2a})^2 \\ \text{ add } c \text{ on both sides } \\ y=a(t+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c\]

- anonymous

I got 141 ?

- anonymous

though I think it may be -3 I'm not sure

- freckles

how did you get 141?

- freckles

are you trying to put your quadratic in vertex from yet?

- anonymous

I'm still using the graph for it

- freckles

oh okay
well don't forget it does say to find the vertex algebraically

- freckles

using a graph to obtain the vertex is not finding it algebraically

- anonymous

oh

- anonymous

so it's not -3

- freckles

are you talking about the x-coordinate of the vertex?

- anonymous

vertex

- freckles

no -3 is not the x-coordinate

- freckles

|dw:1436978591061:dw|

- anonymous

141 ?

- freckles

the vertex is pretty close to (3,142)
or I guess you could say (3,141)
but we don't know the vertex exactly until you find it algebraically

- freckles

it could be (3,144)

- freckles

or (3,143)

- freckles

but to find out we need to put your quadratic in vertex form

- freckles

I gave a general way to do that above

- freckles

|dw:1436979098053:dw|
please look at this graph...
|dw:1436979115849:dw|

- freckles

again this is not finding the vertex algebraically
this is approximating the vertex with the help of visual aid
you can find the exact vertex by putting the quadratic into vertex form
you will need to complete the square

- freckles

\[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+?)+16(?)\]
what do you need to add inside the ( ) to complete the square

- anonymous

1 ?

- freckles

no recall that:
\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]

- freckles

the number in front of t is -6
let's call k, -6
so if k=-6 then k/2=?

- anonymous

it would -6/2

- freckles

\[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2\]
right can you finish putting into vertex form

- freckles

by the way I didn't put the - in because (6/2)^2 is the same as (-6/2)^2
but you could have

- anonymous

so the answer is -6

- freckles

no

- freckles

we already know we are suppose to get something close to (3,142) because this what we found visually

- freckles

so I don't know what -6 is suppose to be

- anonymous

would it 3,2

- freckles

but anyways I was asking if you could finish putting into the vertex form

- freckles

i have to ask
are you playing with me?

- freckles

like you keep giving me these y values that aren't closer to 140

- anonymous

no I'm not i'm really stuck on this equation

- freckles

I have shown you visually the y-value is a little above 140
we know the x-value to be exactly 3
the vertex is definitely somewhere close to (3, 142)

- anonymous

which I had gotten 141

- freckles

\[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2 \\ y=-16(t-3)^2+16(\frac{6}{2})^2 \]
there I completed the square for you

- freckles

as we see the x-coordinate of the vertex is 3
just evaluate 16(6/2)^2 to find the y coordinate of the vertex

- freckles

@Aliypop yes I see you said 141 as approximate but then you asked me if it was 2,14,... and so on...

- anonymous

I confused on what the x value is

- freckles

the x-coordinate of the vertex is 3
we have found this both visually and algebraically now

- freckles

the y-coordinate of the vertex is given by 16(6/2)^2
which we have approximated to be somewhere just above 140

- anonymous

so is it (3,3)

- freckles

no...

- freckles

again 3 is nowhere near 140

- freckles

the x-coordinate of the vertex is 3
the y-coordinate of the vertex is something just above 140
we have found this information visually
we have also found algebraically
the x-coordinate of the vertex is 3
the y-coordinate of the vertex is 16(6/2)^2

- freckles

\[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2 \\ y=-16(t-3)^2+16(\frac{6}{2})^2 \\ \text{ this says the vertex is } (3,16(\frac{6}{2})^2)\]

- freckles

all you have to do is finishing simplifying

- freckles

do you need help doing 16(6/2)^2?

- freckles

6 divided by 2 is 3
3 squared is 9
16*9 is...

- anonymous

144

- freckles

yes \[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2 \\ y=-16(t-3)^2+16(\frac{6}{2})^2 \\ \text{ this says the vertex is } (3,16(\frac{6}{2})^2) =(3,144)\]

- freckles

the vertex is exactly (3,144)

- freckles

as we see this makes sense from our visual answer
because we did say the vertex had x-coordinate 3 and y-coordinate of something just above 140

- anonymous

okay

- anonymous

What is the maximum height of the ball? Write your answer as an integer.
this was the other question i need help on

- freckles

you actually already found this when you found the vertex
the vertex is the highest point
and you know the y-value represents the height

- anonymous

so it would 144

- freckles

|dw:1436980560407:dw|
yes the height is 144
and the time in which the height is 144 is t=3 sec if the time unit is second

- freckles

though usually you would say 144 units
where units is some kind of distance measurement but that isn't really given here

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