anonymous
  • anonymous
f(x) = -16t^2 + 96t what is the x and y intercept
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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freckles
  • freckles
where is the function of x? is t suppose to be x?
freckles
  • freckles
well I mean that could be a function of x it would just be a constant function of x
anonymous
  • anonymous
http://assets.openstudy.com/updates/attachments/53ce7ea6e4b081ba963957d9-coltjoco-1406041829668-paulkicks.jpg this is the graph

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Vocaloid
  • Vocaloid
|dw:1436976311744:dw|
Vocaloid
  • Vocaloid
|dw:1436976350942:dw|
Vocaloid
  • Vocaloid
the x-intercepts are the points where the function is equal to 0 the y-intercept is the point where x = 0
anonymous
  • anonymous
is the x intercept 1,6?
Vocaloid
  • Vocaloid
yes, or more specifically (1,0) and (6,0)
anonymous
  • anonymous
also could you help me with a few more questions
anonymous
  • anonymous
Find the vertex algebraically. Write as an ordered pair.
anonymous
  • anonymous
What is the maximum height of the ball? Write your answer as an integer.
freckles
  • freckles
(1,0) isn't on the graph though :p
anonymous
  • anonymous
so it would 6,0
freckles
  • freckles
(0,0) and (6,0) if we are talking about the function in the graph
anonymous
  • anonymous
okay
freckles
  • freckles
\[f(x) = -16t^2 + 96t \] so I guess that that one x was suppose to be a t :p
anonymous
  • anonymous
oh
anonymous
  • anonymous
Find the vertex algebraically. Write as an ordered pair. would the vertex 6,0 ?
freckles
  • freckles
wait and it looks like they called the function h(t) instead of f(t)
freckles
  • freckles
but whatever
freckles
  • freckles
the vertex would be that peak point
freckles
  • freckles
if you wanted to find it using that graph you could observe that the vertex occurs midway between the x-intercepts
anonymous
  • anonymous
which is 3?
freckles
  • freckles
I will also show you how to find it algebraically since you asked
freckles
  • 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
  • freckles
but anyways algebraically ...
freckles
  • 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
  • anonymous
3o would the vertex be 3,14 or 3,140?
freckles
  • freckles
neither
freckles
  • 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
  • anonymous
160?
freckles
  • freckles
160 is too high
freckles
  • freckles
it is closer to 140 than 160
freckles
  • freckles
|dw:1436977378401:dw|
freckles
  • freckles
|dw:1436977410220:dw|
anonymous
  • anonymous
120
freckles
  • freckles
how is 120 just a little above 140 ?
freckles
  • freckles
you can find the y-coordinate of the vertex exactly by using the steps I gave you above
anonymous
  • anonymous
sorry I was confused
freckles
  • 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
  • freckles
|dw:1436977596987:dw| so you know the vertex is gonna be somewhere close to (3,142)
freckles
  • freckles
you can find the answer algebraically using the method above I suggested
freckles
  • 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
  • anonymous
I got 141 ?
anonymous
  • anonymous
though I think it may be -3 I'm not sure
freckles
  • freckles
how did you get 141?
freckles
  • freckles
are you trying to put your quadratic in vertex from yet?
anonymous
  • anonymous
I'm still using the graph for it
freckles
  • freckles
oh okay well don't forget it does say to find the vertex algebraically
freckles
  • freckles
using a graph to obtain the vertex is not finding it algebraically
anonymous
  • anonymous
oh
anonymous
  • anonymous
so it's not -3
freckles
  • freckles
are you talking about the x-coordinate of the vertex?
anonymous
  • anonymous
vertex
freckles
  • freckles
no -3 is not the x-coordinate
freckles
  • freckles
|dw:1436978591061:dw|
anonymous
  • anonymous
141 ?
freckles
  • 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
  • freckles
it could be (3,144)
freckles
  • freckles
or (3,143)
freckles
  • freckles
but to find out we need to put your quadratic in vertex form
freckles
  • freckles
I gave a general way to do that above
freckles
  • freckles
|dw:1436979098053:dw| please look at this graph... |dw:1436979115849:dw|
freckles
  • 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
  • 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
  • anonymous
1 ?
freckles
  • freckles
no recall that: \[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]
freckles
  • freckles
the number in front of t is -6 let's call k, -6 so if k=-6 then k/2=?
anonymous
  • anonymous
it would -6/2
freckles
  • 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
  • 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
  • anonymous
so the answer is -6
freckles
  • freckles
no
freckles
  • freckles
we already know we are suppose to get something close to (3,142) because this what we found visually
freckles
  • freckles
so I don't know what -6 is suppose to be
anonymous
  • anonymous
would it 3,2
freckles
  • freckles
but anyways I was asking if you could finish putting into the vertex form
freckles
  • freckles
i have to ask are you playing with me?
freckles
  • freckles
like you keep giving me these y values that aren't closer to 140
anonymous
  • anonymous
no I'm not i'm really stuck on this equation
freckles
  • 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
  • anonymous
which I had gotten 141
freckles
  • 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
  • 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
  • freckles
@Aliypop yes I see you said 141 as approximate but then you asked me if it was 2,14,... and so on...
anonymous
  • anonymous
I confused on what the x value is
freckles
  • freckles
the x-coordinate of the vertex is 3 we have found this both visually and algebraically now
freckles
  • 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
  • anonymous
so is it (3,3)
freckles
  • freckles
no...
freckles
  • freckles
again 3 is nowhere near 140
freckles
  • 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
  • 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
  • freckles
all you have to do is finishing simplifying
freckles
  • freckles
do you need help doing 16(6/2)^2?
freckles
  • freckles
6 divided by 2 is 3 3 squared is 9 16*9 is...
anonymous
  • anonymous
144
freckles
  • 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
  • freckles
the vertex is exactly (3,144)
freckles
  • 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
  • anonymous
okay
anonymous
  • 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
  • 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
  • anonymous
so it would 144
freckles
  • 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
  • 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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