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Thread: Quadratic Parabola equations

  1. #1
    New Member LaaLa's Avatar
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    Question Quadratic Parabola equations

    Hi there
    Would like a little help on this problem here if thats ok?
    Your help is greatly appreciated!

    Screen Shot 2019-01-11 at 1.57.30 PM.png Screen Shot 2019-01-11 at 1.55.27 PM.jpg

    I have worked out the equation for the parabola:

    x^2 = -28(y-8.5)

    However, I am puzzled on how to move on from that point onwards.
    Again, it's another study I am stuck on!
    Unfortunately I was only given one example on this particular equation (especially a negative quadratic equation), but it doesn't suit the problem I am trying to solve on this one.
    Here is that example:

    Screen Shot 2019-01-11 at 2.07.51 PM.jpg
    Screen Shot 2019-01-11 at 2.08.07 PM.jpgScreen Shot 2019-01-11 at 2.08.20 PM.jpg

    I am unsure as to what I should do next. I am familiar with factorising and squaring but only worked on normal parabolas that already have their equations given. (The canvas is blank on this round!)
    I feel that I need to substitute 6 metres somehow to give me an answer. I was trying to find a way to calculate everything that I have learnt but it doesn't seem to give me the answer I am looking for.
    Trying not to rely too much on the desmos calculator but maybe I might need to for an answer.
    Who knows? Would love your input on this. Cheers!

  2. #2
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    Quote Originally Posted by LaaLa View Post
    Hi there
    Would like a little help on this problem here if thats ok?
    Your help is greatly appreciated!

    Screen Shot 2019-01-11 at 1.57.30 PM.png Screen Shot 2019-01-11 at 1.55.27 PM.jpg

    I have worked out the equation for the parabola:

    x^2 = -28(y-8.5)

    However, I am puzzled on how to move on from that point onwards.
    Again, it's another study I am stuck on!
    Unfortunately I was only given one example on this particular equation (especially a negative quadratic equation), but it doesn't suit the problem I am trying to solve on this one.
    Here is that example:

    Screen Shot 2019-01-11 at 2.07.51 PM.jpg
    Screen Shot 2019-01-11 at 2.08.07 PM.jpgScreen Shot 2019-01-11 at 2.08.20 PM.jpg

    I am unsure as to what I should do next. I am familiar with factorising and squaring but only worked on normal parabolas that already have their equations given. (The canvas is blank on this round!)
    I feel that I need to substitute 6 metres somehow to give me an answer. I was trying to find a way to calculate everything that I have learnt but it doesn't seem to give me the answer I am looking for.
    Trying not to rely too much on the desmos calculator but maybe I might need to for an answer.
    Who knows? Would love your input on this. Cheers!
    OK, so you think that you have your equation. What exactly are you trying to do next and why are you struck?
    A mathematician is a blind man in a dark room looking for a black cat which isn’t there. - Charles R. Darwin

  3. #3
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    Quote Originally Posted by LaaLa View Post
    Hi there
    Would like a little help on this problem here if thats ok?
    Your help is greatly appreciated!

    Screen Shot 2019-01-11 at 1.57.30 PM.png Screen Shot 2019-01-11 at 1.55.27 PM.jpg

    I have worked out the equation for the parabola:

    x^2 = -28(y-8.5)

    However, I am puzzled on how to move on from that point onwards.
    Again, it's another study I am stuck on!
    Unfortunately I was only given one example on this particular equation (especially a negative quadratic equation), but it doesn't suit the problem I am trying to solve on this one.
    Here is that example:

    Screen Shot 2019-01-11 at 2.07.51 PM.jpg
    Screen Shot 2019-01-11 at 2.08.07 PM.jpgScreen Shot 2019-01-11 at 2.08.20 PM.jpg

    I am unsure as to what I should do next. I am familiar with factorising and squaring but only worked on normal parabolas that already have their equations given. (The canvas is blank on this round!)
    I feel that I need to substitute 6 metres somehow to give me an answer. I was trying to find a way to calculate everything that I have learnt but it doesn't seem to give me the answer I am looking for.
    Trying not to rely too much on the desmos calculator but maybe I might need to for an answer.
    Who knows? Would love your input on this. Cheers!
    I have not checked your equation, but assuming it is correct and assuming that the equation is based on the horizontal line being the x-axis and the vertical line being the y-axis, at what points on the x-axis is the vertical distance between the pool and roof at a minimum?

    If you know x, then how do you find the corresponding y?

    What does that tell you?

  4. #4
    New Member LaaLa's Avatar
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    Quote Originally Posted by Jomo View Post
    OK, so you think that you have your equation. What exactly are you trying to do next and why are you struck?
    Hi Jomo,

    I don't know what to do next with the equation x^2=-28(y-8.5)
    I don't have the answer to the 6metre question. I don't know how to get there first.
    I have been trying to figure it out for 2 days now. Everytime I try and test it on the desmos calculator, the numbers go out unevenly and doesn't seem to fit the equation of the parabola. Can someone give me the answer please and end my pain. I would appreciate it.
    Last edited by LaaLa; 01-10-2019 at 11:53 PM.

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    Quote Originally Posted by LaaLa View Post
    I don't know what to do next with the equation x^2=-28(y-8.5)
    I don't have the answer to the 6metre question. I don't know how to get there first.
    I have been trying to figure it out for 2 days now. Everytime I try and test it on the desmos calculator, the numbers go out unevenly and doesn't seem to fit the equation of the parabola. Can someone give me the answer please and end my pain. I would appreciate it.
    You may want to solve the equation for y, though that's not essential.

    To check the 6 meter requirement, set x to 8 (one end of the pool) and find what y is. Is it at least 6?

    You know the roof is no more than 8.5 m high anywhere.

    You can check for the width being no more than 36 m either by solving for x when y = 0 (the base), or by setting x to 18 (half the maximum width) and seeing that y is negative (that is, it will have reached the ground before then).

    But I don't see what trouble you could be having with Desmos. I pasted your equation in, and it's easy enough to see the answers (though you shouldn't be depending on that). Tell me what is going wrong when you use it. What do you see?

  6. #6
    New Member LaaLa's Avatar
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    Quote Originally Posted by Dr.Peterson View Post
    You may want to solve the equation for y, though that's not essential.

    To check the 6 meter requirement, set x to 8 (one end of the pool) and find what y is. Is it at least 6?

    You know the roof is no more than 8.5 m high anywhere.

    You can check for the width being no more than 36 m either by solving for x when y = 0 (the base), or by setting x to 18 (half the maximum width) and seeing that y is negative (that is, it will have reached the ground before then).

    But I don't see what trouble you could be having with Desmos. I pasted your equation in, and it's easy enough to see the answers (though you shouldn't be depending on that). Tell me what is going wrong when you use it. What do you see?

    Thanks Dr.Peterson for your reply.
    I am not sure what you are meaning in in regards to finding 'y'. Are you meaning 'a'? The distance between the vertex and the focus? (8.5-1.5=7).

    The equation x^2=-28(y-8.5) works on the desmos because I calculated it using the vertex and the focus. Hoping that it is the right equation unless someone kindly corrects me!
    Notice the line I placed at 6 metres...
    Screen Shot 2019-01-11 at 6.19.37 PM.jpgScreen Shot 2019-01-11 at 6.19.52 PM.png

    When I talk about having problems with the desmos, I am trying to make the equation by using the technique to what I have been learning (from the one example I had supplied). When the equation is made however, I then get stuck and don't know where to go from then onwards.
    The form used in the example is:
    y= ax^2 + b
    a = 18m (which is half of the base measurement)
    b = 8.5m (vertex)

    put it together...
    y= 18^2 + 8.5

    now if I work it out the way the example shows...
    18^2 = 324
    8.5/324 = 0.0262

    to make sure this is correct...
    -0.0262 x 18^2 = -8.5

    Now if I put this on desmos alongside the other equation and the 6 metre line. This is what it looks like:
    Screen Shot 2019-01-11 at 6.26.01 PM.jpgScreen Shot 2019-01-11 at 6.26.09 PM.png

    Now I see they look slightly different to each other; the first equation was based on the vertex and focus which has no base measurement, whereas the 2nd equation does.

    What is confusing me is that this equation is based on the measurements given from a design for building a roof. And when it comes to numbers and construction, most of the numbers are usually rounded. I tried to do the same when it came to this particular equation that I'm confused on. when I was in the middle of working out the second equation this is what happened:

    The form used in the example is:
    y= ax^2 + b
    a = 18m (which is half of the base measurement)
    b = 8.5m (vertex)

    put it together...
    y= 18^2 + 8.5

    now if I work it out the way the example shows...
    18^2 = 324

    8.5/324 = 0.0262
    = 0.03 (3s.f)

    to make sure this is correct...
    -0.03 x 18^2 = -9.72

    Okay that didn't work, even though it's 0.01 difference. I will try it a different way:

    18^2 = 324[/I]
    8.5/324 = 0.0262
    = 0.026

    to make sure this is correct...
    -0.026 x 18^2 = -8.424

    Okay that is slightly 1 cm off.

    The only option I see is to stay with the first set of numbers.

    I would really like to use more than one option in regards to answering these questions without relying on using the desmos to show me the answers. It would be easier if I just give all the points and measurements by looking at the graph but this is about using the forms and methods I have already learnt to find the answer.

    I don't know HOW to find out whether or not the height of the roof is at least 6 m above the pool at any point or whether or not the width of the pool roof is less the 36 metres.

    If I make an equation for 6m height and 18m width of the pool, this is what it looks like
    Screen Shot 2019-01-11 at 7.09.10 PM.jpgScreen Shot 2019-01-11 at 7.08.56 PM.jpg

    I have questions!!
    what point from the pool is it actually saying?

    at least 6 m above the pool at any point.
    at least 6 m above any point on the water in the pool.

    Now it said "any point on the water...." does that mean vertically at any point or diagonally as well? What about horizontally??
    Does that mean I have to add 6m to each side of the pool to get a safe answer??

    How can I work this out? I am stuck and I have no idea what to do.

  7. #7
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    Quote Originally Posted by LaaLa View Post
    Thanks Dr.Peterson for your reply.
    I am not sure what you are meaning in in regards to finding 'y'. Are you meaning 'a'? The distance between the vertex and the focus? (8.5-1.5=7).
    I meant exactly what I said: solve the equation for y! If I gave you the equation x^2=-28(y-8.5) and asked you to solve for y, what would you do? (Some people call it "making y the subject".)

    If you don't do that, there will just be a little more work in other steps, but it is not essential, as I said. At some point, you have to find the value of y (the height of the roof) at the ends of the pool.

    Quote Originally Posted by LaaLa View Post
    The equation x^2=-28(y-8.5) works on the desmos because I calculated it using the vertex and the focus. Hoping that it is the right equation unless someone kindly corrects me!
    Yes, the equation is correct. Since they gave you the focus, what you did is exactly what is needed. Not all problems give the same information.

    Quote Originally Posted by LaaLa View Post
    Before we move on to what you say about this, do you see in the graph the answers to the questions? Is it wider than 36 meters at the base? Is it higher than 6 meters everywhere above the pool? This will not be the ultimate way you are supposed to answer those questions, which can be done without graphing, but the graph does help in understanding the questions.

    Quote Originally Posted by LaaLa View Post
    When I talk about having problems with the desmos, I am trying to make the equation by using the technique to what I have been learning (from the one example I had supplied). When the equation is made however, I then get stuck and don't know where to go from then onwards.
    The form used in the example is:
    y= ax^2 + b
    a = 18m (which is half of the base measurement)
    b = 8.5m (vertex)

    put it together...
    y= 18^2 + 8.5
    Why would "a" be half of the base measurement? The example you think you are following is entirely different (it doesn't mention the focus), and it doesn't set a to half the base width, but instead replaces x with that number and solves for a.

    But you didn't even do what you say you're doing; you replaced "ax^2" with "18^2", not with 18x^2.

    Quote Originally Posted by LaaLa View Post
    now if I work it out the way the example shows...
    18^2 = 324
    8.5/324 = 0.0262

    to make sure this is correct...
    -0.0262 x 18^2 = -8.5

    Now if I put this on desmos alongside the other equation and the 6 metre line. This is what it looks like:
    Screen Shot 2019-01-11 at 6.26.01 PM.jpgScreen Shot 2019-01-11 at 6.26.09 PM.png

    Now I see they look slightly different to each other; the first equation was based on the vertex and focus which has no base measurement, whereas the 2nd equation does.
    This formula is wrong, so they shouldn't look the same.

    Quote Originally Posted by LaaLa View Post
    What is confusing me is that this equation is based on the measurements given from a design for building a roof. And when it comes to numbers and construction, most of the numbers are usually rounded. I tried to do the same when it came to this particular equation that I'm confused on. when I was in the middle of working out the second equation this is what happened:

    The form used in the example is:
    y= ax^2 + b
    a = 18m (which is half of the base measurement)
    b = 8.5m (vertex)

    put it together...
    y= 18^2 + 8.5

    now if I work it out the way the example shows...
    18^2 = 324

    8.5/324 = 0.0262
    = 0.03 (3s.f)

    to make sure this is correct...
    -0.03 x 18^2 = -9.72

    Okay that didn't work, even though it's 0.01 difference. I will try it a different way:

    18^2 = 324[/I]
    8.5/324 = 0.0262
    = 0.026

    to make sure this is correct...
    -0.026 x 18^2 = -8.424

    Okay that is slightly 1 cm off.

    The only option I see is to stay with the first set of numbers.
    Again, this didn't work because the method is inappropriate for this problem, and you didn't follow it correctly anyway. Go with what you did correctly.

    Quote Originally Posted by LaaLa View Post
    I would really like to use more than one option in regards to answering these questions without relying on using the desmos to show me the answers. It would be easier if I just give all the points and measurements by looking at the graph but this is about using the forms and methods I have already learnt to find the answer.

    I don't know HOW to find out whether or not the height of the roof is at least 6 m above the pool at any point or whether or not the width of the pool roof is less the 36 metres.

    If I make an equation for 6m height and 18m width of the pool, this is what it looks like
    Screen Shot 2019-01-11 at 7.09.10 PM.jpgScreen Shot 2019-01-11 at 7.08.56 PM.jpg

    I have questions!!
    what point from the pool is it actually saying?

    at least 6 m above the pool at any point.
    at least 6 m above any point on the water in the pool.

    Now it said "any point on the water...." does that mean vertically at any point or diagonally as well? What about horizontally??
    Does that mean I have to add 6m to each side of the pool to get a safe answer??

    How can I work this out? I am stuck and I have no idea what to do.
    I think I told you what to do. Did you try following my suggestions?

    The height above the pool at any point means the value of y for any x within the pool (that is, between -9 and 9). As you can see from the graph, or just know about quadratic equations with negative leading coefficient, y decreases as you move away from the axis, so you only need to check the value of y when x = 9 (or -9), which are the locations of the ends of the pool.

    Height is always measured vertically. As I said, it's what y means in this problem.

    Try again, and just show me the work of using your good equation to answer the questions, so we can keep things simple.

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    It seems to me that you are not asking yourself what things mean. This problem is not hard once you ask about meaning rather than tools. A first step in grasping what the problem means is to write down brief definitions for your variables.

    [tex]x = \text {horizontal distance from middle line of pool.}[/tex]

    [tex]y = \text {vertical distance from pool level to roof.}.[/tex]

    Why do we use vertical distance? Because if the vertical distance from a point on the pool's surface is 6m, the distance in any other direction is more than 6m.

    Your question is about roof height, y, but your equation does not give you y. So the first tool that is going to be handy is to re-arrange the equation so that it gives you y. As Dr. P says, you do not have to do this, but if you are looking for y, it is obviously going to be extremely convenient to have an equation in the form y = f(x). Show us what you get from putting the equation into a form that is most useful for solving this problem.

    Now we think again. We need to know whether the vertical distance between pool surface and roof falls below some number. Consider your diagram. Where is that distance at a minimum? Is it at the midline. Obviously not. The distance is at a minimum at the edges of the pool because the parabola is always falling from its vertex (peak). Please tell us what is the value of x at the edge of the pool?

    Once you know what value to assign to x, what does your equation say is the value of y?

    How does this compare to 6?

    Now if you cannot answer any of those questions, we can help you. But start by trying to answer them.
    Last edited by JeffM; 01-11-2019 at 11:33 AM.

  9. #9
    New Member LaaLa's Avatar
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    Quote Originally Posted by Dr.Peterson View Post
    I meant exactly what I said: solve the equation for y! If I gave you the equation x^2=-28(y-8.5) and asked you to solve for y, what would you do? (Some people call it "making y the subject".)
    Hi Dr.Peterson, thanks for getting back to me. I appreciate the time you are taking to help me out.
    So far on parabolas I have learnt how to identify them using the key information (vertex, focus, centre, directrix and latus rectum ends) and drawing them onto graphs. I have also learnt about parametric equations, implicit differentiation, gradients, tangents, normals and intersections.
    Coming across this exercise in the study has put me to a halt. I am unfamiliar with this type of parabola and problem. The only exercise I have done was this one:

    Screen Shot 2019-01-11 at 2.07.51 PM.jpgScreen Shot 2019-01-11 at 2.08.07 PM.jpgScreen Shot 2019-01-11 at 2.08.20 PM.jpg

    What do you mean when you say 'find the y'? I have not come across an exercise like that in this study before. I have with ellipses and circles but not parabolas. Do you mean I must differentiate the parabola equation?

    Quote Originally Posted by Dr.Peterson View Post
    Why would "a" be half of the base measurement? The example you think you are following is entirely different (it doesn't mention the focus), and it doesn't set a to half the base width, but instead replaces x with that number and solves for a.


    I don't understand when you say this example is entirely different. Are you saying I should be using a different technique??
    The example is the only method I have learnt in regards to problem solving. I will explain how I got the idea...

    EXAMPLE states:
    - HEIGHT of 2.4metres
    - WIDTH of 1.6metres

    If you divide the width by half you get 0.8metres
    It was also stated in the example that the parabola intersects the x-axis at (8,0) and (-8,0).
    0.8metres is exactly what I got by dividing the width.

    The form used: y= ax^2+b
    for 'a' they used 0.8metres (width from centre to end of parabola)
    for 'b' they used 2.4metres (height from centre to top of parabola)

    If I use the same form but give different figures using the design I am working on:
    - HEIGHT of 8.5metres
    - WIDTH of 36metres
    Divide the width by half you get 18metres
    It was also stated that whether or not the width of the pool roof is less the 36 metres - I wouldn't have a clue what the width is, since 36metres was mentioned in the first place, I assume that number is what the width is.

    a= 18metres
    b= 8.5metres


    Quote Originally Posted by Dr.Peterson View Post
    But you didn't even do what you say you're doing; you replaced "ax^2" with "18^2", not with 18x^2.
    Can you explain please how it makes any difference adding an 'x' in there?
    I shall correct it then...

    y= 18x^2 + 8.5

    sorry if this has confused you, but I really need an explanation for everything.



    Quote Originally Posted by Dr.Peterson View Post
    The height above the pool at any point means the value of y for any x within the pool (that is, between -9 and 9). As you can see from the graph, or just know about quadratic equations with negative leading coefficient, y decreases as you move away from the axis, so you only need to check the value of y when x = 9 (or -9), which are the locations of the ends of the pool.
    THANK YOU!! That makes total sense! I wondered maybe those particular points could be used since it was the only information given other than the focus and vertex. I just don't know how to solve this now.
    I don't know what I am doing so I went online to use the symbolab calculator which hopefully has given me a correct answer:

    Y^2=-28(y-8.5)
    EXPAND
    (x^2)(-28y)((-28x-8.5=238)
    x^2=-28y+238

    SIMPLIFY
    x^2= 28-238
    x^2= 210
    x^2= √210 = 14.491
    x= 14.5

    I am hoping this is the answer to the distance from the centre to the x-intercept either side?
    So if I multiply it by 2, the total distance is 29metres?


    I am still not sure where to go from there. I feel that somehow I need to substitute Point (9,0) into the equation but I don't know how to do it.
    Can someone just please make it clear to me because I really don't know what to do. I'm just guessing everything now and I don't know where I'm going and I would like some guidance for the answer so in the future I can remember and use it again. Please, thank you
    Last edited by LaaLa; 01-12-2019 at 04:18 PM.

  10. #10
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    Quote Originally Posted by LaaLa View Post
    Hi Dr.Peterson, thanks for getting back to me. I appreciate the time you are taking to help me out.
    So far on parabolas I have learnt how to identify them using the key information (vertex, focus, centre, directrix and latus rectum ends) and drawing them onto graphs. I have also learnt about parametric equations, implicit differentiation, gradients, tangents, normals and intersections.
    Coming across this exercise in the study has put me to a halt. I am unfamiliar with this type of parabola and problem.

    What do you mean when you say 'find the y'? I have not come across an exercise like that in this study before. I have with ellipses and circles but not parabolas. Do you mean I must differentiate the parabola equation?
    Surely you know how to solve an equation for a single variable!

    What I mean is to rewrite "x^2 = -28(y-8.5)" as "y = ...". You can divide both sides by -28, then add 8.5 to both sides. There is nothing advanced about this.

    Please try.

    Quote Originally Posted by LaaLa View Post
    I don't understand when you say this example is entirely different. Are you saying I should be using a different technique??
    The example is the only method I have learnt in regards to problem solving. I will explain how I got the idea...

    EXAMPLE states:
    - HEIGHT of 2.4metres
    - WIDTH of 1.6metres

    If you divide the width by half you get 0.8metres
    It was also stated in the example that the parabola intersects the x-axis at (8,0) and (-8,0).
    0.8metres is exactly what I got by dividing the width.

    The form used: y= ax^2+b
    for 'a' they used 0.8metres (width from centre to end of parabola)
    for 'b' they used 2.4metres (height from centre to top of parabola)

    If I use the same form but give different figures using the design I am working on:
    - HEIGHT of 8.5metres
    - WIDTH of 36metres
    Divide the width by half you get 18metres
    It was also stated that whether or not the width of the pool roof is less the 36 metres - I wouldn't have a clue what the width is, since 36metres was mentioned in the first place, I assume that number is what the width is.

    a= 18metres
    b= 8.5metres
    I meant what I said: that example doesn't use the focus, so you can't do the same steps.

    But I also meant it when I said your equation was correct. Forget this example!!! Just think about the problem you're working on. You have a correct equation; now just answer the questions about it.

    Quote Originally Posted by LaaLa View Post
    Can you explain please how it makes any difference adding an 'x' in there?
    I shall correct it then...

    y= 18x^2 + 8.5

    sorry if this has confused you, but I really need an explanation for everything.
    Surely you know that 18^2 and 18x^2 are not the same thing! Without the x, it's just a constant.

    Quote Originally Posted by LaaLa View Post
    THANK YOU!! That makes total sense! I wondered maybe those particular points could be used since it was the only information given other than the focus and vertex. I just don't know how to solve this now.
    I don't know what I am doing so I went online to use the symbolab calculator which hopefully has given me a correct answer:

    Y^2=-28(y-8.5)
    EXPAND
    (x^2)(-28y)((-28x-8.5=238)
    x^2=-28y+238

    SIMPLIFY
    x^2= 28-238
    x^2= 210
    x^2= √210 = 14.491
    x= 14.5

    I am hoping this is the answer to the distance from the centre to the x-intercept either side?
    So if I multiply it by 2, the total distance is 29metres?

    I am still not sure where to go from there. I feel that somehow I need to substitute Point (9,0) into the equation but I don't know how to do it.
    Can someone just please make it clear to me because I really don't know what to do. I'm just guessing everything now and I don't know where I'm going and I would like some guidance for the answer so in the future I can remember and use it again. Please, thank you
    I'm not sure what you think you were doing there, but you seem to have solved for x when y = -1. That's irrelevant.

    Please think for yourself. You don't need a computer to do anything for you. This issue here is understanding how questions are related to the equation, which the computer will not do for you.


    Answering these long posts is very difficult. What I want from you is a minimal response, so I can answer just one question at a time.

    Here's the first question. You have your equation, x^2 = -28(y - 8.5). Although you never stated it explicitly, x means the distance from the center of the pool and y is the height above ground. What is the height above the point x=9, which is at one end of the pool? That is, what is y, when x is 9?

    Just answer that one question; don't tell me anything else yet. That will help us stay focused.

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