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Math Help - Precalculus Quadratic Functions

  1. #1
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    Precalculus Quadratic Functions

    For a) how would I get the formula? Please show me the steps.



    Thank you.
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    Quote Originally Posted by nox16 View Post
    For a) how would I get the formula? Please show me the steps.



    Thank you.
    Ok, here's the first clue: notice how the decimals go .00, .01, .04, .09... Look familiar? Use that to determine the squared term. (Remember when a fraction is squared it becomes smaller.)

    Clue #2: Notice how the closest integer values go 1,3,5,7. Also realize that, f(0) will give you the constant term (the last term in the quadratic).

    Please try to do this yourself before you look at the answer. I assure you will feel much more satisfied and will have gained and learned more.

    So, if you're still stuck, or want to check your answer, this is the answer I got which I'm quite sure is right:

    Spoiler:
    y=(\frac{x}{10})^2+2x+1
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    Yeah that answer is correct. In the back of the book it says
    y = 0.01x^2 + 2x + 1. But I still don't get how you got that answer. What did you do? Did you plug in some values into the equation y = ax^2 + bx + c? Was there an elimination process?
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    Quote Originally Posted by nox16 View Post
    Yeah that answer is correct. In the back of the book it says
    y = 0.01x^2 + 2x + 1. But I still don't get how you got that answer. What did you do? Did you plug in some values into the equation y = ax^2 + bx + c? Was there an elimination process?
    Oh I see, that's the main problem.

    Okay look at the clues I gave you. That was basically my thought process. Remember 7th grade Math when you learned about basic sequences? Well, when I saw the 1,3,5,7... I saw that the difference between each term was roughly 2. So I assumed that the bx + c part would be 2x+1 [c=1 because c = y(0)]. I also noticed the 0,1,4,9 pattern in the decimals. These perfect squares were very familiar. This is basically all the needed information. Since the decimal parts are 100th of the perfect squares, this gives the final formula of y = 0.01x^2 + 2x + 1.

    Unfortunately, there isn't a systematic method that will always work. The key is practice. Learn to notice patterns like this.

    Hope that helped. If you have any more questions to clarify, feel free to ask.
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    Thank you so much for clarifying that. I have understand how to calculate for b and c now. But for a I'm not so sure about. I have noticed the pattern 0,1,4,9 in the decimals but what do you mean by "the decimal parts are 100th of the perfect squares" and how did you get that 0.01. Again I thank you for your patience and persistence in helping me.
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  6. #6
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    Quote Originally Posted by nox16 View Post
    Thank you so much for clarifying that. I have understand how to calculate for b and c now. But for a I'm not so sure about. I have noticed the pattern 0,1,4,9 in the decimals but what do you mean by "the decimal parts are 100th of the perfect squares" and how did you get that 0.01. Again I thank you for your patience and persistence in helping me.
    No problem.

    You want me to explain how I got the value for a?

    Well I noticed the 0,1,4,9 perfect squares pattern.

    Normally, if y=x^2, then for x=0,1,2,3... the y-values will be 0,1,4,9...

    However, I don't want to add 0,1,4,9... to 2x+1 because I only need to add 0, 0.01, 0.04, 0.09.... So, x^2 will be exactly 100 times too big (bigger than what I want). This means that I need to make the squared term 100 times smaller to make the pattern work. So, I will change the first term to 0.01x^2. Now it is correct.

    Do you understand? Don't hesitate to ask for more help!
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    Quote Originally Posted by mathemagister View Post
    No problem.

    You want me to explain how I got the value for a?

    Well I noticed the 0,1,4,9 perfect squares pattern.

    Normally, if y=x^2, then for x=0,1,2,3... the y-values will be 0,1,4,9...

    However, I don't want to add 0,1,4,9... to 2x+1 because I only need to add 0, 0.01, 0.04, 0.09.... So, x^2 will be exactly 100 times too big (bigger than what I want). This means that I need to make the squared term 100 times smaller to make the pattern work. So, I will change the first term to 0.01x^2. Now it is correct.

    Do you understand? Don't hesitate to ask for more help!
    Haha yeah I got it now thank you so much! However just to clarify, let's say this is a new table.

    x: 0, 1, 2, 3
    y: 3, 5.4, 7.6, 9.8

    Would we do the same thing over again? Meaning the quadratic formula for this equation would be 0.01x^2 + 2x + 3. Or did I do something wrong?
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    Quote Originally Posted by nox16 View Post
    Haha yeah I got it now thank you so much! However just to clarify, let's say this is a new table.

    x: 0, 1, 2, 3
    y: 3, 5.4, 7.6, 9.8

    Would we do the same thing over again? Meaning the quadratic formula for this equation would be 0.01x^2 + 2x + 3. Or did I do something wrong?
    You are very welcome!

    Something went wrong. Your table doesn't work with a simple quadratic with easy numbers.

    y = 0.01x^2 + 2x + 3 would give:
    x: 0, 1, 2 ,3
    y: 3, 5.01, 7.04, 9.09

    I think you just did a silly miscalculation. Okay, I have to sleep now (it is almost 2am here).

    I hope I helped you well today, I will be back tomorrow to clarify any more doubts you have.
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    Sorry if I kept you up late at night just for that. Hope tomorrow isn't a school day or something for you. I really appreciate your help in trying to help me learn this.

    Alright so I notice the pattern goes like this, 0, 4, 6, 8. To me this doesn't seem like it would fit into any perfect square(except for the 0 and the 4). So in this case how would we solve this problem?
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    Quote Originally Posted by nox16 View Post
    Sorry if I kept you up late at night just for that. Hope tomorrow isn't a school day or something for you. I really appreciate your help in trying to help me learn this.

    Alright so I notice the pattern goes like this, 0, 4, 6, 8. To me this doesn't seem like it would fit into any perfect square(except for the 0 and the 4). So in this case how would we solve this problem?
    Don't worry about it; it isn't a school day. So about this function, I don't think there is an easy quadratic function that works with
    x: 0, 1, 2, 3
    y: 3, 5.4, 7.6, 9.8
    So, I don't think you will find a question like that. I put it in a graphing calculator, and it only gives a crazy quadratic with long decimals as coefficients. Did you just make those numbers up? If you didn't that's a really hard question...
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    As usual thanks for answering my question. Yeah I did make up that question. Alright so I got through all the problem except e). Do you happen to know how to do that? I don't get how to find the answer how that they diff by less than 0.5.
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    Quote Originally Posted by nox16 View Post
    As usual thanks for answering my question. Yeah I did make up that question. Alright so I got through all the problem except e). Do you happen to know how to do that? I don't get how to find the answer how that they diff by less than 0.5.
    Okay, phew that those numbers were just made up. As you can see, a nice, easy quadratic will not fit for any random numbers.

    For some reason, the picture isn't working for me anymore. Can you send me a link if you have one?
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    Quote Originally Posted by nox16 View Post
    As usual thanks for answering my question. Yeah I did make up that question. Alright so I got through all the problem except e). Do you happen to know how to do that? I don't get how to find the answer how that they diff by less than 0.5.
    Oh right, pressing quote lets me see the link url. Sorry for being a bit stupid there.

    Well, there's a way to do e) quite handily, but you're allowed to use a graphing calculator, so that makes it easy. So, about the 0.05 part, I'm assuming you got the linear function right because you said you get the other parts of the question. You have to find the interval where the function y and the linear function (L) you created have values which are less than 0.05. So solve for y(x) - L(x) < 0.05 Is that what you didn't know how to do? Do you get it now? Do you need anymore help?
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    Thanks for helping me so far. Alright so I know the linear function is:
    y = 2.03x + 0.98 and the quadratic formula is 0.01x^2 + 2x + 1. So as you said y(x) - L(x) < 0.5 should be (2.03x + 0.98) - (0.01x^2 + 2x + 1) which should come out to be -0.01x^2 + 0.03x - 0.02 < 5. So now what I need to do?

    Also the answer in the back of the book says it is -0.791 < x < 3.791. Note: The < sign is a less than or equal to
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  15. #15
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    Quote Originally Posted by nox16 View Post
    Thanks for helping me so far. Alright so I know the linear function is:
    y = 2.03x + 0.98 and the quadratic formula is 0.01x^2 + 2x + 1. So as you said y(x) - L(x) < 0.5 should be (2.03x + 0.98) - (0.01x^2 + 2x + 1) which should come out to be -0.01x^2 + 0.03x - 0.02 < 5. So now what I need to do?

    Also the answer in the back of the book says it is -0.791 < x < 3.791. Note: The < sign is a less than or equal to
    Oh sorry, it should technically be |y(x) - L(x)| < 0.05, but it doesn't matter because y(x) - L(x) will always be positive.
    To solve it, just use a graphing calculator or a computer program, the question asks you to do it that way. Do you have a good graphing calculator or a computer program like Autograph?

    By the way, you keep saying 0.5 instead of 0.05; make sure you solve for the right thing.

    If you want me to help you on how to use the calculator/program to solve the equation, I would be glad to help.
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