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How to find constants of quadratic equation given the coordinates. Watch

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    I tried solving this as simultaneous equation but the biggest problem for me is at the point where x=0 which makes the whole equation 0 and then i cannot solve it simultaneously.
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    (Original post by Jyashi)
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    I tried solving this as simultaneous equation but the biggest problem for me is at the point where x=0 which makes the whole equation 0 and then i cannot solve it simultaneously.
    Not sure I understand you? At (0, -1) we have x= 0 and y= -1, so:

    -1 = a(0^2) + b(0) + c \Rightarrow -1 = 0 + 0 + c \Rightarrow -1 = c
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    (Original post by Zacken)
    Not sure I understand you? At (0, -1) we have x= 0 and y= -1, so:

    -1 = a(0^2) + b(0) + c \Rightarrow -1 = 0 + 0 + c \Rightarrow -1 = c
    Hey Zacken. That is exactly correct. According to the question that is what i am getting for (0,-1) so how do we even find the constants?
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    (Original post by Jyashi)
    Hey Zacken. That is exactly correct. According to the question that is what i am getting for (0,-1) so how do we even find the constants?
    So you've found the value for c, before we move on are you happy with what I've done and you realise what you were doing wrong?
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    (Original post by Zacken)
    Not sure I understand you? At (0, -1) we have x= 0 and y= -1, so:

    -1 = a(0^2) + b(0) + c \Rightarrow -1 = 0 + 0 + c \Rightarrow -1 = c
    (Original post by Jyashi)
    Name:  Screenshot_2016-03-02-18-09-21.jpg
Views: 175
Size:  227.9 KB

    I tried solving this as simultaneous equation but the biggest problem for me is at the point where x=0 which makes the whole equation 0 and then i cannot solve it simultaneously.
    (Original post by Jyashi)
    Name:  Screenshot_2016-03-02-18-09-21.jpg
Views: 175
Size:  227.9 KB

    I tried solving this as simultaneous equation but the biggest problem for me is at the point where x=0 which makes the whole equation 0 and then i cannot solve it simultaneously.
    (Original post by Zacken)
    Not sure I understand you? At (0, -1) we have x= 0 and y= -1, so:

    -1 = a(0^2) + b(0) + c \Rightarrow -1 = 0 + 0 + c \Rightarrow -1 = c
    I think he forgot to subsitute the -1 for y or perhaps forgot c was there. Remember x is always the first coordinate and y the second coordinate . All you have to do is subsitute those values in as is done above

    (Original post by Jyashi)
    Hey Zacken. That is exactly correct. According to the question that is what i am getting for (0,-1) so how do we even find the constants?
    The constant is c which is the value Zacken worked out to be -1 .
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    (Original post by Zacken)
    So you've found the value for c, before we move on are you happy with what I've done and you realise what you were doing wrong?
    Well i had already gone so far before i posted this question. I also got to the point where at the coords (1,9) i got the equation 10= a + b. But after that i hit a wall.
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    (Original post by Jyashi)
    Well i had already gone so far before i posted this question. I also got to the point where at the coords (1,9) i got the equation 10= a + b. But after that i hit a wall.
    Time to use the next bit of information! We know that the derivative at (0, -1) is equal to 8.

    So \frac{dy}{dx} = 8, that should get you another equation in a and b that you can then use to solve simultaneously with your 10 = a+ b.
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    (Original post by Zacken)
    Time to use the next bit of information! We know that the derivative at (0, -1) is equal to 8.

    So \frac{dy}{dx} = 8, that should get you another equation in a and b that you can then use to solve simultaneously with your 10 = a+ b.
    Ahhh cant believe i forgot to differentiate.

    8= 2ax + b
    Since x=0 therefore b =8
    World makes sense again...
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    (Original post by Jyashi)
    Ahhh cant believe i forgot to differentiate.

    8= 2ax + b
    Since x=0 therefore b =8
    World makes sense again...
    :hat2:
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    (Original post by Zacken)
    :hat2:
    Thanks Zacken you saved my ass again.
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    (Original post by Jyashi)
    Thanks Zacken you saved my ass again.
    Glad I helped!
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    (Original post by Zacken)
    :hat2:
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    While we are at the question of calculus do you think you can help with the above question?
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    (Original post by Jyashi)
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    While we are at the question of calculus do you think you can help with the above question?
    Sure thing. y = \sqrt{x} = x^{1/2}, so using the power rule:

    \displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2} \times x^{1/2 - 1} = \frac{1}{2} \times x^{-1/2} = \frac{1}{2x^{-1/2}} = \frac{1}{2\sqrt{x}} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2\sqrt{x}}

    Now try multiplying both sides by 2x and making use of the indices rule: \frac{x}{\sqrt{x}} = \frac{x^1}{x^{1/2}} = x^{1 - 1/2} = \sqrt{x} = y
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    (Original post by Zacken)
    Sure thing. y = \sqrt{x} = x^{1/2}, so using the power rule:

    \displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2} \times x^{1/2 - 1} = \frac{1}{2} \times x^{-1/2} = \frac{1}{2x^{-1/2}} = \frac{1}{2\sqrt{x}} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2x}

    Now try multiplying both sides by 2x and making use of the indices rule: \frac{x}{\sqrt{x}} = \frac{x^1}{x^{1/2}} = x^{1 - 1/2} = \sqrt{x} = y
    I had managed to do till 1÷ 2 root x after which to get 1÷ 2x i am assuming you multiplied 2 root x with root x to get 2x.

    After that you seem to have done x ÷ y or x÷root x . I dont know how that got formed. Also finally doesnt the question ask to prove 2x times dy÷dx is equal to y ? Or am i misreading the question.
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    (Original post by Jyashi)
    I had managed to do till 1÷ 2 root x after which to get 1÷ 2x i am assuming you multiplied 2 root x with root x to get 2x.

    After that you seem to have done x ÷ y or x÷root x . I dont know how that got formed. Also finally doesnt the question ask to prove 2x times dy÷dx is equal to y ? Or am i misreading the question.
    Sorry, that was typo in my post, look at it now?
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    (Original post by Zacken)
    Sorry, that was typo in my post, look at it now?
    That's ok i think i had a sudden eurika moment. Check out my version of solution.

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    (Original post by Jyashi)
    That's ok i think i had a sudden eurika moment. Check out my version of solution.

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    That doesn't work, unfortunately! :no: You've only proven it for a specific value of x, but not for all x > 0 like they want you to!

    Look at my post again, which part are you tripping up over?
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    (Original post by Zacken)
    That doesn't work, unfortunately! :no: You've only proven it for a specific value of x, but not for all x > 0 like they want you to!

    Look at my post again, which part are you tripping up over?
    Pretty much everything after 1÷2x bit. I dont know how x÷root x came to be exactly. And just multiplying both sides by 2x doesnt make it clear for me. Sorry
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    (Original post by Jyashi)
    Pretty much everything after 1÷2x bit. I dont know how x÷root x came to be exactly. And just multiplying both sides by 2x doesnt make it clear for me. Sorry
    Do you agree that \frac{ dy }{dx} = \frac{1}{2\sqrt{x}}?
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    (Original post by Zacken)
    Do you agree that \frac{ dy }{dx} = \frac{1}{2\sqrt{x}}?
    Yes i do.
    Lol feel like im getting married.
 
 
 
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