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Math Help - finding constants of function

  1. #1
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    finding constants of function

    The function f(x)=ax^3+bx^2+cx+d has turning point (2, -4) and cuts the y axis at (0, 2). Find constants a, b, c, and d.

    can someone check my work please?

    a=0
    b=3/2
    c=-6
    d=2
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  2. #2
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    if you have 2 equations and 4 constants, certainly you'll find more than 1 solution.

    how did you get to that answer?

    edit:I notice now about the 'turning' point. I dont know if s some mathematical term in English that provides more info.
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  3. #3
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    what's that supposed to mean?
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  4. #4
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    it means I dont know what turning means

    if its something like \frac{df(x)}{dx}=0, you get an extra euqation. its still not enough to provide only one solution.

    but if it means that \frac{df(x)}{dx}=0 and \frac{d^2f(x)}{dx^2}=0 (this is only an example, as I said I dont know what it means) you get for equations and 4 constants you can have a system and solve it for only one solution.

    if it doesnt mean nothing you still have a system but you'll get more than 1 solution.
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  5. #5
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    Quote Originally Posted by Haytham View Post
    it means I dont know what turning means

    if its something like \frac{df(x)}{dx}=0, you get an extra euqation. its still not enough to provide only one solution.

    but if it means that \frac{df(x)}{dx}=0 and \frac{d^2f(x)}{dx^2}=0 (this is only an example, as I said I dont know what it means) you get for equations and 4 constants you can have a system and solve it for only one solution.

    if it doesnt mean nothing you still have a system but you'll get more than 1 solution.
    oh, well my teacher meant by max/min point = turning point.. so I took it as f'(x)

    Can you check my answers with that being said now?
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  6. #6
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    considering that, and only that, your solution is incomplete. that is only one solution, you have to find all.


    with the f(x)' info you get 3 equations

    2=d

    -4=8a+4b+2c+d

    0=12a+4b+c


    you need to solve those 3. you will get something like
    d=2

    c=4a-6

    b=\frac{3}{2}-4a

    a  \epsilon  R

    this means, you can choose any value for a, and calculate the values of b,c and d, with that a and get different solutions.
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  7. #7
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    Quote Originally Posted by skeske1234 View Post

    d=2
    This is correct

    Quote Originally Posted by skeske1234 View Post

    a=0
    b=3/2
    c=-6
    How did you get these values?
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  8. #8
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    Here is my work, please check:

    f'(x)=0 when
    0=12a+4b+c

    f(2)=-4
    -4=24a+4b+2c+d

    2=d

    -12a-4b=c

    -4=24a+4b+2(-12a-2b)+2
    3/2=b
    a=0

    -4(3/2)=c
    6=c

    Is this ok?
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  9. #9
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    Quote Originally Posted by skeske1234 View Post
    The function f(x)=ax^3+bx^2+cx+d has turning point (2, -4) and cuts the y axis at (0, 2). Find constants a, b, c, and d.
    This gives the following

    f(2) = -4

    f(0) = 2 \Rightarrow d = 2

    f'(2)=0


    Quote Originally Posted by skeske1234 View Post

    f'(x)=0 when
    0=12a+4b+c

    You mean f'(2)=0 \Rightarrow 0=12a+4b+c


    Quote Originally Posted by skeske1234 View Post

    f(2)=-4
    -4=24a+4b+2c+d

    Given f(x) = ax^3+bx^2+cx+2

    f(2) = 2^3a+2^2b+2c+2

    -4 = 2^3a+2^2b+2c+2

    -4 = 8a+4b+2c+2
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