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Math Help - turning point of a curve

  1. #1
    Newbie Zharif93's Avatar
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    turning point of a curve

    Given ( 1 , 4 ) is a turning point of the curve y = ax^3 + bx^2 - 6x - 1/2 .

    Find the values of a and b

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  2. #2
    Senior Member eumyang's Avatar
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    If (1, 4) is on the curve y = ax^3 + bx^2 - 6x -  1/2, plug in x and y to get
    \begin{aligned}<br />
4 &= a(1)^3 + b(1)^2 - 6(1) -  1/2 \\<br />
4 &=  a + b - 13/2 \\<br />
a + b &=  21/2 \\<br />
\end{aligned}

    If (1, 4) is a turning point, then the derivative of y is 0 at that point:
    \begin{aligned}<br />
y' &= 3ax^2 + 2bx - 6 \\<br />
0 &=  3a(1)^2 + 2b(1) - 6 \\<br />
0 &= 3a + 2b - 6 \\<br />
3a + 2b &= 6<br />
\end{aligned}

    You're left with a system of equations
    \begin{aligned}<br />
a + b &=  21/2 \\<br />
3a + 2b &= 6<br />
 \end{aligned}

    Solve for a and b.
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