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Math Help - finding tangent

  1. #1
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    finding tangent

    Show that the tangent to P: y= a(x-h)^2 + k at point T, where x= alpha, is y=2a(alpha-h)x+k-a(alhpa^2-h^2).

    Thank you for any help
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  2. #2
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    Hint: If

    y=a(x-h)^2+k,

    then

    \begin{aligned}<br />
y'&=\frac{d}{dx}(a(x-h)^2+k)\\<br />
&=\frac{d}{dx}(a(x-h)^2)+\frac{d}{dx}(k)\\<br />
&=2a(x-h)\cdot \frac{d}{dx}(x-h)+0\;\;\;\;\;\;\;\;\;\;\mbox{(Chain Rule)}\\<br />
&=2a(x-h)\cdot 1\\<br />
&=2a(x-h).<br />
\end{aligned}
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  3. #3
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    Quote Originally Posted by slaypullingcat View Post
    Show that the tangent to P: y= a(x-h)^2 + k at point T, where x= alpha, is y=2a(alpha-h)x+k-a(alhpa^2-h^2).

    Thank you for any help
    Substitute x = \alpha into the given equation to get the y-coordinate. So you have a point on the tangent.

    Calculate the value of dy/dx at x = \alpha to get the gradient of the tangent.

    Substitute all of the above into y - y_1 = m (x - x_1) to get the equation.
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  4. #4
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    I am using m=2a(x-h), x1=alpha and y1=a(alpha-h)^2 +k. I'm getting fairly close but its still not correct. Are my substitution values right? If so I guess I am simplifying incorrectly
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  5. #5
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    Quote Originally Posted by slaypullingcat View Post
    I am using m=2a(x-h), x1=alpha and y1=a(alpha-h)^2 +k. I'm getting fairly close but its still not correct. Are my substitution values right? If so I guess I am simplifying incorrectly
    Your trouble will be somewhere in the algebra.
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