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Thread: Tangent line help..

  1. #1
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    Tangent line help..

    Find the equations of the 2 tangent lines to the graph of f that pass through the indicated point.

    f(x)= 4x-x^2 point is (2, 5)

    The farthest I have gotten with this is writting this:

    (a^2-5)/(a-2)


    The next thing is I'm trying to figure out the equal thingy..

    a^2-5 = ???
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by elpermic View Post
    Find the equations of the 2 tangent lines to the graph of f that pass through the indicated point.

    f(x)= 4x-x^2 point is (2, 5)

    The farthest I have gotten with this is writting this:

    (a^2-5)/(a-2)


    The next thing is I'm trying to figure out the equal thingy..

    a^2-5 = ???
    What you need to do is first find $\displaystyle f'(x)$:

    $\displaystyle f'(x)=4-2x$

    Thus, the slope at the point (2,5) is 4-2(2)=0.

    So the tangent line is horizontal.

    The equation of the tangent would then be $\displaystyle y-5=0(x-2)\implies \color{red}\boxed{y=5}$

    Does this make sense?

    --Chris
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  3. #3
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    Nope.. Actually, the line isn't horizontal at all..

    I got a^2-5/a-2 and took the derivative and that is 2a..

    Now what do I do?
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  4. #4
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    Actually, scratch that, I can't believe how stupid I am.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by elpermic View Post
    Nope.. Actually, the line isn't horizontal at all.
    I think I see what may need to be done...

    Let $\displaystyle (x_0,f(x_0))$ be a point on the function.

    Then, $\displaystyle f'(x)=4-2x\implies f'(x_0)=4-2x_0$

    Thus, the tangent line has the form $\displaystyle y-f(x_0)=f'(x_0)(x-x_0)$

    So we see that $\displaystyle y-(4x_0-x_0^2)=(4-2x_0)(x-x_0)$

    This line passes through the point $\displaystyle (2, 5)$, so our equation becomes:

    $\displaystyle 5-(4x_0-x_0^2)=(4-2x_0)(2-x_0)$

    So we get $\displaystyle x_0^2-4x_0+5=8-8x_0+2x_0^2$

    Getting all the terms on side, we see that $\displaystyle x_0^2-4x_0+3=0$

    So $\displaystyle x_0^2-4x_0+3=0\implies (x_0-3)(x_0-1)=0\implies \color{red}\boxed{x_0=1~or~x_0=3}$

    Thus, $\displaystyle f(1)=3$ and $\displaystyle f(3)=3$

    Therefore, we get two different tangent lines:

    $\displaystyle y=2(x-1)+3\implies \color{red}\boxed{y=2x+1}$

    $\displaystyle y=-2(x-3)+3\implies \color{red}\boxed{y=-2x+9}$

    Does this make sense?

    The graph verifies this:



    --Chris
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  6. #6
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    Yup, those 2 are the answers..

    The stupid thing is though, I just had to take the slope from (2,5) and the other 2 points on the graph, and I would've gotten the answer..

    The question would've only taken me a minute, and it ends up taking 20 minutes trying to find stuff, lol.
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