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Math Help - Equations of the tangent that pass through a point

  1. #1
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    Equations of the tangent that pass through a point

    I'm trying to determine the tangents to the curve f(x)=2x^2+3 that pass through the point (2,3)

    I got the first equation right, but the second one is apparently wrong (according to the book).

    What I did was set the derivative equal to the slope of two points. Points I used were (x,f(x)) and (2,3) which which would leave the slope to be  (2x^2)/(2-3)

    4x=(2x^2)/(2-3)
    0=2x(3x-4)

    x=0, 4/3

    First equation is the tangent that passes through f(0)=3 so (0,3). Finding the equation by setting the derivative at f(0) equal to the slope like this:
    0=(y-3)/(x-0)
    y=3

    But I'm getting a different answer for the second equation:/

    The book has the second equation as: 16x-y-29=o
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  2. #2
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    I think there's an error in your slope calculation. The slope of the tangent line through the points (2,3) and (x,2x^2+3) is \frac{y_2-y_1}{x_2-x_1}=\frac{3-(2x^2+3)}{2-x}. The derivative of the function is 4x, and should be equal to slope at the corresponding value of x. So you have:

    4x=\frac{-2x^2}{2-x}

    You should obtain a quadratic from this.

    Edit: Note there should be two tangents, since you'll probably find two real roots.
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  3. #3
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    Quote Originally Posted by adkinsjr View Post
    I think there's an error in your slope calculation. The slope of the tangent line through the points (2,3) and (x,2x^2+3) is \frac{y_2-y_1}{x_2-x_1}=\frac{3-(2x^2+3)}{2-x}. The derivative of the function is 4x, and should be equal to slope. So you have:

    4x=\frac{-2x^2}{2-x}


    You should obtain a quadratic from this.
    Wouldn't y_2 be 2x^2+3?

    Regardless, that's pretty much what I did and ended up with the roots of the quadratic as 0 and 4/3. Can you tell me what I would do next?
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  4. #4
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    If that were the case, then the denominator would also change to x-2. Here I'm defining the points (x_1,y_1)=(x,2x^2+3) and (x_2,y_2)=(2,3). It's an arbitrary decision since we're only interested in the change between the values. You get the same slope either way. Just muliply by -1 to get \frac{2x^2}{x-2}
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  5. #5
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    Quote Originally Posted by adkinsjr View Post
    If that were the case, then the denominator would also change to x-2. Here I'm defining the points (x_1,y_1)=(x,2x^2+3) and (x_2,y_2)=(2,3). It's an arbitrary decision since we're only interested in the change between the values. You get the same slope either way. Just muliply by -1 to get \frac{2x^2}{x-2}
    Okay, so what would you do next in regards to the original question? Would you agree the first tangent's equation is y=0? If so, do you know how they got the second equation?
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  6. #6
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    I solved the quadratic:

    8x-4x^2=-2x^2

    x(8-2x)=0

    x=0 and x=4

    Basically we've reduced the problem down to: find the equation of the tangent line to the curve f(x)=2x^2+3 at x=0 and x=4.

    So just use the standard procedure. Find the derivative:

    f'(x)=4x

    The lines will have the form:

    y-y_o=f'(x_o)(x-x_o)

    Evaluate the derivative at x=0 and x=4

    f'(0)=0 and  f'(4)=16

    The points, which will be our (x_o,y_o), on the curve are (0, f(0))=(0,3) and (4,f(4))=(4,35)

    The first one is easy because the slope is zero, so it's a constant:

    y-3=0(x-0)

    y=3

    For the other equation I get:

    y-35=16(x-4)

    y=16x-64+35
    y=16x-29
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  7. #7
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    Ah, thanks a lot for the very detailed explanation! I see where I went wrong.

    Thanks again!
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