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Thread: 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 $\displaystyle f(x)=2x^2+3$ that pass through the point $\displaystyle (2,3)$

    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 $\displaystyle (x,f(x))$ and $\displaystyle (2,3)$ which which would leave the slope to be $\displaystyle (2x^2)/(2-3)$

    $\displaystyle 4x=(2x^2)/(2-3)$
    $\displaystyle 0=2x(3x-4)$

    $\displaystyle x=0, 4/3$

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

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

    The book has the second equation as: $\displaystyle 16x-y-29=o$
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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 $\displaystyle (2,3)$ and $\displaystyle (x,2x^2+3)$ is $\displaystyle \frac{y_2-y_1}{x_2-x_1}=\frac{3-(2x^2+3)}{2-x}$. The derivative of the function is $\displaystyle 4x$, and should be equal to slope at the corresponding value of x. So you have:

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

    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 $\displaystyle (2,3)$ and $\displaystyle (x,2x^2+3)$ is $\displaystyle \frac{y_2-y_1}{x_2-x_1}=\frac{3-(2x^2+3)}{2-x}$. The derivative of the function is $\displaystyle 4x$, and should be equal to slope. So you have:

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


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

    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 $\displaystyle x-2$. Here I'm defining the points $\displaystyle (x_1,y_1)=(x,2x^2+3)$ and $\displaystyle (x_2,y_2)=(2,3)$. 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 $\displaystyle -1$ to get $\displaystyle \frac{2x^2}{x-2}$
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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 $\displaystyle x-2$. Here I'm defining the points $\displaystyle (x_1,y_1)=(x,2x^2+3)$ and $\displaystyle (x_2,y_2)=(2,3)$. 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 $\displaystyle -1$ to get $\displaystyle \frac{2x^2}{x-2}$
    Okay, so what would you do next in regards to the original question? Would you agree the first tangent's equation is $\displaystyle y=0$? 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:

    $\displaystyle 8x-4x^2=-2x^2$

    $\displaystyle x(8-2x)=0$

    $\displaystyle x=0$ and $\displaystyle x=4$

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

    So just use the standard procedure. Find the derivative:

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

    The lines will have the form:

    $\displaystyle y-y_o=f'(x_o)(x-x_o)$

    Evaluate the derivative at $\displaystyle x=0$ and $\displaystyle x=4$

    $\displaystyle f'(0)=0$ and $\displaystyle f'(4)=16$

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

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

    $\displaystyle y-3=0(x-0)$

    $\displaystyle y=3$

    For the other equation I get:

    $\displaystyle y-35=16(x-4)$

    $\displaystyle y=16x-64+35$
    $\displaystyle y=16x-29$
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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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