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Math Help - Equation of line that is tangent to graph of f and parallel

  1. #1
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    Equation of line that is tangent to graph of f and parallel

    I need to find an equation of the line that is tangent to the graph and parallel to the given line:

    function:
    f(x)=x^2-x

    line:
    x+2y-6=0

    How can I do this problem? (step by step always works best for me )

    Thanks so much for reading!!
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  2. #2
    Senior Member I-Think's Avatar
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    Good day AMaccy. Step by step it is then

    Step 1. Rearrange the line equation into the form y=mx+c
               y=\frac{-x}{2}+3

    Step 2. Remember that your teacher said,for our line to be parallel to the given line, it must have the same gradient.

    Step 3. Gain insight/remember for a line to be tangent to a curve, it must possess the same gradient as the curve at one particular point.

    Step 4. Knowing these 2 facts, we state that the gradient of our wanted line is m=\frac{-1}{2}<br />
    Step 5. Differentiate the curve to discover where it's gradient is \frac{-1}{2}
    \frac{dy}{dx}=2x-1, Let \frac{dy}{dx}=2x-1=\frac{-1}{2}
    x=\frac{1}{4}<br />
    Step 6. Obtain the y-coordinate at x=
    f(\frac{1}{4})=(\frac{1}{4})^2-\frac{1}{4}
    =\frac{-3}{16}

    Step 7. We now have the point at which the curve and line touch and our line's gradient, calculate our line's y-intercept.
    y=mx+c
    \frac{-3}{16}=\frac{-1}{2}(\frac{1}{4})+c
    c=\frac{-1}{16}

    So the equation of the line is y=\frac{-1}{2}x-\frac{-1}{16}
    Last edited by I-Think; June 1st 2009 at 05:31 PM.
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  3. #3
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    Quote Originally Posted by AMaccy View Post
    I need to find an equation of the line that is tangent to the graph and parallel to the given line:

    function:
    f(x)=x^2-x

    line:
    x+2y-6=0

    How can I do this problem? (step by step always works best for me )

    Thanks so much for reading!!
    1. Re-write the equation of the line:

    y=-\dfrac12 x +3 . Thus the slope is m = -\dfrac12

    2. The slope of the tangent equals first derivation of f at the tangent point. Therefore you need the first derivation:

    f'(x)= 2x-1

    Solve for x: m = f'(x):
    -\dfrac12=2x-1~\implies~x=\dfrac14

    The tangent point has the coordinates T\left(\dfrac14\ ,\ -\dfrac3{16}\right) because T is a point of the graph of f.

    3. Use the point-slope-formula to get the equation of the tangent:

    y-\left(-\dfrac3{16}\right)=-\dfrac12 \left(x-\dfrac14\right) Solve for y:

    y=-\dfrac12 x -\dfrac1{16}
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  4. #4
    MHF Contributor

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    "British English", "gradient".
    "American English", "slope". Both mean the same thing.
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