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Math Help - Equation of tangent to parabola

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    Equation of tangent to parabola

    Got a Prelim on monday , please help, thanks

    Q: Find the equation of the tangent to the parabola with equation
    y=6+x-x(squared) at the point (1,6 ) .



    If possible please explain to me how to do this question thanks .
    Last edited by mr fantastic; January 16th 2010 at 07:45 PM. Reason: Changed post title
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    Quote Originally Posted by goguy123 View Post
    Got a Prelim on monday , please help, thanks

    Q: Find the equation of the tangent to the parabola with equation
    y=6+x-x(squared) at the point (1,6 ) .



    If possible please explain to me how to do this question thanks .
    If you've done (or supposed to have done) calculus take the derivative and find f'(1)

    In this case f'(1) = -1

    Now you know the gradient use the straight line equation y=mx+c to find c using the ordered pair (1,6)

    In this case c = 7

    So the equation is y = -x + 7 or y= 7-x
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    Quote Originally Posted by goguy123 View Post
    Got a Prelim on monday , please help, thanks

    Q: Find the equation of the tangent to the parabola with equation
    y=6+x-x(squared) at the point (1,6 ) .



    If possible please explain to me how to do this question thanks .
    The slope of the line tangent to f(x) at the point (1.6) is m=f'(1).

    So, recall that the equation to a line is (y_2-y_1)=m(x_2-x)
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    Thanks a lot for the help.
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  5. #5
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    Quote Originally Posted by goguy123 View Post
    Got a Prelim on monday , please help, thanks

    Q: Find the equation of the tangent to the parabola with equation
    y=6+x-x(squared) at the point (1,6 ) .



    If possible please explain to me how to do this question thanks .
    Since you did not post this question in the Calculus subforum I assume a non-calculus approach is required ....?

    Since the point (1, 6) lies on the tangent line, the equation of the tangent is y - 6 = m(x - 1) \Rightarrow y = mx - m + 6 where m is the gradient.

    Also, the tangent (by definition) only intersects the parabola once. So the equations:

    y = 6 + x - x^2 .... (1)

    y = mx - m + 6 .... (2)

    have only one solution. Solving (1) and (2) simultaneously:

    6 + x - x^2 = mx - m + 6 \Rightarrow x^2 + (m - 1)x - m = 0.

    Since there's only one solution, the discriminant is equal to zero:

    (m - 1)^2 + 4m = 0.

    Solve this equation for m and substitute the value into y = mx - m + 6 to get the final equation of the tangent.
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