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Math Help - Finding tangent equations

  1. #1
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    Finding tangent equations

    Finding tangent equations of both lines through the point (2,-3) that are tangent to the parabola y=x^2 + x.

    I found the derivative which is 2x + 1 then I'm stuck.
    Any help/advice are greatly appreciated ! Thanks !
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  2. #2
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    Quote Originally Posted by letzdiscuss View Post
    Finding tangent equations of both lines through the point (2,-3) that are tangent to the parabola y=x^2 + x.

    I found the derivative which is 2x + 1 then I'm stuck.
    Any help/advice are greatly appreciated ! Thanks !

    Equation of a line in point slope form

     <br />
y-y_0=m(x-x_0)<br />

    You are given a point and the other point is (x,x^2+x)

     <br /> <br />
x^2+x-(-3)=(2x+1)(x-2)<br />
is what you get when you substitute all the given info into the point slope form

    Solving that will give you what x values make a tangent line that go through your point
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  3. #3
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    Quote Originally Posted by letzdiscuss View Post
    Finding tangent equations of both lines through the point (2,-3) that are tangent to the parabola y=x^2 + x.

    I found the derivative which is 2x + 1 then I'm stuck.
    Any help/advice are greatly appreciated ! Thanks !
    The equation of a line tangent to the parabola at (a, a^2 + a) is given by y = (2a + 1)(x - a) + a^2 + a = (2a + 1)x - a^2. This line goes through (2, -3) iff (2a + 1)*2 - a^2 = 3. So you need to solve the quadratic equation a^2 - 4a + 1 = 0 for a and plug the answers into the equation for the line.
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  4. #4
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    Quote Originally Posted by rn443 View Post
    The equation of a line tangent to the parabola at (a, a^2 + a) is given by y = (2a + 1)(x - a) + a^2 + a = (2a + 1)x - a^2. This line goes through (2, -3) iff (2a + 1)*2 - a^2 = 3. So you need to solve the quadratic equation a^2 - 4a + 1 = 0 for a and plug the answers into the equation for the line.
    You mean (2a + 1)*2 - a^2 = -3
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  5. #5
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    Quote Originally Posted by artvandalay11 View Post
    You mean (2a + 1)*2 - a^2 = -3
    Ah, whoops, yes.
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