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Math Help - [SOLVED] Anyone know ?

  1. #1
    jayson81
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    [SOLVED] Anyone know ?

    The curves y + ax^2 +b and y = 2x^2 +cx have a common tangent at the point (-1,0).
    Find a, b and c.
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  2. #2
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    Quote Originally Posted by jayson81
    The curves y + ax^2 +b and y = 2x^2 +cx have a common tangent at the point (-1,0).
    Find a, b and c.
    Firstly, (-1,0) is a common point thus.
    Substitute that into the second function,
    0=2(-1)^2+c(-1)
    Thus,
    0=2-c solve, c=2
    Thus, the second function has equation,
    y=2x^2+2x.

    Since we say that these two function have a common tangent means the equation of the tangent through point (-1,0) is the same. Remember from analytic geomtery that two lines having same slope at same point are the same. Same thing here the tangent lines share a point thus they need to have the same slope.

    The slope at x=-1 for the first function is,
    y'=2ax at x=-1 is -2a.
    The slope for the second is,
    y'=4x+2 at x=-1 is -2
    Thus, -2=-2a thus, a=1
    All we need is to find "b" in,
    y=x^2+b since (-1,0) is on the curve we have,
    0=(-1)^2+b thus, b=-1
    Thus,
    (a,b,c)=(1,-1,2)
    Which is not true.
    Thus, either you erred in the formulation of your question or there is no such numbers.
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  3. #3
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    Hello, jayson81!

    I got the same result as ThePerfectHacker, but I find that the solution works . . .

    The curves y = ax^2 +b and y = 2x^2 +cx have a common tangent at the point (-1,0).
    Find a, b and c.
    Since (-1,0) is on both curves, we have:
    . . a(-1)^2 + b \,=\,0\quad\Rightarrow\quad a + b \,=\,0
    . . 2(-1)^2 + c(-1)\,=\,0\quad\Rightarrow\quad c = 2

    Since the slopes are equal at (-1,0), we have:
    . . 2ax\,=\,4x + c\quad\Rightarrow\quad -2a\,=\,-4 + c
    Since c = 2, we get: a = 1 . . . and hence: b = -1

    Therefore: . (a,b,c)\,=\,(1,-1,2)


    The parabolas are: . y\,=\,x^2 - 1 and y\,=\,2x^2 + 2x
    They share the common point (-1,0) where they both have slope -2.
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