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Math Help - geometrical interpretations of the solutuon of equations- HELP PLEASE

  1. #1
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    Cool geometrical interpretations of the solutuon of equations- HELP PLEASE

    Hey this is probably really simple but I can't get my head around it.

    1) Find the possible values of a for which y = ax+1 is a tangent to y = 3x-4x+4

    2) Find the possible values of k for which y = x-8 is a tangent to y= 4x+ kx + 1

    Would be really grateful if someone could just get me started and explain what to do. Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lra11 View Post
    Hey this is probably really simple but I can't get my head around it.

    1) Find the possible values of a for which y = ax+1 is a tangent to y = 3x-4x+4
    I'll get you started with the first. the second should be similar.

    what do we know about the line: y = ax + 1?

    it has slope a and y-intercept 1

    how do we find the slope? by the derivative, so let's find the derivative. i will call the slope m.

    m = 6x - 4 ..................(1)

    ok, now we know that the given line passes through (0,1) and that all points on the curve are given by \left( x, 3x^2 - 4x + 4 \right)

    thus, the tangent line must pass through the points ( \underbrace{0}_{x_1} , \underbrace{1}_{y_1}) and (  \underbrace{x}_{x_2} , \underbrace{3x^2 - 4x + 4}_{y_2} )

    so another formula from the slope is (from precalc):

    m = \frac {y_2 - y_1}{x_2 - x_1} = \frac {3x^2 - 4x + 3}{x} ....................(2)

    but we are talking about the same lines here, the slopes must be equal. thus we equate (1) and (2) to obtain:

    6x - 4 = \frac {3x^2 - 4x + 3}x

    solving this, we find that x = \pm 1

    thus, the points any tangent line would go through are:

    (0,1) and (1,3) ....tangent line 1

    or

    (0,1) and (-1,11) ....tangent line 2

    find the two lines that pass through those points and you will find the two values for a
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  3. #3
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    Quote Originally Posted by lra11 View Post
    Hey this is probably really simple but I can't get my head around it.

    1) Find the possible values of a for which y = ax+1 is a tangent to y = 3x-4x+4
    For a parabola "being tangent" is the same as saying "only one intersection point".

    Thus,
    ax+1 = 3x^2 - 4x+4 has only a single solution.
    Thus,
    3x^2 - (4+a)x+3 = 0
    To have exactly one solution we require that,
    (4+a)^2 - 4(3)(3) = 0
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    For a parabola "being tangent" is the same as saying "only one intersection point".

    Thus,
    ax+1 = 3x^2 - 4x+4 has only a single solution.
    Thus,
    3x^2 - (4+a)x+3 = 0
    To have exactly one solution we require that,
    (4+a)^2 - 4(3)(3) = 0
    well, that was easy. i need a Staples easy button
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  5. #5
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    thanks

    thanks for the help both of you
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