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Thread: 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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    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: $\displaystyle y = ax + 1$?

    it has slope $\displaystyle a$ 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.

    $\displaystyle m = 6x - 4$ ..................(1)

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

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

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

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

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

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

    solving this, we find that $\displaystyle x = \pm 1$

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

    $\displaystyle (0,1)$ and $\displaystyle (1,3)$ ....tangent line 1

    or

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

    find the two lines that pass through those points and you will find the two values for $\displaystyle a$
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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,
    $\displaystyle ax+1 = 3x^2 - 4x+4$ has only a single solution.
    Thus,
    $\displaystyle 3x^2 - (4+a)x+3 = 0$
    To have exactly one solution we require that,
    $\displaystyle (4+a)^2 - 4(3)(3) = 0$
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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,
    $\displaystyle ax+1 = 3x^2 - 4x+4$ has only a single solution.
    Thus,
    $\displaystyle 3x^2 - (4+a)x+3 = 0$
    To have exactly one solution we require that,
    $\displaystyle (4+a)^2 - 4(3)(3) = 0$
    well, that was easy. i need a Staples easy button
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    thanks

    thanks for the help both of you
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