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  1. #1
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    Find a

    If f(x) = 3x^2 - 7 and g(x) = 2x + a, find a so that the graph of f(g(x)) crosses the y-axis at 68.
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  2. #2
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    Quote Originally Posted by blueridge View Post
    If f(x) = 3x^2 - 7 and g(x) = 2x + a, find a so that the graph of f(g(x)) crosses the y-axis at 68.
    $\displaystyle f(g(x)) = 3(2x+a)^2 - 7$

    The y-intercept is when x=0.
    So when x=0 we have $\displaystyle f(g(0)) = 68$
    That means,
    $\displaystyle 3(2\cdot 0 + a)^2 - 7 = 68$
    Thus,
    $\displaystyle 3(2a)^2 - 7 = 68$

    $\displaystyle 3(2a)^2 = 75 \Rightarrow (2a)^2 = 25$
    Thus,
    $\displaystyle 2a = \pm 5$
    Thus,
    $\displaystyle a= \pm 2.5$
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  3. #3
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    curious...

    I'm curious, but what would be the answer if g(f(x)) crosses the x-axis at 68?
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  4. #4
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    Quote Originally Posted by blueridge View Post
    I'm curious, but what would be the answer if g(f(x)) crosses the x-axis at 68?
    So you're trying to find: $\displaystyle g(f(68))$
    Since x = 68

    First find g(f(x)):..$\displaystyle 2(3x^2 - 7) + a$..$\displaystyle \Rightarrow$.. $\displaystyle 6x^2 - 14 + a$

    Since it crosses the x - axis at 68, the y coordinate is zero.

    $\displaystyle g(f(x))$ is the same as the y coordinate:..$\displaystyle 0 = 6(68)^2 - 14 + a$..$\displaystyle \Rightarrow$..$\displaystyle a = -27730$
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