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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- Jun 12th 2007, 12:31 PMblueridgeFind 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.

- Jun 12th 2007, 01:09 PMThePerfectHacker
$\displaystyle f(g(x)) = 3(2x+a)^2 - 7$

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,

$\displaystyle a= \pm 2.5$ - Jun 12th 2007, 01:55 PMblueridgecurious...
I'm curious, but what would be the answer if g(f(x)) crosses the x-axis at 68?

- Jun 12th 2007, 02:27 PMJonboy
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:..$\displaystyle 0 = 6(68)^2 - 14 + a$..$\displaystyle \Rightarrow$..$\displaystyle a = -27730$