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,
$\displaystyle a= \pm 2.5$
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:..$\displaystyle 0 = 6(68)^2 - 14 + a$..$\displaystyle \Rightarrow$..$\displaystyle a = -27730$