# Find a

• Jun 12th 2007, 12:31 PM
blueridge
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.
• Jun 12th 2007, 01:09 PM
ThePerfectHacker
Quote:

Originally Posted by blueridge
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.

$f(g(x)) = 3(2x+a)^2 - 7$

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

$3(2a)^2 = 75 \Rightarrow (2a)^2 = 25$
Thus,
$2a = \pm 5$
Thus,
$a= \pm 2.5$
• Jun 12th 2007, 01:55 PM
blueridge
curious...
I'm curious, but what would be the answer if g(f(x)) crosses the x-axis at 68?
• Jun 12th 2007, 02:27 PM
Jonboy
Quote:

Originally Posted by blueridge
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: $g(f(68))$
Since x = 68

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

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

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