# Functions

• Oct 4th 2009, 10:54 AM
shane99
Functions
1. a) Describe the transformations from x^4-x^3+x^2 to
-3((0.5(x+4))^4-(0.5(x+4))^2)
c) Factor each function in a) to find the x-intercepts

2. A farm has a sale on eggs seling 13 eggs for the usual price of a dozen eggs. As a result, the price of eggs is reduced by 24 cents a dozen. What was the original price for the dozen eggs?
A: I guessed \$37.00 I dont know if that is right.

[ ] = means that whatever's inside appears in the bottom of f and in small letters, like the reverse of ^2.
3. Given f[0](x) = x^2 and f[n-1] = f[0](f[n](x)), where n is any natural number
a) determine f[1](x), f[2](x), and f[3](x)
b) determine a formula for f[n](x) in terms of n
• Oct 4th 2009, 11:54 AM
apcalculus
Quote:

Originally Posted by shane99
1. a) Describe the transformations from x^4-x^3+x^2 to
-3((0.5(x+4))^4-(0.5(x+4))^2)
c) Factor each function in a) to find the x-intercepts

2. A farm has a sale on eggs seling 13 eggs for the usual price of a dozen eggs. As a result, the price of eggs is reduced by 24 cents a dozen. What was the original price for the dozen eggs?
A: I guessed \$37.00 I dont know if that is right.

[ ] = means that whatever's inside appears in the bottom of f and in small letters, like the reverse of ^2.
3. Given f[0](x) = x^2 and f[n-1] = f[0](f[n](x)), where n is any natural number
a) determine f[1](x), f[2](x), and f[3](x)
b) determine a formula for f[n](x) in terms of n

x^4-x^3+x^2
to

f(x)= 3((0.5(x+4))^4-(0.5(x+4))^2

Note that, generally, f[(x+4)] is a horizontal translation of f(x) to the left by 4 units. The multiplication by 3 and -0.5 of the two terms is a vertical stretch and a shrink, respectively, except it's applied to the terms separately. But you have a cubic term in the original function which makes me wonder if something is missing in the 'transformed' function when you typed it.

You could of course expand the new function but it seems kind of pointless... or maybe I am not seeing it. I am sure other team members may have good ideas.

Good luck!