• May 23rd 2006, 04:15 PM
takkun0486
write an equation of a periodic function that has:

a.) an amplitude of 200 units.

b.) been moved to the left 20 degrees

b.) been moved down 15 units

d.) a period of Pi/4

A parabola has intercepts at (-3,0), (1,0), and (0,6). find the exact equation for this parabola. Write the equation in both standard and graphing form

I just placed $1000 in an account which earns 8%per year compounded annually. ow much money will be in the account in 10 years? how long will it take for this account to have$5000 in it?

I cant seem to figure these out...

Nvm i got it
• May 23rd 2006, 06:30 PM
ThePerfectHacker
Quote:

Originally Posted by takkun0486

A parabola has intercepts at (-3,0), (1,0), and (0,6). find the exact equation for this parabola. Write the equation in both standard and graphing form

The equation of parabola is,
$y=ax^2+bx+c$.
The points tell you that,
$0=a(-3)^2+b(-3)+c$
$0=a(1)^2+b(1)+c$
$6=a(0)^2+b(0)^2+c$
From here we have,
$\left\{ \begin{array}{c}9a-3b+c=0\\a+b+c=0\\c=6$
Hence, since you know the value of "c" you can use the first two equations,
$\left\{ \begin{array}{c}9a-3b+6=0\\a+b+6$
From these equation we can simply to,
$\left\{ \begin{array}{c}3a-b=-2\\a+b=-6$
$4a=-8$ thus, $a=-2$
Finally, throw that value for "a" into second equation,
$-2+b=-6$ thus, $b=-4$
From here we have,
$y=-2x^2-4x+6$
• May 23rd 2006, 06:35 PM
ThePerfectHacker
Quote:

Originally Posted by takkun0486
write an equation of a periodic function that has:

a.) an amplitude of 200 units.

b.) been moved to the left 20 degrees

b.) been moved down 15 units

d.) a period of Pi/4

First convert 20 degrees into radians, $\frac{\pi}{9}$, also remember when you move something to the right you subtract and to the left you add.

These are either,
$f(x)=200\cos \left( \frac{\pi}{4}x+\frac{\pi}{9}\right)-15$
$f(x)=200\sin \left( \frac{\pi}{4}x+\frac{\pi}{9} \right)-15$
• May 23rd 2006, 06:41 PM
ThePerfectHacker
Quote:

Originally Posted by takkun0486

I just placed $1000 in an account which earns 8%per year compounded annually. ow much money will be in the account in 10 years? how long will it take for this account to have$5000 in it?

The formula is,
$A(t)=1000(1.08)^t$
In ten years gives,
$A(10)=1000(1.08)^{10}\approx 2,159$.

For the second part solve for t,
$1000(1.08)^t=5000$
Divide by 1000,
$(1.08)^t=5$
Take logarithms,
$\log 1.08^t=\log 5$
By exponents rule bring down the exponent,
$t \log 1.08=\log 5$
Thus,
$t=\frac{\log 5}{\log 1.08}\approx 20.91$
This means that 20 years is a little below, thus the next years is good. Thus answer is 21 years.