Question

I need help with 4 5 and 6

Answer 1

4. Create a table, write an equation, AND show all your work.

Currently the stock is $62 and has been decreasing 3% each month. So, if we call
`x=0` the first month then:

`at \ x=0 \ y=62 \\ \\ at \ x=1 \ y=62-62* 3\%=60.14 \\ \\ at \ x=2 \ y=60.14-60.14* 3\%=58.3358 \\ \\ at \ x=3 \ y=58.3358-58.3358* 3\%=56.5857 \\ \\ at \ x=4 \ y=56.5857-56.5857* 3\%=54.8881 \\ \\ at \ x=5 \ y=54.8881-54.8881* 3\%=53.2414`

So the Table is:

`\left[\begin{array}{cc}x & y\\0 & 62\\1 & 60.14\\2 & 58.3358\\3 & 56.5857\\4 & 54.8881\\5 & 53.2414\end{array}\right]`

This is an exponential function because it has a constant ratio.

From our values, we can conclude that the equation is:

`y=62(100%-3%)^(x) \\ \\ \boxed{y=62(0.97)^(x)}`

In five months, the stock will be $53.2414

5. Write the exponential function

We are given two points:

`P_(1)(0,5) \\ \\ P_(2)(3,135)`

We know that the exponential function is given by the form:

`y=ab^(x)`

So:

`\bullet \ If \ x=0 \ then \ y=5 \\ \\ 5=ab^(0) \therefore a=5 \\ \\ \\ \bullet \ If \ x=3 \ y=135 \\ \\ 135=ab^(3) \therefore ab^(3)=135 \\ \\ \\ Finding \ b: \\ \\ (ab^(3))/(a)=(135)/(5) \\ \\ b^3=27 \therefore b=\sqrt[3]{27} \therefore \boxed{b=3}`

Then:

`\boxed{y=5(3)^(x)}`

6. Slope, length and midpoint

Here we are given two points:

`A(4,9) \ and \ B(-2,3)`

So, the slope can be found as:

`m=(Change \ in \ y)/(Change \ in \ x) \\ \\ m=(y_(2)-y_(1))/(x_(2)-x_(1)) \\ \\ m=(3-9)/(-2-4) \\ \\ \boxed{m=1}`

Using the distance formula we can find the length:

`d=\sqrt{(x_(2)-x_(1))^2+(y_(2)-y_(1))^2} \\ \\ d=√((-2-4)^2+(3-9)^2) \\ \\ d=√(36+36) \\ \\ d=√(72) \\ \\ \boxed{d=6√(2)}`

To find the midpoint, we have to use the midpoint formula:

`Midpoint=((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2)) \\ \\ Midpoint=((4-2)/(2),(9+3)/(2)) \\ \\ \boxed{Midpoint=(1,6)}`