Review Packet Questions 6-8

I know that we were supposed to do two but since 7 and 8 are almost the same thing, why not. so...

6a. Find a relative minimum of the function O(t)= (100t^2+100)/(t^2+4t+1) on the interval [0, +∞) and describe its significance.
If you graph this on your calculator, the only relative minimum on this interval is at t=1. This means that after 1 hour, the stream has reached its lowest concentration of oxygen.

6b. Find the average rate of change of O(t) on the interval [2,4] and describe its significance.

Since the ARC is the slope of a certain interval, it can be found by taking the coordinates for O(2) & O(4) and using the slope formula:

Δy/Δx = 51.5-38.5/4-2 = 6.5

This means that from the 2nd hour after the waste has been in the stream to the 4th hour, the oxygen concentration has increased by an average of 6.5% an hour.

7. Since the graph in #7 is only translations, you only have to add or subtract from the x and y variables.

The graph is translated to the left by 3 so x--> x+3 (when x-->x+a, translated to the left by a; when x-->x-a, translated to the right by a)

It is also translated up by 4 so y-->y-4 (when y-->y+a, translated down by a; when y-->y-a, translated up by a)

since the vertex of the graph lies on the x-axis, the function equals zero

so: abs(x+3) + abs(y-4) = 0

8. This graph was only scaled, not translated so the x and y were multiplied.

The graph was scaled in the x-direction by 3 so x--> 1/3*x (when x-->a*x, scaled by 1/a)

Since the neither the highest nor lowest points of the graph lie on the x-axis but instead lies on 1 and -1, the function equals 1

so: abs(1/3*x) + abs(y) = 1