Problem Number 1

Problem:  Obtain a quadratic depth vs. clock time model if depths of 67.39103 cm, 52.27462 cm and 42.65076 cm are observed at clock times t = 9.731353, 19.46271 and 29.19406 seconds.

This means to plug the numbers into the form of a quadratic and solve the 3 simultaneous equations for a, b, c; then plug the parameters a, b and c back into the model.

Problem: The quadratic depth vs. clock time model corresponding to depths of 67.39103 cm, 52.27462 cm and 42.65076 cm at clock times t = 9.731353, 19.46271 and 29.19406 seconds is depth(t) = .029 t2 + -2.4 t + 88. Use the model to determine the clock time at which depth is 63.89874 cm.

depth(t) = .029 t2 + -2.4 t + 88.

Find the clock time at which depth is 63.89874 cm.

Plug in 63.9 cm for depth(t) to get the equation

63.9 = .029 t2 + -2.4 t + 88

and solve the equation (use the quadratic formula).

Problem Number 2

Problem:  Sketch a graph of the basic exponential function y = 2 x. Sketch the graph of this function stretched vertically by factor -2.72 then shifted -1.78 units vertically.  Show that the graph is different than that obtained if the vertical shift precedes the vertical stretch.  Give the algebraic form of the resulting function.

Vertical stretch by factor -2.72 multiplies all y values by -2.72, making the function y = -2.72 * 2^x. 

Use basic points like (-1, 1/2), (0, 1) and (1, 2) to sketch a picture of the graph then vertically stretch each point, and use these points as a basis for your new graph. 

Vertical shift -1.78 lowers each point 1.78 units, giving us the function y = -2.72 * 2^x - 1.78.

Problem:  At clock times t = 6, 12, 18 and 24 seconds, the depth of water in a uniform cylinder was observed to be 34.336, 30.544, 28.624 and 28.576 cm. At what average rate was the depth changing during each of the three time intervals?

For each interval calculate change in depth / change in clock time, corresponding to rise / run of the graph.

There are 3 intervals and you need to calculate for each, meaning you'll calculate 3 rates.

 Look at the rates you have calculated and predict what the next average rate would be. If your prediction is correct, then what will be the depth at t = 30 seconds?

There will be a pattern to these rates and it will be easy to predict the next rate.

Then using the rate you can calculate the change in depth over the next 6-second interval.

Problem Number 3

Sketch a graph of y = x^2, from x = -3 to x = 3. Then sketch a graph of y = 4 x^2 over the same domain.

Sketch the second graph shifted -.25 units in the x direction and -1.25 units in the y direction.

What are the three basic points of this graph?

If f(x) = x^2, then what are

and what are the basic points of each?

f(x - -.25) gives us (x - -.25)^2 = (x + .25)^2, which shifts the graph -.25 units in the horizontal direction, giving basic points (-1.25,1), (-.25, 0) and (.75, 1).

f(x) + -1.25 gives us x^2 - 1.25, which shifts the y=x^2 graph vertically -1.25 units, giving us basic points (1, -.25), (0, -1.25) and (1, -.25).

etc.

Problem Number 4

Make tables and sketch graphs of the power functions y = x 2 , y = x 3 , y = x -2 and y = x -3 , for x = -3 to 3.

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Problem Number 5

If water depths of 63.7, 56.3, 52.2 and 51.3 cm are observed at clock times 14.6, 21.9, 29.2 and 36.5 sec, then at what average rate does the depth change during each time interval?

This is as before.

Sketch a graph of this data set and use a sketch to explain why the slope of this graph between 21.9 and 29.2 sec represents the average rate at which depth changes during this time interval.

Here you need to sketch the graph, explain what the rise between the two points means, what the run means, and what the slope therefore means.

 If f(x) = x2, give the vertex and the three basic points of the graphs of f(x--.25), f(x) - 1.85, 2 f(x) and 2 f(x--.25) + 1.85. Quickly sketch each graph.

Completely analogous to the above.