040217 Practice Major Quiz

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Practice Major Quiz

Problem Number 1

Problem:  Obtain a quadratic depth vs. clock time model if depths of 74.44157 cm, 67.70271 cm and 63.78343 cm are observed at clock times t = 6.855136, 13.71027 and 20.56541 seconds.

Problem: The quadratic depth vs. clock time model corresponding to depths of 74.44157 cm, 67.70271 cm and 63.78343 cm at clock times t = 6.855136, 13.71027 and 20.56541 seconds is

depth(t) = .03 t2 + -1.6 t + 84.

Use the model to determine the clock time at which depth is 74.18898 cm.

Problem Number 2

Problem:  Sketch a graph of the basic exponential function y = 2 x. If this function is stretched vertically by factor -1.41 then shifted 2.46 units vertically, what is the algebraic form of the function obtained?  Sketch a graph of this new function, and show that the graph is different than that obtained if the vertical shift is performed before the vertical stretch.

Problem:  Water depths of 67.536, 50.544, 39.024 and 32.976 cm are observed at clock times t = 12, 24, 36 and 48 seconds.  What is the average rate of depth change during each of the three time intervals? Predict what the next average rate would be, and use this result to predict the depth at t = 60 seconds.

Problem Number 3

Give the full statement of the Quadratic Formula.

Problem Number 4

Sketch a graph of y = x^2, from x = -3 to x = 3. Then sketch a graph of y = .25 x^2 over the same domain.

Sketch the second graph shifted 1.75 units in the x direction and 1.75 units in the y direction.

What are the three basic points of this graph?

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

Problem Number 5

At clock times 49.4, 74.10001, 98.8 and 123.5 sec, we observe water depths of -57, -93.2, -117.2 and -129 cm.

Sketch a graph of depth vs. clock time.

If f(x) = x2, what are the vertex and the three basic points of the graphs of f(x- 1.25), f(x) - .85, 5 f(x) and 5 f(x- 1.25) + .85. Quickly sketch each graph.

Problem Number 1

Problem:  Obtain a quadratic depth vs. clock time model if depths of 74.44157 cm, 67.70271 cm and 63.78343 cm are observed at clock times t = 6.855136, 13.71027 and 20.56541 seconds.

This means plug the coordinates into the form y = a t^2 + b t + c and get three simultaneous equations in a, b and c.  Then solve the equations to get a, b and c.   Finally plug a, b and c into the form to get your function.

The function will look something like y = .02 t^2 - 3 t + 80, but with different numbers.

Problem: The quadratic depth vs. clock time model corresponding to depths of 74.44157 cm, 67.70271 cm and 63.78343 cm at clock times t = 6.855136, 13.71027 and 20.56541 seconds is

depth(t) = .03 t2 + -1.6 t + 84.

Use the model to determine the clock time at which depth is 74.18898 cm.

To get the clock time at which depth is 74.2 cm (you can round off to 3 significant figures) reason as follows:

Problem Number 2

Problem:  Sketch a graph of the basic exponential function y = 2 x. If this function is stretched vertically by factor -1.41 then shifted 2.46 units vertically, what is the algebraic form of the function obtained?  Sketch a graph of this new function, and show that the graph is different than that obtained if the vertical shift is performed before the vertical stretch.

The graph is easily sketched using its basic points (0, 1) and (1, 2) and knowing the basic shape of the exponential function (asymptotic to the negative x axis, increasing at an increasing rate, etc.).

The vertical stretch takes the basic points to (0, -1.41) and (1, -2.82).  The asymptote will still be the negative x axis.  y values have been multiplied by -1.41 so the function is now y = -1.41 * 2^x.

Vertically shifting 2.46 units moves every point 2.46 units in the vertical direction.   The basic points become (0, -1.41 + 2.46) = (0, 1.05) and (1, -2.82 + 2.46) = (1, -.36).  The horizontal asymptote rises from the negative x axis (y = 0) to the horizontal line y = 2.46.  The equation of the function is now y = -1.41 * 2^x + 2.46.

If the vertical shift is performed first

The equation of the function at this point will be y = 2^x + 2.46.

All y values will have been multiplied by -1.41 so the equation is now y = -1.41 * (2^x + 2.46) = -1.41 * 2^x - 3.47.

Problem:  Water depths of 67.536, 50.544, 39.024 and 32.976 cm are observed at clock times t = 12, 24, 36 and 48 seconds.  What is the average rate of depth change during each of the three time intervals? Predict what the next average rate would be, and use this result to predict the depth at t = 60 seconds.

The key word here is 'each'.  We need to calculate the average rate over each of the three intervals, meaning we will get three average rates.

To get the average rate over an interval we find the change in depth and the change in clock time over the interval and use these values to calculate average rate = change in depth / change in clock time.

The average rate will change from one interval to the next.   Writing down the three average rates we can then identify a pattern.  For example if the average rates change by the same amount as we move from the first interval to the second, then from the second to the third, we will predict that the average rate will change by the same amount as we move to the next interval and we can use this pattern to predict the next average rate.

Having predicted the next average rate of depth change over the next interval we can predict the change in the depth over that interval.  Since ave rate = change in depth / change in clock time, it follows that change in depth = average rate * change in clock time.  The change in clock time for the new interval is again 12 seconds so it will be easy to make the prediction.

 

clock time (sec) depth (cm) rise = change in depth slope = change in depth / change in clock time change in slope

12

67.5

24

50.5

-17

-1.417

36

39

-11.5

-0.958

0.458

48

33

-6

-0.5

0.458

60

? ? ? ?

The next slope will be .458 greater than the previous, which makes the next slope -0.5 + .458 = -.042.

The next rise will therefore be

The next depth (the t = 60 sec depth) will therefore be

Problem Number 3

Give the full statement of the Quadratic Formula.

Use the full statement, including the conditions and the 'if and only if' logic.

Problem Number 4

Sketch a graph of y = x^2, from x = -3 to x = 3. Then sketch a graph of y = .25 x^2 over the same domain.

We vertically stretch the y = x^2 graph by factor .25 to get the graph of y = .25 x^2.  The basic points (-1, 1), (0, 0) and (1, 1) become (-1, .25), (0, 0) and (1, .25).

Sketch the second graph shifted 1.75 units in the x direction and 1.75 units in the y direction.

What are the three basic points of this graph?

The three basic points (-1, .25), (0, 0) and (1, .25) will

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

We get

Problem Number 5

At clock times 49.4, 74.10001, 98.8 and 123.5 sec, we observe water depths of -57, -93.2, -117.2 and -129 cm.

This is done in the same manner as #2.

Sketch a graph of depth vs. clock time.

This requires a sketch indicating the graph points (49.4 sec , -57 cm) and (74.1 sec, -93.2 cm).  The rise and run from one point to the other should also be indicated.   The explanation goes something like this:

If f(x) = x2, what are the vertex and the three basic points of the graphs of f(x- 1.25), f(x) - .85, 5 f(x) and 5 f(x- 1.25) + .85. Quickly sketch each graph.

The three basic points of the y = x^2 graph are (-1, 1), (0, 0) and (1, 1).

The graph is then vertically stretched by factor 5, transforming the basic points (.25, 1), (1.25, 0) and (2.25, 1) to (.25, 5), (1.25, 0) and (2.25, 5).

The graph is finally shifted -.85 units in the x direction, transforming the basic points (.25, 5), (1.25, 0) and (2.25, 5) to (.25, 4.15), (1.25, -.85) and (2.25, 4.15).

The function y = x^2 transforms to y = (x-1.75)^2, then to y = 5 ( x-1.75)^2 and finally to y = 5 ( x - 1.75)^2 - .85.