Class 040922

To check your test grades:

• Go to bb.vccs.edu
• log in
• select Supervised Study Current Semester
• click on Tools > Check Grade

At the computer:

Using the program h:\shares\physics\mcg run the program.  When asked for course number enter the word 'quizzes'.

If you get an endless loop, hit ctrl-alt-del and choose Task Manager, highlight the program and close it, then start it again.

Run the program and do as it says.  Type in your VCCS login name, if you know it, and if not just type in your first initial and last name (no punctuation).  Then type in quizzes for your course. 

The sequence of buttons is Next Question, Display Choices, Enter Answer and See Results, or something very much like that.

On paper:

The three points depicted in the figure below are the 'basic points' of the basic quadratic function y = x^2.  If each of the three points are shifted 2 units to the right then 1 unit down, sketch the resulting 3 points and the resulting parabola.

The figure below depicts the three basic points of the basic quadratic function. 

• If  the basic quadratic function is vertically stretched by a factor of 3, what are the coordinates of the resulting three points?  If the function is then vertically shifted -4 units, what are the coordinates of the resulting three points? If the function is finally shifted horizontally -1 unit, what are the coordinates of the resulting three points?

The vertical stretch by factor 3 moves each point 3 times further from the x axis, moving the original points from (-1, 1), (0, 0) and (1, 1) to (-1, 3), (0, 0) and (1, 3).  The vertical stretch multiplies all y values by 3 so the function, which was originally y = x^2, becomes y = 3 x^2.

The vertical shift of -2 units changes all y coordinates by -4, moving the three points (-1, 3), (0, 0) and (1, 3) to (-1, -1), (0, -4) and (1, -1).  The y values have all been changed by -2 so the function changes from y = 3 x^2 to y = 3 x^2 - 4.

The horizontal shift of -1 unit changes all x coordinates by -1, moving the three points(-1, -1), (0, -4) and (1, -1) to (-2, -1), (-1, -4) and (0, -1).  The x values have all been changed by -1 so the function changes from y = 3 x^2 - 4 to y = 3 ( x - (-1) ) ^ 2 - 4, or y = 3 ( x + 1 ) - 4.

You should be able to identify each of these graphs in the figure at right, each of which you should have constructed based on just the behavior of its three basic points.

• Sketch the three points at every stage of the transformation process, and sketch the parabola that results at each stage.
• Label each parabola with the formula of the corresponding quadratic function.

The figure below depicts the three basic points of the basic quadratic function. 

• To what points would these three points be transformed by the function y = 3x^2?  Sketch these points on your graph.
• To what points would these points then be transformed if the function was changed to y = 3 ( x - 2) ^ 2?    Sketch these points on your graph.
• To what points would these points then be transformed if the function was changed to y = 3 ( x - 2) ^ 2 - 4?    Sketch these points on your graph.
• Sketch each of the corresponding parabolas and label with the appropriate function.

The figure below depicts the three basic points of the basic quadratic function. 

• What are the coordinates of the vertex of the parabola defined by y = x ^ 2 + 4 x - 3?

The vertex is at x = - b / (2 a) = -4 / (2 * 1) = -2.  At this point y = (-2)^2 + 4 * (-2) - 3 =-7.

• How far would we have to shift the three basic points of the basic quadratic, shown on the graph below, to obtain the three basic points of the given parabola, given the information that for this particular function the actual shape of the parabola does not change?

The vertex of the y = x^2 parabola is at (0, 0), corresponding the the middle of the three basic points shown below.  To move the vertex to (-2, -7) would require a horizontal shift of 2 units and a vertical shift of -7 units.

• Sketch the resulting basic points and use them as a basic for sketching your parabola. 

The three basic points (-1, 1), (0, 0) and (1, 1) would shift to

• (-1 - 2, 1 - 7) = (-1, -6), 
• (0 - 2, 0 - 7 ) = (-2, -7) and
• (1 - 2, 1 - 7) = (-1, -6).

Since the parabola would not change shape, it parabola would pass through these points, as giving us the parabola shown to the right of the basic points graph.

• What is the function, in the form y = (x - h ) ^ 2 + k, that describes the new parabola?

The function that horizontally shifts the y = x^2 parabola -1 units and vertically shifts it -7 units is y = ( x - (-1) ) ^ 2 - 7 = (x+1)^2 - 7.

The figure below depicts the three basic points of the basic quadratic function. 

• What are the coordinates of the vertex of the parabola defined by y = 2 x ^ 2 + 4 x - 3?
• For this parabola we require before shifting a vertical stretch by factor a = 2.  What are the basic points after this stretch?  Sketch these basic points and the resulting parabola.

The basic points become (-1,2), (0,0) and (1, 2).

• How far would we have to shift the three basic points of this stretched parabola to obtain the three basic points of the given parabola y = 2 x ^ 2 + 4 x - 3?  Sketch the resulting basic points and use them as a basis for sketching a graph of the new parabola.

The vertex is at x = - b / ( 2 a) = -4 / (2 * 2) = -1.  Substituting into the equation of the function we get y = -5.

So we have to shift the original vertex (0, 0) through -1 unit in the horizontal direction and -5 units in the vertical.

Our basic points therefore move from (-1,2), (0,0) and (1, 2) to (-2, -3), (1, -5) and (0, -3).

Using these points we construct the graph depicted below.

• What is the function, in the form y = a (x - h ) ^ 2 + k, that describes the new parabola?

The new function is y = (x - (-1) ) ^ 2 - 5, or y = (x + 1) ^ 2 - 5.

The graph below depicts a function y = f(x).  Sketch on the same graph the graphs of

• y = 2 f(x)
• y = f ( x + 2) and
• y = f(x) + 2

and label each graph accordingly.

y = 2 f(x) multiplies all y values by 2, vertically stretching the graph by factor 2, i.e., moving all points twice as far from the x axis.  Points on the x axis will therefore not be moved ( 2 * 0 = 0, after all), so the graph will pass through the x axis at the same points as the original.  The peak and the valley of the graph will end up directly above and below the peak and valley of the original but twice as far from the x axis.  You should be able to easily pick this graph out from those shown below.

y = f ( x + 2 ) will have the same y values as the original, but for x values which are 2 less than for the original.  This shifts the graph 2 units to the left.  The points where the graph intersects the x axis, and the peak and the valley, will all move 2 units to the left, as will all other points.  You should be able to easily pick this graph out from those shown below.

y = f(x) + 2 adds 2 to all the y values of the original.  The points where the graph intersects the x axis, and the peak and the valley, will all move 2 units to up, as will all other points.  You should be able to easily pick this graph out from those shown below.