TRIGONOMETRIC FUNCTIONS Graphing And Analysis (Page 11 of 13)

To decide on an optimal window for the graph, we need the period, vertical shift and phase shift (horizontal shift). You should be able to determine (from the function) that the period is 4, the vertical shift is 3 units downward, and the phase shift is 1 to the left. From what we know about the period, the function will complete one cycle when -90 <= 45x+45 <= 90. Solving for x, this means -3 <= x <= 1.

Set the left edge of the window (Xmin) to -3 and the right edge (Xmax) to 1. A scale that will place four tick marks within one cycle is desirable, so set Xscl to 1. Because of the shift downward by 3 units, the center point of the graph will occur at y = -3. Therefore, set the lower edge (Ymin) to -3-5 and the upper edge (Ymax) to -3+5. A scale of one unit on the y-axis will work nicely, so be sure Yscl is set to 1.

 WINDOW key.

 Negation key. 3 ENTER key. 1 ENTER key. 1 ENTER key. Negation key. 8 ENTER key. 2 ENTER key. 1
 ENTER key.

 Calculator screen image.
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