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Answers To The Exercises: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14
Lesson 9
 


9.1.
d).

9.2.
You can calculate the value sin(/6) = ½. Because /6 < ½ , the point (/6, ½) lies above the line y = x.

9.3.
It is easy to see that sin(/2 – ) = sin(/2 +) since its left and right parts are equal to cos. The points (/2 –) and (/2 +) are symmetric on the number line over the point /2. The identity says that the values of sin x at that points are equal. It means that the graph of sine is symmetric over the line x = /2.

9.4.
In problems 1) – 5) take the graph of the parabola y = x2 and make the following transformations:

1) Shift three units up;
2) Shift three units down;
3) Shift three units to the left;
4) Shift three units to the right;
5) Stretch three times along the y-axis.
In problems 6) and 7), take the graph of the function y = sin x and make the following transformations:
6) Shift two units to the right;
7) Shift two units down;

9.5.
The graph has the shape of the sine with the amplitude of 2 (y runs from – 2 to 2), has the period of 2/3 2.1, starts at the point (0, 2sin1) (0, 1.7) and ends at the point (10, 2sin31) (10, – 0.8).

 

 

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