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Unformatted text preview: , we seek the y -intercept. Notice that x = 0 falls in the domain x < 1. Thus f (0) = 4 − 02 = 4 yields the y -intercept (0, 4). As far as symmetry is concerned, you can check that the equation y = 4 − x2 is symmetric about the y -axis; unfortunately, this equation (and its symmetry) is valid only for x < 1. You can also verify y = x − 3 possesses none of the symmetries discussed in the Section 1.3. When plotting additional points, it is important to keep in mind the restrictions on x for each piece of the function. The sticking point for this function is x = 1, since this is where the equations change. When x = 1, we use the formula f (x) = x − 3, so the point on the graph (1, f (1)) is (1, −2). However, for all values less than 1, we use the formula f (x) = 4 − x2 . As we have discussed earlier in Section 1.2, there is no real number which immediately precedes x = 1 on the number line. Thus for the values x = 0.9, x = 0.99, x = 0.999, and so on, we find the corresponding y values usin...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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