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The points (x, y) and (y, x) are mirror images of each other across the line y = x. Therefore, by interchanging x and y in the equation y = ax² + bx + c, we obtain the equation of its mirror image across the line y = x; namely, x = = y? + 2, ay² + by + c. Thus to graph x we can first graph y = x² + 2 and then reflect it across the line y = x, as indicated in Figure 8.22. Graph each of the following parabolas. x = y? (c) x = y² – 1 (е) х (а) х %— (b) х — — у? = -y? (d) x = -) -y² + 3 Зу? -2) (g) x = y + 4y + 7 (f) х — (h) x = y² – 2y – 3 y4(1, 3). y =x² + 2 -(-1, 3) (3, 1) -(0, 2). (3, – 1)- (2, 0)HH| |x = y² +2 Figure 8.22

The points (x, y) and (y, x) are mirror images of each other across the line y = x. Therefore, by interchanging x and y in the equation y = ax² + bx + c, we obtain the equation of its mirror image across the line y = x; namely, x = = y? + 2, ay² + by + c. Thus to graph x we can first graph y = x² + 2 and then reflect it across the line y = x, as indicated in Figure 8.22. Graph each of the following parabolas. x = y? (c) x = y² – 1 (е) х (а) х %— (b) х — — у? = -y? (d) x = -) -y² + 3 Зу? -2) (g) x = y + 4y + 7 (f) х — (h) x = y² – 2y – 3 y4(1, 3). y =x² + 2 -(-1, 3) (3, 1) -(0, 2). (3, – 1)- (2, 0)HH| |x = y² +2 Figure 8.22

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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The points (x, y) and (y, x) are mirror images of each other across the line y = x. Therefore, by interchanging x and y in the equation y = ax² + bx + c, we obtain the equation of its mirror image across the line y = x; namely, x = = y? + 2, ay² + by + c. Thus to graph x we can first graph y = x² + 2 and then reflect it across the line y = x, as indicated in Figure 8.22. Graph each of the following parabolas. x = y? (c) x = y² – 1 (е) х (а) х %— (b) х — — у? = -y? (d) x = -) -y² + 3 Зу? -2) (g) x = y + 4y + 7 (f) х — (h) x = y² – 2y – 3 y4(1, 3). y =x² + 2 -(-1, 3) (3, 1) -(0, 2). (3, – 1)- (2, 0)HH| |x = y² +2 Figure 8.22
Transcribed Image Text:The points (x, y) and (y, x) are mirror images of each other across the line y = x. Therefore, by interchanging x and y in the equation y = ax² + bx + c, we obtain the equation of its mirror image across the line y = x; namely, x = = y? + 2, ay² + by + c. Thus to graph x we can first graph y = x² + 2 and then reflect it across the line y = x, as indicated in Figure 8.22. Graph each of the following parabolas. x = y? (c) x = y² – 1 (е) х (а) х %— (b) х — — у? = -y? (d) x = -) -y² + 3 Зу? -2) (g) x = y + 4y + 7 (f) х — (h) x = y² – 2y – 3 y4(1, 3). y =x² + 2 -(-1, 3) (3, 1) -(0, 2). (3, – 1)- (2, 0)HH| |x = y² +2 Figure 8.22
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