Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Test - Page 412: 14

Answer

$x = 4t-3$ $y = t^2$ $t$ in $[-3,4]$ We can see the graph below:
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Work Step by Step

$x = 4t-3$ $y = t^2$ $t$ in $[-3,4]$ When $t = -3$: $x = 4(-3)-3 = -15$ $y = (-3)^2 = 9$ When $t = -2$: $x = 4(-2)-3 = -11$ $y = (-2)^2 = 4$ When $t = -1$: $x = 4(-1)-3 = -7$ $y = (-1)^2 = 1$ When $t = 0$: $x = 4(0)-3 = -3$ $y = (0)^2 = 0$ When $t = 1$: $x = 4(1)-3 = 1$ $y = (1)^2 = 1$ When $t = 2$: $x = 4(2)-3 = 5$ $y = (2)^2 = 4$ When $t = 3$: $x = 4(3)-3 = 9$ $y = (3)^2 = 9$ When $t = 4$: $x = 4(4)-3 = 13$ $y = (4)^2 = 16$ We can see the graph below:
Small 1530868202
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