C 5 Video Example <> 1 EXAMPLE 1 Find an equation of the tangent line to the function y = 5x2 at the point P(1, 5). SOLUTION We will be able to find an equation of the tangent line t as soon as we know its slope m. The difficulty is that we know only one point, P, on t, whereas we need two points to compute the slope. But observe that we can compute an approximation to m by choosing a nearby point Q(x, 5x2) on the graph (as in the figure) and computing the slope mpo of the secant line PQ. [A secant line, from the Latin word secans, meaning cutting, is a line that cuts (intersects) a curve more than once.] We choose x # 1 so that Q # P. Then, 5x2 - 5 X-1 For instance, for the point Q(1.5, 11.25) we have ✓ 6.25 11.25 1.5 MpQ = -5 - 1 .5 The tables below show the values of mpo for several values of x close to 1. The closer Q is to P, the closer x is to 1 and, it appears from the tables, the closer mpo is to 1.5 X. This suggests that the slope of the tangent line t should be m= 12.5 X. MpQ = X MpQ 0 5 mpQ 2 15 1.5 12.5 1.1 10.5 1.01 10.050 .99 9.950 .5 7.5 .9 9.500 1.001 10.005 .999 9.995 We say that the slope of the tangent line is the limit of the slopes of the secant lines, and we express this symbolically by writing lim_ mpQ = m and Q-P X = 12.5 lim 5x² - 5 = 10 x-1 X-1 . Assuming that this is indeed the slope of the tangent line, we use the point-slope form of the equation of a line (see Appendix B) to write the equation of the tangent line through (1, 5) as y- 5 = 12.5 x (x-1) or y = 10 ✔x-1 x The graphs below illustrate the limiting process that occurs in this example. As Q approaches P along the graph, the corresponding secant lines rotate about P and approach the tangent line t.