Answer
(a) (i) 2 (ii) 1.111111 (iii) 1.010101 (iv) 1.001001 (v) 0.666667 (vi) 0.909091 (vii) 0.990099 (viii) 0.999001
(b) 1
(c) y = x - 3
Work Step by Step
(a) Slope of secant line = $\frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}$. We can use a calculator to evaluate the x- and y-coordinates of Q for each value of x and plug in the values into a calculator to find the slope.
(i) 2
(ii) 1.111111
(iii) 1.010101
(iv) 1.001001
(v) 0.666667
(vi) 0.909091
(vii) 0.990099
(viii) 0.999001
(b) It appears that the slope approaches 1 as the x-coordinate of Q approaches 1.
(c) The slope at P is 1. Using point-slope form, we obtain
$y + 1 = 1(x-2)$
$y = x - 3$
So, $y = x - 3$ is the equation of the tangent line to the curve at P.
