Let f(x) = 6x/x-1. Find the slope of secant lines on the graph

To find the slope, we have this formula: $\frac{y_2 - y_1}{x_2 - x_1}$.
The given points are P(3, 9) and Q(x, $\frac{6x}{x - 1}$)

So, slope = $\frac{\frac{6x}{x \, - \, 1} - 9}{x \, - \, 3}$ = $\frac{6x \, - \, 9(x \, - \, 1)}{(x \, - \, 1)(x \, - \, 3)}$ = $\frac{9 \, - \, 3x}{(x \, - \, 1)(x \, - \, 3)}$ = $\frac{-3}{x \, - \, 1}$

a) x = 4
slope = $\frac{-3}{3}$ = 1

b) x = 3.1
slope = $\frac{-3}{2.1}$ (Leave it in fraction because it is stated in the question that answers must be exact).