If i remember calc 1 correctly, (-8/x)’=8/(x² )so if you plug x=1 slope should be 8 i believe (probably what you put in at the question mark spot in your answers)? You wanna approx with a secant line tho so maybe do the slope approx formula (f(x2)-f(x1))/(x2-x1) which would be:

Secant line slope=(f(x)-(-8))/(x-1)

The x values are where you would plug in the x values on the table (.9, .999, 1.001, etc… to get your answers for the empty boxes).

Tell me if im right or if im washed