So we know that f(x)=3-5x, and we need to find the slope of the tangent line at (-1,8).

Two Ways to Solve this Problem:

1. We can derive this function:

f'(x)=0-5

and then plug in the point into this equation we found to give us what the slope of the tangent line would be:

f'(-1)=0-5=-5

So the slope of the tangent line would be -5

OR

2. We know that this function is just a straight line since we would have f(x)=-5x+3 written in Slope-Intercept form (y=mx+b).

So the slope (m) of this line is -5

We also know that the a tangent line of a line is just a coincident line to the original. Meaning that it is the same line.

Therefore, we know that the slope of the tangent line has to be equal to the slope of the original line:

m=-5

The given function is:-

f(x) = 3 - 5x

differentiating both sides w.r.t 'x' we get

f'(x) = 0 - 5

Now, slope of the tangent at the point (-1,8) = f'(-1) = -5