Find the equation(s) of the tangent line(s) to the graph of the curve y = x3 − 8x through the point (1, −8) not on the graph. (Enter your answers as a comma-separated list of equations.)

Answer 1

Let (x,x³-8x) represent a point on the given function. The slope of the line passing through that point and (1,-8), a point not on the curve, is then Δy=x³-8x-(-8) = x³-8x+8 over Δx=x-1. We want to determine when that expression is equal to the derivative of f(x), i.e. f'(x) = 3x²-8.

Solve this equation for x. There will be two values of x:

(x³-8x+8)/(x-1) = 3x²-8.

Once you have those x coordinates, determine the corresponding y coordinates and the slope of the tangent line(s), then use point-slope (or slope-intercept) to write the equations of the two tangent lines.

Check your work here:

desmos.com/calculator/ofesdn8ple

Answer 2

f(x) = x³ - 8x

Determine f'(1) this is slope of tangent line.

Use point-slope form (Algebra 1) to determine equation of tangent line.