Find the slope of the tangent line to the curve y = (x^2 -8)^6 at x = 3, and

write the equation of the tangent line.

Take the derivative.

y'=6(x^2-8)^5 * 2x

at x=3, put that in, and compute y'

equation: y=y' x + b
you have y'. you know x, and can compute y (original equation), so can compute b from that equation.

To find the slope of the tangent line to the curve y = (x^2 - 8)^6 at x = 3, we need to take the derivative of the function and evaluate it at x = 3.

1. Start by finding the derivative of the function:
- Use the chain rule: dy/dx = 6(x^2 - 8)^5 * 2x
- Simplify: dy/dx = 12x(x^2 - 8)^5

2. Evaluate the derivative at x = 3:
- Substituting x = 3 into the derivative gives us: dy/dx = 12(3)(3^2 - 8)^5 = 12(3)(9 - 8)^5 = 12(3)(1)^5 = 36

The slope of the tangent line to the curve y = (x^2 - 8)^6 at x = 3 is 36.

To write the equation of the tangent line, we'll use the point-slope form:
y - y1 = m(x - x1)

3. Substitute the values into the equation:
- x1 = 3 (the x-coordinate of the given point)
- y1 = (3^2 - 8)^6 = (-1)^6 = 1 (the y-coordinate of the given point)
- m = 36 (the slope we found earlier)

The equation of the tangent line is:
y - 1 = 36(x - 3)

Simplifying further, we get:
y = 36x - 107