Verified answer

f(x) = 9x² - x³

Get the first derivative of the polynomial. For each term, the new coefficient is the product of the old coefficient and old exponent and the new exponent is the old exponent minus 1. Do that for all terms:

f'(x) = 18x - 3x²

Now solve for f'(1) by substituting 1 for x and simplifying:

f'(1) = 18(1) - 3(1)²

f'(1) = 18(1) - 3(1)

f'(1) = 18 - 3

f'(1) = 15

The slope of the tangent line at the point (1, 8) is 15.

We have the slope, an x, and a y, so we can solve for the intercept:

y = mx + b

8 = 15(1) + b

8 = 15 + b

-7 = b

The equation of the tangent line is:

y = 15x - 7

-

f (x) = 9x² - x³

f ' (x) = 18x - 3x²

f ' (1) = 18 - 3 = 15

y - 8 = 15 ( x - 1 )

y - 8 = 15x - 15

y = 15x - 7______equation of tangent line at (1,8)

Formula

equation of the tangent line to the curve y = f(x) at the point (a,f(a)) is

y - f(a) = f '(a) (x-a)