Find the slope of the curve at the given point P and an equation of the tangent line at P.?

y = x³ - 9x ← this is a curve, i.e. a function

y(1) = 1 - 9 = - 8 → the representative curve of the function passes through the point P (1 ; - 8)

y' = 3x² - 9 ← this is the derivative

…but the derivative is too the slope of the tangent line to the curve at x

y'(1) = 3 - 9 = - 6 ← this is the slope of the tangent line to the curve at P (1 ; - 8)

The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept

The slope of the tangent line to the curve is - 6, so the equation of this tangent line is: y = - 6x + y₀

The tangent line passes through P (1 ; - 8), so the coordinates of this point must verify the equation of the tangent line.

y = - 6x + y₀

y₀ = y + 6x → you substitute x and y by the coordinates of the point P (1 ; - 8)

y₀ = - 8 + 6

y₀ = - 2

The equation of the tangent line to the curve at P (1 ; - 8) is: y = - 6x - 2

f'(x) => [f(x + h) - f(x)]/h

f(x + h) = (x + h)³ => x³ + 3x²h + 3xh² + h³ - 9(x + h)

Hence, f(x + h) - f(x) = x³ + 3x²h + 3xh² + h³ - 9x - 9h - x³ + 9x

i.e. 3x²h + 3xh² + h³ - 9h 

Then, [f(x + h) - f(x)]/h => (3x²h + 3xh² + h³ - 9h)/h

Hence, 3x² + 3xh + h² - 9

As h --> 0, 3x² + 3xh + h² - 9 --> 3x² - 9

So, gradient function is 3x² - 9

Then, at x = 1 we have:

f '(1) = -6

At the point (1, -8) we have:

y - (-8) = -6(x - 1)

or, y + 8 = -6x + 6

i.e. y = -6x - 2

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