Find the slope of a tangent line?

Question

f(x)=(-5x^2-5x-2)(-4x-5), x=0

Answer 1

f(x) = 20x³ + 25x² + 20x² + 25x + 8x + 10
f(x) = 20x³ + 45x² + 33x + 10
f `(x) = 60x² + 90x + 33
f `(0) = 33 is slope of tangent line at x = 0

Answer 2

First get the derivative function f'(x).
f'(x) =(-5x^2 - 5x - 2)(-4) + (-4x-5)(-10x - 5)
Then plug in x=0 for all the x's in f'(x) and get:
f'(0) = (-2)(-4) + (-5)(-5)
f'(0) = 8 + 25
f'(0) = 33

Answer 3

given f(x)=(-5x^2-5x-2)(-4x-5),
the slope is its derivative (calculus I), given by:
f(x)=(-5x^2-5x-2)(-4x-5)=(5x^2+5x+2)(4x+5)
= 20x^3 + 20x^2 + 8x + 25x^2 + 25x + 10
= 20x^3 + 45x^2 + 33x + 10
slope = f'(x) = d/dx [f(x0] = 60x^2 + 50x +25
slope at x=0 is 25

Ans =25