Find an equation for the tangent line to the given curve at the point where x=x0: y=(5x-1)(4+3x); x0=0.?

F(x) = x^3 - 4x^2 f(a) = a^3 - 4a^2 f'(x) = 3x^2 - 8x f'(a) = 3a^2 - 8a So the equation of the tangent lines is: y - y0 = m(x - x0); m = 3a^2 - 8a, (x0, y0) = (-1, 4) y - 4 = (3a^2 - 8a)(x - (-1)) y = (3a^2 - 8a)(x + 1) + 4 This line must go through the point (a, f(a)) = (a, a^3 - 4a^2), so plug in x = a and y = a^3 - 4a^2: y = (3a^2 - 8a)(x + 1) + 4; x = a, y = a^3 - 4a^2 a^3 - 4a^2 = (3a^2 - 8a)(a + 1) + 4 a^3 - 4a^2 = 3a^3 + 3a^2 - 8a^2 - 8a + 4 a^3 - 4a^2 = 3a^3 - 5a^2 - 8a + 4 -2a^3 + a^2 + 8a - 4 = 0 This factors into: -2a^3 + a^2 + 8a - 4 = 0 -a^2*(2a - 1) + 4(2a - 1) = 0 (-a^2 + 4)(2a - 1) = 0.

Dy / dx = 3x^2 - 8x At x = -1, the slope = -5 y = -5x + b. When x = -1, be = -1. That's the only tangent line I find at (-1, 4).

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