Question 1123515
Find the tangent line equation for the function f(x)=(2x+1)(x^2-3) at the point x=0.
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{{{f(x)=(2x+1)(x^2-3) = 2x^3 + x^2 - 6x - 3}}}
f(0) = -3 --> the point (0,-3)
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f'(x) = 6x^2 + 2x - 6
f'(0) = -6
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m = -6 thru (0,-3)