# Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 5

The equation of the tangent line $l$ at point $A(2,-4)$ is $$(l):y=-8x+12$$

#### Work Step by Step

$$y=f(x)=4x-3x^2$$ Given point $A(2,-4)$ According to definition, the slope of the tangent line $l$ at point $A(2, -4)$ is $$m_l=\lim\limits_{x\to2}\frac{f(x)-f(2)}{x-2}$$$$m_l=\lim\limits_{x\to2}\frac{(4x-3x^2)-(-4)}{x-2}$$$$m_l=\lim\limits_{x\to2}\frac{-3x^2+4x+4}{x-2}$$$$m_l=\lim\limits_{x\to2}\frac{-3(x-2)(x+2/3)}{x-2}$$$$m_l=\lim\limits_{x\to2}[-3(x+\frac{2}{3})]$$$$m_l=-3\times(2+\frac{2}{3})=-8$$ Therefore, the equation of the tangent line $l$ would have the following form: $$(l): y=-8x+b$$ Since $l$ passes through point $A(2,-4)$, we have $$-8\times2+b=-4$$$$-16+b=-4$$$$b=12$$ In conclusion, the equation of the tangent line $l$ at point $A(2,-4)$ is $$(l):y=-8x+12$$

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