# Chapter 2 - Differentiation - 2.3 Exercises: 64

Tangent Line: $y=3x-8$

#### Work Step by Step

$f(x)=(x-2)(x^{2}+4), (1,-5)$ a. Use the Product Rule to Find f'(x): $f'(x)=f_{1}(x)f_{2}'(x)+f_{2}(x)f_{1}'(x)$ $f'(x)=(x-2)(2x)+(x^{2}+4)(1)$ $f'(x)=(2x^{2}-4x)+(x^{2}+4)$ Plug in Given x Value to Find Slope: $f'(1)=(2[1]^{2}-4[1])+([1]^{2}+4)=(2-4)+(1+4)=3$ Plug Found Slope and Given Point Values into Point Slope Form: $y-y_{1}=(slope)(x-x_{1})$ $y-[-5]=3(x-[1])$ $y+5=3x-3$ Equation of Tangent Line: $y=3x-8$ b. Can be found using a graphing calculator c. This answer was confirmed using a graphing calculator

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