## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 33

#### Answer

The equation of the tangent line to the curve at $(1,3)$ is $$(l):y=4x-1$$

#### Work Step by Step

$$y=2x^3-x^2+2$$ 1) First, find the derivative of $y$ $$y'=\frac{d}{dx}(2x^3)-\frac{d}{dx}(x^2)+\frac{d}{dx}(2)$$ $$y'=2\times3x^2-2x+0$$ $$y'=6x^2-2x$$ 2) Find the slope of the tangent line to the curve at the point $(1,3)$, or in fact, $y'(1)$: $$y'(1)=6\times1^2-2\times1=4$$ 3) The tangent line to the curve at the point $(1,3)$ would have the following equation: $$(l):y=y'(1)x+b$$ $$(l):y=4x+b$$ Since the point $(1,3)$ also lies in the tangent line $(l)$, we can apply that point to the equation of $(l)$ to find $b$. In detail, $$4\times1+b=3$$$$4+b=3$$$$b=-1$$ Therefore, the equation of the tangent line to the curve at $(1,3)$ is $$(l):y=4x-1$$

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