Chapter 3.1, Problem 67E

To determine

To find:

The equation of the tangent and parallel to the line

Answer to Problem 67E

The equation of the line is $m=-\frac{2}{3}$

Explanation of Solution

Given:

The tangent to the curve

  $y=a{x}^3+b{x}^2+cx+d$

Concept used:

The equation is in slope −intercept form, $y=mx+c$

An equation for the line through the point $\left(x_1,y_1\right)$ with slope m is

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The function

  $y=a{x}^3+b{x}^2+cx+d\mathrm{.......................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}=m=\frac{x_2-x_1}{y_2-y_1}\end{array}$

Differentiating the equation (1) with respect to $x$

  $\begin{array}{l}y=a{x}^3+b{x}^2+cx+d\\ y^{\prime }=3a{x}^2+2bx+c\mathrm{........................}\left(2\right)\end{array}$

The derivative is slope of the tangent line so in order to the slope of the tangent line

The derivative of constant is zero

  $\begin{array}{l}m=\frac{x_2-x_1}{y_2-y_1}\\ m=\frac{-2-2}{6-0}\\ m=\frac{-4}{6}\\ m=-\frac{2}{3}\mathrm{..............................}\left(3\right)\end{array}$