Chapter 2.3, Problem 40E

To determine

To calculate: The $x$ intercept and $y$ intercept of the equation of tangent.

Answer to Problem 40E

The $x$ intercept and $y$ intercept are $-8$ and $16$ respectively.

Explanation of Solution

Given information:

The equation of the curve is $y=x^3$ , $\left(x_1,y_1\right)=\left(-2,-8\right)$

Concept used:

The formulae used are $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

Calculation:

Slope of tangent line.

  $\begin{array}{c}m=\frac{dy}{dx}\\ =\frac{d}{dx}\left(x^3\right)\\ =3x^2\end{array}$

Solve further by substituting $-2$ for $x$ in $m=3x^2$

  $\begin{array}{c}m=3x^2\\ =3\times {\left(-2\right)}^2\\ =3\times 4\\ =12\end{array}$

The slope of tangent line is $12$

Substitute the values in the standard equation of line $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{l}y-\left(-8\right)=12\left(x-\left(-2\right)\right)\\ y=-8+12\left(x+2\right)\end{array}$

The equation of tangent line is $y=-8+12\left(x+2\right)$.

Substitute $0$ for $x$ to find the $y$ intercept.

  $\begin{array}{c}y=-8+12\left(0+2\right)\\ y=-8+12\left(2\right)\\ =-8+24\\ =16\end{array}$

Substitute $0$ for $y$ to find the $x$ intercept.

  $\begin{array}{c}0=-8+12\left(x+2\right)\\ 8=12\left(x+2\right)\\ x+2=\frac{8}{12}\\ x=\frac{8}{12}-2\end{array}$

Solve further.

  $\begin{array}{c}x=\frac{8-24}{2}\\ =-\frac{16}{2}\\ =-8\end{array}$

Conclusion: The $x$ intercept is $-8$ and $y$ intercept is $16$