Chapter 3.3, Problem 40E

To determine

To calculate: The x and y -intercepts of the line that is tangent to the curve at the point $\left(-2,-8\right)$ .

Answer to Problem 40E

The required x -intercept is $-\frac{4}{3}$ and the y -intercept is 16.

Explanation of Solution

Given information:

The curve: $y=x^3$

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Point-slope form of a line:

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The given curve is $y=x^3$ and the given point is $\left(-2,-8\right)$ .

Differentiate the given curve.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^3\right)\\ =3x^{\left(3-1\right)}\\ =3x^2\end{array}$

Substitute $-2$ for x in the above derivative.

  $\begin{array}{c}\frac{dy}{dx}=3{\left(-2\right)}^2\\ =3\left(4\right)\\ =12\end{array}$

Thus, the slope of the line is 12.

Now, substitute $-2$ for $x_1$ , $-8$ for $y_1$ and 12 for m in the point-slope form of the equation of the line that is, $y-y_1=m\left(x-x_1\right)$ .

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

To find the x -intercept, substitute 0 for y in the above equation.

  $\begin{array}{c}0=12x+16\\ \frac{\cancel{12}x}{\cancel{12}}=-\frac{16}{12}\\ x=-\frac{4}{3}\end{array}$

To find the y -intercept, substitute 0 for x in the above equation.

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

Hence, the required x -intercept is $-\frac{4}{3}$ and the y -intercept is 16.