in exercises 59 - 64, find the equations of the tangent line and the normal line to the graph of the…

Explanation:

Step1: Find the derivative of the function

The derivative of $y=(x + 3)^3$ using the chain - rule. Let $u=x + 3$, then $y = u^3$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=3u^2$ and the derivative of $u$ with respect to $x$ is $\frac{du}{dx}=1$. So, by the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=3(x + 3)^2$.

Step2: Evaluate the slope of the tangent line at the given point

Substitute $x=-2$ into the derivative. $\frac{dy}{dx}\big|_{x = - 2}=3(-2 + 3)^2=3(1)^2 = 3$. The slope of the tangent line $m_t=3$.

Step3: Find the slope of the normal line

The slope of the normal line $m_n$ is the negative reciprocal of the slope of the tangent line. So, $m_n=-\frac{1}{3}$.

Step4: Use the point - slope form to find the equation of the tangent line

The point - slope form is $y - y_1=m(x - x_1)$. Using the point $(-2,1)$ and $m_t = 3$, we have $y-1=3(x + 2)$. Expand to get $y-1=3x+6$, or $y = 3x+7$.

Step5: Use the point - slope form to find the equation of the normal line

Using the point $(-2,1)$ and $m_n=-\frac{1}{3}$, we have $y - 1=-\frac{1}{3}(x + 2)$. Expand to get $y-1=-\frac{1}{3}x-\frac{2}{3}$, or $y=-\frac{1}{3}x+\frac{1}{3}$.

Answer:

Tangent line: $y = 3x+7$ Normal line: $y=-\frac{1}{3}x+\frac{1}{3}$