Question

(c) find f(0).

1. (8 points) find an equation of the line that is tangent to the graph of y = (x + 1)^3 at x = -2. use exact values in your answer.

Explanation:

Step1: Find the derivative of $y=(x + 1)^3$

Use the chain - rule. If $u=x + 1$, 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 $\frac{du}{dx}=1$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=3(x + 1)^2$.

Step2: Find the slope of the tangent line at $x=-2$

Substitute $x=-2$ into the derivative. When $x=-2$, $m = 3(-2 + 1)^2=3(-1)^2=3$.

Step3: Find the y - coordinate of the point of tangency

Substitute $x=-2$ into the original function $y=(x + 1)^3$. When $x=-2$, $y=(-2 + 1)^3=-1$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is the point of tangency and $m$ is the slope. Here, $x_1=-2$, $y_1=-1$, and $m = 3$. So $y+1=3(x + 2)$.

Step5: Simplify the equation

Expand the right - hand side: $y+1=3x+6$. Then subtract 1 from both sides to get $y = 3x+5$.

Answer:

$y = 3x + 5$