2. Find the equation of the tangent line to the curve. $$ y=x^{3}-2 x \quad \text { at the point when } x=-1 $$

Question

UpStudy AI · Accepted Answer

To find the equation of the tangent line to the curve $$ y = x^3 - 2x $$ at the point where $$ x = -1 $$, follow these steps:

1. **Find the $$ y $$-coordinate at $$ x = -1 $$:**

$$
   y = (-1)^3 - 2(-1) = -1 + 2 = 1
   $$
   
   So, the point of tangency is $$ (-1, 1) $$.

2. **Find the derivative $$ \frac{dy}{dx} $$ to determine the slope of the tangent line:**

$$
   \frac{dy}{dx} = \frac{d}{dx}(x^3 - 2x) = 3x^2 - 2
   $$
   
   Evaluate the derivative at $$ x = -1 $$:

$$
   \left. \frac{dy}{dx} \right|_{x = -1} = 3(-1)^2 - 2 = 3(1) - 2 = 1
   $$
   
   So, the slope $$ m $$ of the tangent line is $$ 1 $$.

3. **Use the point-slope form of the equation of a line:**

$$
   y - y_1 = m(x - x_1)
   $$
   
   Plugging in the values:

$$
   y - 1 = 1(x - (-1)) \
   y - 1 = x + 1 \
   y = x + 2
   $$

**Final Answer:**
$$
y = x + 2
$$
The equation of the tangent line is $$ y = x + 2 $$.

2. Find the equation of the tangent line to the curve. \[ y=x^{3}-2 x \quad \text { at the point when } x=-1 \]