To find the equation of the tangent line to the function $f(x) = x^2 + 2$ at the point where $x = -1$, follow these steps:

1. Find the derivative $f'(x)$

The derivative of $f(x)$ gives the slope of the tangent line at any point $x$.

$f(x) = x^2 + 2$ $f'(x) = \frac{d}{dx}(x^2 + 2) = 2x$

2. Evaluate the derivative at $x = -1$

Substitute $x = -1$ into the derivative to find the slope of the tangent line at that point.

$f'(-1) = 2(-1) = -2$

So, the slope of the tangent line at $x = -1$ is $-2$.

3. Find the coordinates of the point on the curve

Substitute $x = -1$ into the original function to find the corresponding $y$-coordinate.

$f(-1) = (-1)^2 + 2 = 1 + 2 = 3$

So, the point on the curve is $(-1, 3)$.

4. Write the equation of the tangent line

Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point of tangency:

$y - 3 = -2(x + 1)$

Simplifying:

$y - 3 = -2x - 2$ $y = -2x + 1$

Final Answer

The equation of the tangent line at $x = -1$ is:

$y = -2x + 1$

Would you like more details or have any questions?

---

Relative Questions:

1. How would you find the tangent line for the function $f(x) = \sqrt{x}$ at $x = 4$?
2. What is the geometric significance of the slope of the tangent line at a given point on the curve?
3. How would the process change if you were asked to find the normal line instead of the tangent line?
4. What happens to the equation of the tangent line if the function $f(x)$ is changed to $f(x) = x^3 + 2x + 1$?
5. How can you find the points where the tangent to the curve is horizontal?

Tip: The derivative of a function at a specific point not only gives the slope of the tangent line but also indicates the rate of change of the function at that point.