The problem in the image is in Indonesian and translates to:

"Determine the equation of the tangent line for $f(x) = x^2 - 1$ at the point $x = -1$."

Let's solve it step-by-step.

Step 1: Differentiate $f(x)$

To find the slope of the tangent line, we need the derivative of $f(x) = x^2 - 1$.

$f'(x) = \frac{d}{dx}(x^2 - 1) = 2x$

Step 2: Calculate the Slope at $x = -1$

Now, substitute $x = -1$ into $f'(x)$ to find the slope of the tangent line at that point.

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

So, the slope of the tangent line is $-2$.

Step 3: Find the Point on the Curve at $x = -1$

We also need the coordinates of the point on the curve where $x = -1$.

$f(-1) = (-1)^2 - 1 = 1 - 1 = 0$

Thus, the point of tangency is $(-1, 0)$.

Step 4: Use the Point-Slope Form for the Tangent Line

The equation of a line with slope $m$ passing through a point $(x_1, y_1)$ is given by:

$y - y_1 = m(x - x_1)$

Substituting $m = -2$, $x_1 = -1$, and $y_1 = 0$:

$y - 0 = -2(x + 1)$

Simplify this equation:

$y = -2x - 2$

Final Answer

The equation of the tangent line is:

$y = -2x - 2$

Would you like further explanation on any step?

Here are some related questions you might find useful:

1. What is the importance of finding the tangent line in calculus?
2. How do you find the derivative of higher-degree polynomial functions?
3. What are some applications of tangent lines in real-world problems?
4. How can we determine if a line is perpendicular to a curve at a given point?
5. How would the solution differ if we had a different function, such as $f(x) = x^3 - 1$?

Tip: The derivative at a point gives the slope of the tangent line, which provides an approximation of the function’s behavior near that point.