For the function f, of, x, equals, x, squared, minus, 12*f*(*x*)=*x*2−12, find the equation of the tangent line at x, equals, minus, 10*x*=−10.

To find the equation of the tangent line to the function \( f(x) = x^2 - 12 \) at \( x = -10 \), we need to follow these steps:

### Step 1: Find the derivative of \( f(x) \)
The derivative of \( f(x) \), denoted as \( f'(x) \), gives the slope of the tangent line at any point \( x \).

Since \( f(x) = x^2 - 12 \), we differentiate it with respect to \( x \):

\[
f'(x) = \frac{d}{dx}(x^2 - 12) = 2x
\]

### Step 2: Evaluate the derivative at \( x = -10 \)
Now, substitute \( x = -10 \) into \( f'(x) \) to find the slope of the tangent line at this point:

\[
f'(-10) = 2(-10) = -20
\]

So, the slope of the tangent line at \( x = -10 \) is \( -20 \).

### Step 3: Find the point on the function at \( x = -10 \)
To find the equation of the tangent line, we also need the point on the curve where \( x = -10 \). We can find the \( y \)-coordinate by substituting \( x = -10 \) into the original function \( f(x) = x^2 - 12 \):

\[
f(-10) = (-10)^2 - 12 = 100 - 12 = 88
\]

So, the point of tangency is \( (-10, 88) \).

### Step 4: Use the point-slope form of the equation of a line
The equation of a line with slope \( m \) passing through a point \( (x_1, y_1) \) is given by the point-slope form:

\[
y - y_1 = m(x - x_1)
\]

Here, \( m = -20 \), \( x_1 = -10 \), and \( y_1 = 88 \). Substituting these values into the equation gives:

\[
y - 88 = -20(x - (-10))
\]
\[
y - 88 = -20(x + 10)
\]
\[
y = -20(x + 10) + 88
\]
\[
y = -20x - 200 + 88
\]
\[
y = -20x - 112
\]

### Final Answer:
The equation of the tangent line to the curve \( f(x) = x^2 - 12 \) at \( x = -10 \) is:

\[
y = -20x - 112
\]

Would you like more details on any step or have further questions?

Here are 5 related questions:
1. What is the geometric interpretation of a tangent line to a curve?
2. How do we find the second derivative of \( f(x) = x^2 - 12 \) and what does it represent?
3. How can you use derivatives to find points of inflection on a curve?
4. What happens to the tangent line equation if the function has a higher degree (e.g., cubic)?
5. How does the chain rule apply if the function were more complex, like \( f(x) = (x^2 - 12)^2 \)?

**Tip:** The slope of a tangent line at a point reflects the instantaneous rate of change of the function at that point.

For the function f(x) = x^2 - 12, find the equation of the tangent line at x = -10.

Learn how to find the equation of the tangent line to the quadratic function f(x) = x^2 - 12 at x = -10. This problem involves key calculus concepts like differentiation and the first derivative to find the slope of the tangent line. Suitable for students in Grades 10-12.

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