To find the equation of the tangent line to the curve $f(x) = -2x^2 + 1$ at the point $(2, -7)$, we need to follow these steps:

1. Find the derivative of $f(x)$: This represents the slope of the tangent line at any point on the curve.
2. Evaluate the derivative at the given point: This gives the slope of the tangent line at $x = 2$.
3. Use the point-slope form of the equation of a line: With the slope and a point on the tangent line, we can find the equation.

Step 1: Find the Derivative

The function given is: $f(x) = -2x^2 + 1$

The derivative $f'(x)$ is found using the power rule: $f'(x) = \frac{d}{dx}(-2x^2 + 1) = -4x$

Step 2: Evaluate the Derivative at $x = 2$

Now, plug $x = 2$ into the derivative to find the slope of the tangent line at this point: $f'(2) = -4(2) = -8$

The slope of the tangent line at $x = 2$ is $-8$.

Step 3: Use the Point-Slope Form

The point-slope form of a line equation is: $y - y_1 = m(x - x_1)$

Here, $m = -8$, $x_1 = 2$, and $y_1 = -7$. Plugging in these values gives: $y - (-7) = -8(x - 2)$

Simplify the equation: $y + 7 = -8x + 16$

Subtract 7 from both sides: $y = -8x + 9$

Conclusion

The equation of the tangent line to the curve $f(x) = -2x^2 + 1$ at the point $(2, -7)$ is: $\boxed{y = -8x + 9}$

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

Here are some related questions you might find useful:

1. How do you find the derivative of a polynomial function?
2. What is the power rule in calculus?
3. How do you apply the point-slope form to find the equation of a line?
4. What is the significance of the slope of the tangent line at a point on a curve?
5. How do you verify that a point lies on the tangent line to a curve?

Tip: Remember that the slope of the tangent line represents the instantaneous rate of change of the function at a specific point.