We are tasked with finding the equation of the tangent line to the function $f(x) = -2x^2 + 4x - 4$ at the point where $x = 3$, and express it in the form $y = mx + b$.

Steps:

1. Find the derivative of the function $f(x)$ to determine the slope of the tangent line. The derivative of $f(x)$ is: $f'(x) = \frac{d}{dx} \left( -2x^2 + 4x - 4 \right)$ Using power rule: $f'(x) = -4x + 4$

2. Evaluate the slope at $x = 3$: $f'(3) = -4(3) + 4 = -12 + 4 = -8$ So, the slope of the tangent line $m = -8$.

3. Find the corresponding $y$-value by plugging $x = 3$ into the original function $f(x)$: $f(3) = -2(3)^2 + 4(3) - 4 = -2(9) + 12 - 4 = -18 + 12 - 4 = -10$ Thus, the point of tangency is $(3, -10)$.

4. Use the point-slope form of a line equation: $y - y_1 = m(x - x_1)$ Substituting $m = -8$, $x_1 = 3$, and $y_1 = -10$: $y - (-10) = -8(x - 3)$ Simplifying: $y + 10 = -8x + 24$ $y = -8x + 14$

So, the equation of the tangent line is: $y = -8x + 14$

Would you like further clarification or additional details on this process?

Related Questions:

1. How is the derivative of a function interpreted in terms of tangent lines?
2. What is the geometric significance of a tangent line to a curve?
3. How does changing the point of tangency affect the slope of the tangent line?
4. How can we find the tangent line for other types of functions, like trigonometric or exponential ones?
5. How do we verify if the tangent line intersects the curve at more than one point?

Tip:

Always ensure you compute both the slope (derivative) and the point on the curve accurately before forming the tangent line equation.