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

Step 1: Differentiate the function

The derivative of $y = x^2 - 2x + 2$ gives the slope of the tangent line at any point $x$.

$\frac{dy}{dx} = \frac{d}{dx}(x^2 - 2x + 2)$ $\frac{dy}{dx} = 2x - 2$

Step 2: Find the slope at the point $(1,1)$

Now, substitute $x = 1$ into the derivative to find the slope of the tangent line at $x = 1$.

$\frac{dy}{dx} \Bigg|_{x=1} = 2(1) - 2 = 2 - 2 = 0$

The slope of the tangent line at $x = 1$ is 0.

Step 3: Use the point-slope form of the line equation

The point-slope form of a line equation is given by:

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

where $m$ is the slope and $(x_1, y_1)$ is the given point on the curve. In this case, $m = 0$, and the point is $(1, 1)$.

Substitute these values into the equation:

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

Simplifying:

$y - 1 = 0$ $y = 1$

Final Answer:

The equation of the tangent line at $(1, 1)$ is $y = 1$.

Would you like more details or have any questions?

Here are 5 relative questions that expand on this:

1. How do we differentiate higher-order polynomial functions?
2. What is the significance of the slope of the tangent line?
3. How can we find the normal line to the curve at a given point?
4. How do you apply the derivative to solve real-world rate-of-change problems?
5. Can the tangent line be vertical? If so, how do you find its equation?

Tip: The derivative of a function at a point gives the slope of the tangent line to the curve at that point.