Find the equation to the tangent line to y=x^2-2x+2 at the point (1,1)

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.

Find the equation to the tangent line to y=x^2-2x+2 at the point (1,1).

Find the equation of the tangent line to the curve y = x^2 - 2x + 2 at the point (1,1). This problem covers key calculus concepts such as differentiation and the point-slope form. Ideal for students in Grades 10-12.

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