find the equation of the tangent and normal line as the points indicated y=3x²-2x+1 at (1,2) 

To find the equations of the tangent and normal lines to the curve \( y = 3x^2 - 2x + 1 \) at the point \((1, 2)\), follow these steps:

### 1. Verify the Point on the Curve
First, confirm that the point \((1, 2)\) lies on the curve.

Substitute \( x = 1 \) into the equation \( y = 3x^2 - 2x + 1 \):

\[
y = 3(1)^2 - 2(1) + 1 = 3 - 2 + 1 = 2
\]

Since \( y = 2 \), the point \((1, 2)\) is indeed on the curve.

### 2. Find the Derivative of the Function
The derivative of the function \( y = 3x^2 - 2x + 1 \) gives the slope of the tangent line at any point \( x \).

\[
\frac{dy}{dx} = \frac{d}{dx} (3x^2 - 2x + 1) = 6x - 2
\]

### 3. Calculate the Slope at \( x = 1 \)
Substitute \( x = 1 \) into the derivative to find the slope of the tangent line at \( (1, 2) \):

\[
\text{Slope of tangent line} = 6(1) - 2 = 6 - 2 = 4
\]

### 4. Equation of the Tangent Line
The equation of a line can be written in point-slope form:

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

where \( m \) is the slope, and \( (x_1, y_1) \) is the point on the curve. Substituting the slope \( m = 4 \) and the point \( (1, 2) \):

\[
y - 2 = 4(x - 1)
\]

Simplify this to get the equation of the tangent line:

\[
y = 4x - 4 + 2
\]
\[
y = 4x - 2
\]

So, the equation of the tangent line is:

\[
\boxed{y = 4x - 2}
\]

### 5. Equation of the Normal Line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is \( 4 \), the slope of the normal line is:

\[
\text{Slope of normal line} = -\frac{1}{4}
\]

Using the point-slope form for the equation of the line, substitute \( m = -\frac{1}{4} \) and the point \( (1, 2) \):

\[
y - 2 = -\frac{1}{4}(x - 1)
\]

Simplify to get the equation of the normal line:

\[
y - 2 = -\frac{1}{4}x + \frac{1}{4}
\]
\[
y = -\frac{1}{4}x + \frac{1}{4} + 2
\]
\[
y = -\frac{1}{4}x + \frac{9}{4}
\]

So, the equation of the normal line is:

\[
\boxed{y = -\frac{1}{4}x + \frac{9}{4}}
\]

Would you like more details or have any questions? 

Here are some related questions:
1. How do you find the equation of a tangent line at any point on a curve?
2. What is the geometric significance of the normal line?
3. How can you determine the point of intersection between a curve and its tangent line?
4. How do tangent and normal lines relate to each other geometrically?
5. What role does differentiation play in finding tangent lines?

**Tip:** When finding the slope of the tangent line, always calculate the derivative of the function first.

Learn how to find the equations of tangent and normal lines to the curve y = 3x² - 2x + 1 at the point (1, 2). This detailed solution covers calculus concepts such as derivatives, point-slope form of a line, and their application to find tangent and normal lines.

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