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### **1. Find the equation of the tangent to:**
   **(a) \( y = x - 2x^2 + 3 \) at \( x = 2 \):**
   
   The equation of the tangent is found by:
   
   1. Finding the derivative \( y' \) of the function \( y = x - 2x^2 + 3 \) to get the slope.
   
   \[
   y' = \frac{d}{dx}(x - 2x^2 + 3) = 1 - 4x
   \]
   
   2. Evaluate \( y' \) at \( x = 2 \) to get the slope at that point:
   
   \[
   y'(2) = 1 - 4(2) = 1 - 8 = -7
   \]
   
   3. Find the value of \( y \) at \( x = 2 \):
   
   \[
   y(2) = 2 - 2(2^2) + 3 = 2 - 8 + 3 = -3
   \]
   
   4. The equation of the tangent line is:
   
   \[
   y - y_1 = m(x - x_1)
   \]
   
   Substituting \( x_1 = 2 \), \( y_1 = -3 \), and \( m = -7 \):
   
   \[
   y + 3 = -7(x - 2)
   \]
   
   Simplifying:
   
   \[
   y = -7x + 14 - 3
   \]
   
   \[
   y = -7x + 11
   \]

This is the tangent line equation for part **1(a)**.

Do you want to continue with part **1(b)** or any other part, or would you like further details on this solution? Let me know!

Here are 5 follow-up questions for review:
1. How is the derivative used to find the slope of the tangent line?
2. What does the tangent line represent geometrically?
3. Why do we substitute the point \( x = 2 \) into both the function and its derivative?
4. How can we verify the correctness of the tangent line equation?
5. What changes in the process if we want to find the normal line instead of the tangent?

**Tip:** The slope of the tangent line is always equal to the value of the derivative of the function at the point of tangency.

Find the equation of the tangent to y = x - 2x^2 + 3 at x = 2.

Learn how to find the tangent line equation to the quadratic function y = x - 2x^2 + 3 at x = 2. This problem explores key calculus concepts like differentiation and tangent lines, with a detailed step-by-step solution. Suitable for Grades 11-12 students.

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