Find the points on the curve 
y = x3 + 3x2 − 9x + 4
 where the tangent is horizontal.
smaller x-value	
(x, y)
=	

 
  

larger x-value	
(x, y)
=	

 
  



To find the points on the curve \( y = x^3 + 3x^2 - 9x + 4 \) where the tangent is horizontal, we need to determine where the derivative of the function is equal to zero. 

1. **Step 1: Find the derivative of \( y = x^3 + 3x^2 - 9x + 4 \)**  
   The derivative represents the slope of the tangent line at any point on the curve. A horizontal tangent occurs when this slope is zero. Let's differentiate the function with respect to \( x \):

   \[
   \frac{dy}{dx} = 3x^2 + 6x - 9
   \]

2. **Step 2: Set the derivative equal to zero to find critical points**  
   We now solve the equation:

   \[
   3x^2 + 6x - 9 = 0
   \]

   Simplifying this equation by dividing everything by 3:

   \[
   x^2 + 2x - 3 = 0
   \]

3. **Step 3: Solve the quadratic equation**  
   Factor the quadratic equation:

   \[
   (x - 1)(x + 3) = 0
   \]

   Therefore, the solutions are:

   \[
   x = 1 \quad \text{or} \quad x = -3
   \]

4. **Step 4: Find the corresponding y-values**  
   To find the points on the curve, we substitute these \( x \)-values back into the original equation \( y = x^3 + 3x^2 - 9x + 4 \).

   - For \( x = 1 \):

     \[
     y = (1)^3 + 3(1)^2 - 9(1) + 4 = 1 + 3 - 9 + 4 = -1
     \]

     So, the point is \( (1, -1) \).

   - For \( x = -3 \):

     \[
     y = (-3)^3 + 3(-3)^2 - 9(-3) + 4 = -27 + 27 + 27 + 4 = 31
     \]

     So, the point is \( (-3, 31) \).

### Conclusion:
- Smaller \( x \)-value: \( (-3, 31) \)
- Larger \( x \)-value: \( (1, -1) \)

Would you like more details or have any questions?

Here are 5 related questions for deeper understanding:
1. What does a horizontal tangent line signify in a graph?
2. How would the procedure differ if the equation were more complex?
3. Can you find the second derivative at these points to analyze concavity?
4. What happens to the tangent at points where the derivative is undefined?
5. How can we apply this method to real-world problems involving rates of change?

**Tip:** When factoring quadratic equations, always check if a common factor can simplify the equation first.

Find the points on the curve y = x^3 + 3x^2 − 9x + 4 where the tangent is horizontal. Smaller x-value (x, y) = Larger x-value (x, y) =

Discover how to find points where the tangent to the curve y = x^3 + 3x^2 − 9x + 4 is horizontal. This step-by-step solution covers calculus concepts such as derivatives, solving quadratic equations, and applying these to find specific points on the curve. Suitable for students in Grades 10-12.

Math Problem Statement

Find the points on the curve y = x3 + 3x2 − 9x + 4 where the tangent is horizontal. smaller x-value (x, y)

larger x-value (x, y)

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