Find the point(s), if any, at which the graph of *f* has a horizontal tangent line.

*f*(*x*) = 

*x*2

*x* − 6

smaller *x*-value(*x*, *y*)

= 
 
 
larger *x*-value(*x*, *y*)

= 
 
 

To find the point(s) where the graph of a function \( f(x) \) has a horizontal tangent line, we need to:

1. Compute the derivative \( f'(x) \), which gives the slope of the tangent line at any point on the graph.
2. Set the derivative equal to 0 to find where the slope is zero, indicating a horizontal tangent line.
3. Solve for \( x \) to find the corresponding points.

The given function is:

\[
f(x) = x^2(x - 6)
\]

### Step 1: Compute the derivative

First, expand the function:

\[
f(x) = x^3 - 6x^2
\]

Now, differentiate \( f(x) \):

\[
f'(x) = \frac{d}{dx}(x^3 - 6x^2) = 3x^2 - 12x
\]

### Step 2: Set the derivative equal to zero

To find the points where the tangent line is horizontal, set the derivative equal to zero:

\[
3x^2 - 12x = 0
\]

Factor the equation:

\[
3x(x - 4) = 0
\]

### Step 3: Solve for \( x \)

Now, solve for \( x \):

\[
x = 0 \quad \text{or} \quad x = 4
\]

### Step 4: Find the corresponding \( y \)-values

Substitute these \( x \)-values back into the original function to find the corresponding \( y \)-coordinates:

- For \( x = 0 \):
  
  \[
  f(0) = 0^3 - 6(0^2) = 0
  \]

  So, one point is \( (0, 0) \).

- For \( x = 4 \):
  
  \[
  f(4) = 4^3 - 6(4^2) = 64 - 96 = -32
  \]

  So, the second point is \( (4, -32) \).

### Final Answer

- Smaller \( x \)-value point: \( (0, 0) \)
- Larger \( x \)-value point: \( (4, -32) \)

Would you like further details or have any questions?

Here are 5 related questions for more practice:
1. How do you find the vertical tangent lines of a curve?
2. What is the geometric interpretation of a derivative?
3. Can a function have multiple horizontal tangent lines?
4. How do you check if a critical point is a maximum or minimum?
5. How do higher-order derivatives affect the shape of a graph?

**Tip:** Remember, horizontal tangent lines occur where the derivative equals zero, meaning the function's slope is flat at those points.

Find the point(s), if any, at which the graph of f has a horizontal tangent line. f(x) = x^2(x - 6)

Discover how to find the horizontal tangent points for the polynomial function f(x) = x^2(x - 6). Learn how to compute the derivative, set it to zero, and solve for the points where the tangent line is horizontal. Perfect for students in Grades 11-12 or early college studying calculus.

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