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.