https://file.aiquickdraw.com/mathbot/file/1727314267141ivf61w5n.jpg

Let's break down the problem from the image you uploaded:

The function provided is:

\[
f(x) = 12\sqrt{x} - 2x
\]

### Part (a): Finding points where the tangent line is horizontal.

1. The slope of the tangent line is given by the derivative of \( f(x) \), which is \( f'(x) \).
2. To find where the tangent line is horizontal, we need to set \( f'(x) = 0 \).

Let’s compute the derivative of \( f(x) \):

\[
f(x) = 12\sqrt{x} - 2x = 12x^{1/2} - 2x
\]

Using the power rule for differentiation:

\[
f'(x) = 12 \cdot \frac{1}{2} x^{-1/2} - 2 = 6x^{-1/2} - 2 = \frac{6}{\sqrt{x}} - 2
\]

Now, set \( f'(x) = 0 \) to find the points where the tangent line is horizontal:

\[
\frac{6}{\sqrt{x}} - 2 = 0
\]

Solving for \( x \):

\[
\frac{6}{\sqrt{x}} = 2
\]
\[
\sqrt{x} = 3
\]
\[
x = 9
\]

Now, plug \( x = 9 \) back into the original function \( f(x) \) to find the corresponding \( y \)-coordinate:

\[
f(9) = 12\sqrt{9} - 2(9) = 12(3) - 18 = 36 - 18 = 18
\]

Thus, the point where the tangent line is horizontal is \( (9, 18) \).

### Part (b): Finding points where the tangent line has slope \( -\frac{1}{2} \).

Again, we use the derivative \( f'(x) = \frac{6}{\sqrt{x}} - 2 \), and now set it equal to \( -\frac{1}{2} \):

\[
\frac{6}{\sqrt{x}} - 2 = -\frac{1}{2}
\]

Solving for \( x \):

\[
\frac{6}{\sqrt{x}} = -\frac{1}{2} + 2 = \frac{3}{2}
\]
\[
\sqrt{x} = \frac{6}{\frac{3}{2}} = 4
\]
\[
x = 16
\]

Now, plug \( x = 16 \) back into the original function to find the corresponding \( y \)-coordinate:

\[
f(16) = 12\sqrt{16} - 2(16) = 12(4) - 32 = 48 - 32 = 16
\]

Thus, the point where the tangent line has slope \( -\frac{1}{2} \) is \( (16, 16) \).

### Final Answers:
- **Part (a)**: The point where the tangent line is horizontal is \( (9, 18) \).
- **Part (b)**: The point where the tangent line has slope \( -\frac{1}{2} \) is \( (16, 16) \).

Would you like further details on any of the steps or a deeper explanation?

Here are some follow-up questions:
1. How do we calculate the derivative of functions with radicals?
2. Why does setting the derivative equal to zero give points where the tangent line is horizontal?
3. How can you determine the concavity of the function from its second derivative?
4. What is the geometric meaning of finding a tangent line with a specific slope?
5. What other applications are there for finding where the tangent line has a certain slope?

**Tip**: When solving for horizontal tangents, always set the derivative to zero and check for any constraints on the domain of the function.

Let f(x) = 12√x − 2x. Find all points on the graph of f where (a) the tangent line is horizontal, (b) the tangent line has slope −1/2.

Solve for points where the tangent line is horizontal and has slope −1/2 for f(x) = 12√x - 2x. Includes differentiation steps, solving for x, and finding the y-coordinates of points of interest. Suitable for students in Grades 11-12 studying calculus concepts like derivatives and slopes of tangent lines.

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