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Let's tackle the questions from the image one by one.

### Question 9: Pendulum Period Function
The function provided for the period \( T \) of a pendulum is:

\[
T = 2\sqrt{\ell}
\]
where \( \ell \) is the length of the pendulum in meters.

#### Part a: Domain and Range of \( T \)
- **Domain**: Since the length \( \ell \) must be non-negative (you can't have a negative length), the domain is \( \ell \geq 0 \), or in interval notation:
  \[
  [0, \infty)
  \]
- **Range**: Since the period \( T \) is dependent on the square root of \( \ell \), \( T \) will always be positive (since the square root function is positive for all non-negative \( \ell \)). The smallest value of \( T \) occurs when \( \ell = 0 \), giving \( T = 0 \). Hence, the range is:
  \[
  [0, \infty)
  \]

#### Part b: Sketching a Graph
- The graph of the function \( T = 2\sqrt{\ell} \) is a square root function, which starts at \( T = 0 \) when \( \ell = 0 \) and increases gradually as \( \ell \) increases.
- The curve is shaped like the right half of a parabola that opens sideways, with the general behavior that \( T \) grows slowly at first, then faster as \( \ell \) increases.

#### Part c: Is the relation a function?
Yes, the relation is a function because for each value of \( \ell \), there is exactly one corresponding value of \( T \). A function requires that every input (in this case, \( \ell \)) has a unique output (here, \( T \)).

### Question 10: Mapping Diagram

The ordered pairs based on the diagram would be:

\[
(1, 2), (2, 1), (3, 2), (4, 3), (7, 3)
\]

- **Is this a function?** No, this is not a function because in a function, each input must correspond to exactly one output. In this case, both \( 1 \) and \( 3 \) in the domain map to \( 2 \), and both \( 4 \) and \( 7 \) map to \( 3 \). Therefore, some inputs have multiple outputs, so it violates the definition of a function.

### Question 11: Parabola Analysis

The function given is:

\[
f(x) = -\frac{1}{2}x^2 + 4x + 3
\]

#### Part a: Finding the Vertex
The formula for the x-coordinate of the vertex for a parabola in the form \( f(x) = ax^2 + bx + c \) is:

\[
x = \frac{-b}{2a}
\]

For \( f(x) = -\frac{1}{2}x^2 + 4x + 3 \), we have \( a = -\frac{1}{2} \) and \( b = 4 \), so:

\[
x = \frac{-4}{2(-\frac{1}{2})} = \frac{-4}{-1} = 4
\]

To find the y-coordinate of the vertex, substitute \( x = 4 \) into the equation for \( f(x) \):

\[
f(4) = -\frac{1}{2}(4)^2 + 4(4) + 3 = -\frac{1}{2}(16) + 16 + 3 = -8 + 16 + 3 = 11
\]

Thus, the vertex is \( (4, 11) \).

#### Part b: Is the vertex a minimum or maximum?
Since the coefficient of \( x^2 \) (i.e., \( a = -\frac{1}{2} \)) is negative, the parabola opens downwards. This means the vertex represents a **maximum** point.

#### Part c: Number of x-intercepts
To find the x-intercepts, set \( f(x) = 0 \):

\[
-\frac{1}{2}x^2 + 4x + 3 = 0
\]

Multiply through by \( -2 \) to eliminate the fraction:

\[
x^2 - 8x - 6 = 0
\]

Now, use the quadratic formula:

\[
x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-6)}}{2(1)}
\]
\[
x = \frac{8 \pm \sqrt{64 + 24}}{2} = \frac{8 \pm \sqrt{88}}{2} = \frac{8 \pm 2\sqrt{22}}{2}
\]
\[
x = 4 \pm \sqrt{22}
\]

Since \( \sqrt{22} \) is a real number, there are two real solutions. Therefore, the function has **two x-intercepts**.

---

Would you like any further clarification or more details on any part?

Here are 5 related questions:
1. How do you determine the range of a square root function in general?
2. What is the general form of the vertex formula for a parabola?
3. How does the sign of the leading

The time needed for a pendulum to make one complete swing is called the period of the pendulum. The period T, in seconds, for a pendulum of length l, in meters, can be approximated using the function T = 2√l.

Explore the mathematical concepts behind the pendulum period formula T = 2√l, including domain, range, and graphing. Also, solve a quadratic function to find its vertex, intercepts, and determine if it's a maximum or minimum. Suitable for high school students.

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