You are being asked to graph the function $f(x) = \sqrt{2x + 1}$. Let me break down how to approach this:

Key Steps:

1. Determine the Domain:

   • Since the function involves a square root, the expression inside the root must be non-negative: $2x + 1 \geq 0 \quad \Rightarrow \quad x \geq -\frac{1}{2}$
   • Therefore, the domain is $x \geq -\frac{1}{2}$.

2. Find the Leftmost Point:

   • The smallest $x$-value is $x = -\frac{1}{2}$, which is where the function starts.
   • At $x = -\frac{1}{2}$, the function value is: $f\left(-\frac{1}{2}\right) = \sqrt{2 \cdot -\frac{1}{2} + 1} = \sqrt{0} = 0$
   • So, the leftmost point is $\left( -\frac{1}{2}, 0 \right)$.

3. Choose Additional Points:

   • Now pick other $x$-values greater than or equal to $-\frac{1}{2}$. Let's try these values:
     • For $x = 0$: $f(0) = \sqrt{2(0) + 1} = \sqrt{1} = 1 \quad \text{Point: } (0, 1)$
     • For $x = 1$: $f(1) = \sqrt{2(1) + 1} = \sqrt{3} \approx 1.732 \quad \text{Point: } (1, 1.732)$
     • For $x = 2$: $f(2) = \sqrt{2(2) + 1} = \sqrt{5} \approx 2.236 \quad \text{Point: } (2, 2.236)$

Points to Plot:

1. $\left( -\frac{1}{2}, 0 \right)$
2. $(0, 1)$
3. $(1, 1.732)$
4. $(2, 2.236)$

You can now plot these points on the graph and connect them to sketch the curve of the function. Let me know if you need further clarification or more points.

Questions for Further Exploration:

1. How does the shape of $f(x) = \sqrt{2x + 1}$ differ from $f(x) = \sqrt{x}$?
2. What happens to the graph of $f(x) = \sqrt{2x + 1}$ if the constant inside the square root changes (e.g., $\sqrt{2x - 3}$)?
3. What is the range of the function $f(x) = \sqrt{2x + 1}$?
4. How would the graph look if the function was $f(x) = -\sqrt{2x + 1}$?
5. What is the effect of multiplying the entire function by a constant (e.g., $f(x) = 3\sqrt{2x + 1}$)?

Tip: Always check the domain when dealing with square roots or logarithmic functions.