The question asks whether the given relation defines $y$ as a function of $x$, and also asks for the domain and range of the relation.

The given equation is: $y = \sqrt{8x + 3}$

Step 1: Is it a function?

A relation is a function if for every input $x$, there is exactly one output $y$.

For this equation, the square root function $\sqrt{}$ always returns only the non-negative square root (the principal root). Therefore, for every valid $x$, there will be only one value of $y$, making this relation a function. Thus, the answer to the question "Does the relation define a function?" is Yes.

Step 2: Domain

The domain of the function refers to the set of all possible $x$-values that make the expression inside the square root non-negative. In this case, we need: $8x + 3 \geq 0$

Solving for $x$: $8x \geq -3 \quad \Rightarrow \quad x \geq -\frac{3}{8}$

Thus, the domain in interval notation is: $\left[ -\frac{3}{8}, \infty \right)$

Step 3: Range

Since the square root function only gives non-negative outputs, the smallest value for $y$ is 0 (which happens when $x = -\frac{3}{8}$). The range is therefore all non-negative numbers: $[0, \infty)$

Final Answers:

1. Does the relation define a function? Yes.
2. Domain: $\left[ -\frac{3}{8}, \infty \right)$
3. Range: $[0, \infty)$

Would you like further details or have any other questions?

Related Questions:

1. How do you find the domain of more complex functions?
2. What happens if the equation includes a denominator or logarithm?
3. Can a square root function ever have negative values in its range?
4. How can we graph $y = \sqrt{8x + 3}$ to confirm its domain and range?
5. How does the concept of a function apply to real-world examples?

Tip:

Always check for restrictions such as square roots, denominators, and logarithms when determining the domain of a function.