Question

question 2
for the following exercise, determine whether the relation represents y as a function of x.
$y = \sqrt{4 - x^2}$
select an answer
you must explain why this is or is not a function on your worksheet in order to receive full credit.
question 3
for the following exercise, find the domain of the function using interval notation.
$f(x) = \frac{4x + 2}{\sqrt{3 - x}}$

Explanation:

Step1: Apply vertical line test (Q2)

A relation is a function if every $x$-value maps to exactly one $y$-value. For $y=\sqrt{4-x^2}$, the square root returns only non-negative values. For each valid $x$, there is one corresponding $y$.

Step2: Identify domain constraints (Q3)

For $f(x)=\frac{4x+2}{\sqrt{3-x}}$, two rules apply:

1. The radicand must be non-negative: $3-x \geq 0$
2. The denominator cannot be zero: $\sqrt{3-x}

eq 0 \implies 3-x
eq 0$
Combining these, $3-x > 0$.

Step3: Solve inequality (Q3)

Rearrange $3-x > 0$:
$x < 3$
In interval notation, this is $(-\infty, 3)$.

Answer:

Question 2: Yes, $y=\sqrt{4-x^2}$ represents $y$ as a function of $x$. Each valid $x$ has exactly one non-negative $y$-value, satisfying the definition of a function.
Question 3: $(-\infty, 3)$