Question

for each of the following, determine whether the equation defines y as a function of x.
$4 = |y| + x^2$ $circ$ function $circ$ not a function
$36 + y^2 = x^2$ $circ$ function $circ$ not a function
$5x = y^3$ $circ$ function $circ$ not a function
$y = 6|x| - 3$ $circ$ function $circ$ not a function

Explanation:

Step1: Rewrite first equation

Rearrange $4 = |y| + x^2$ to isolate $|y|$:
$|y| = 4 - x^2$
This gives $y = \pm(4 - x^2)$ for valid $x$, so one $x$ maps to two $y$-values.

Step2: Rewrite second equation

Rearrange $36 + y^2 = x^2$ to isolate $y^2$:
$y^2 = x^2 - 36$
This gives $y = \pm\sqrt{x^2 - 36}$ for valid $x$, so one $x$ maps to two $y$-values.

Step3: Rewrite third equation

Rearrange $5x = y^3$ to isolate $y$:
$y = \sqrt[3]{5x}$
Each $x$ maps to exactly one $y$-value.

Step4: Analyze fourth equation

Given $y = 6|x| - 3$, each $x$ maps to exactly one $y$-value.

Answer:

1. $4 = |y| + x^2$: Not a function
2. $36 + y^2 = x^2$: Not a function
3. $5x = y^3$: Function
4. $y = 6|x| - 3$: Function