Question

determine if the following relations define y as a function of x.
part 1 of 2
(a) ( x = 9y^2 )
( x = 9y^2 ) select a function.
part 2 of 2
(b) ( y = 9x^2 )
( y = 9x^2 ) select a function.

Explanation:

Step1: Rewrite for y in terms of x

Starting with $x=9y^2$, solve for y:
$\frac{x}{9}=y^2$
$y=\pm\sqrt{\frac{x}{9}}=\pm\frac{\sqrt{x}}{3}$

Step2: Test function definition

For a positive x-value (e.g., $x=9$), $y=\frac{\sqrt{9}}{3}=1$ and $y=-\frac{\sqrt{9}}{3}=-1$. One x maps to two y-values, so it is not a function.

Step3: Analyze $y=9x^2$

For any real x, substitute into the equation:
$y=9x^2$ gives exactly one y-value per x-value.

Step4: Confirm function definition

Each x has only one corresponding y, so this is a function.

Answer:

(a) $x=9y^2$ is not a function.
(b) $y=9x^2$ is a function.