Question

which equation is a function of x?
$x^{2}=y^{2}+16$
$x = 5$
$x^{2}=y$
$x=y^{2}+9$
pre-algebra ic sem 1 fall 2025
analytic geometry

Explanation:

Step1: Recall the definition of a function

A relation is a function if for each input \(x\) there is exactly one output \(y\). We can use the vertical - line test (for equations in \(x\) and \(y\), if we can solve for \(y\) such that for each \(x\) value there is at most one \(y\) value).

Step2: Analyze \(x^{2}=y^{2}+16\)

Solve for \(y\): \(y^{2}=x^{2}-16\), then \(y = \pm\sqrt{x^{2}-16}\). For a non - zero \(x\) value (e.g., \(x = 5\), \(y=\pm3\)), there are two \(y\) values for a single \(x\) value. So it is not a function of \(x\).

Step3: Analyze \(x = 5\)

This is a vertical line. For \(x = 5\), \(y\) can be any real number. So for the input \(x = 5\), there are infinitely many \(y\) values. It is not a function of \(x\).

Step4: Analyze \(x^{2}=y\)

Solve for \(y\): \(y=x^{2}\). For each real - number value of \(x\), there is exactly one value of \(y\) (since squaring a real number gives a unique result). By the definition of a function (for each \(x\) in the domain, there is exactly one \(y\) in the range), \(y=x^{2}\) (or \(x^{2}=y\)) is a function of \(x\).

Step5: Analyze \(x=y^{2}+9\)

Solve for \(y\): \(y^{2}=x - 9\), then \(y=\pm\sqrt{x - 9}\). For a given \(x>9\) (e.g., \(x = 13\), \(y=\pm2\)), there are two \(y\) values for a single \(x\) value. So it is not a function of \(x\).

Answer:

\(x^{2}=y\)