Question

comparing graphs of radical functions
which of the following graphs are identical? choose two correct answers.
$y = \sqrt{x}$ $y = -\sqrt3{x}$ $y = \sqrt{-x}$
$y = -\sqrt{x}$ $y = \sqrt3{x}$ $y = \sqrt3{-x}$

Explanation:

To determine which graphs are identical, we analyze the properties of radical functions, specifically cube root functions (since the domain of cube root functions includes all real numbers, and they have odd symmetry) and square root functions (with restricted domain and non - negative range).

Step 1: Recall properties of cube root functions

For a cube root function \(y = \sqrt[3]{x}\), we know that \(\sqrt[3]{-x}=-\sqrt[3]{x}\) (by the property of cube roots: \(\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}\), so \(\sqrt[3]{-x}=\sqrt[3]{-1\cdot x}=\sqrt[3]{-1}\cdot\sqrt[3]{x}= - \sqrt[3]{x}\)). Let's check the functions \(y =-\sqrt[3]{x}\) and \(y=\sqrt[3]{-x}\).
If we substitute \(x\) into \(y = \sqrt[3]{-x}\), we get \(y=\sqrt[3]{-x}=-\sqrt[3]{x}\) (using the property of cube roots \(\sqrt[3]{-a}=-\sqrt[3]{a}\) for any real number \(a\)). So the functions \(y =-\sqrt[3]{x}\) and \(y=\sqrt[3]{-x}\) are equivalent, which means their graphs are identical.

Step 2: Analyze square root functions (for elimination)

• For \(y = \sqrt{x}\), the domain is \(x\geq0\), and the range is \(y\geq0\).
• For \(y=-\sqrt{x}\), the domain is \(x\geq0\), and the range is \(y\leq0\). So it is a reflection of \(y = \sqrt{x}\) over the \(x\) - axis, not identical.
• For \(y=\sqrt{-x}\), the domain is \(x\leq0\), which is different from the domain of \(y = \sqrt{x}\) (\(x\geq0\)), so they are not identical.

Answer:

The two functions with identical graphs are \(y =-\sqrt[3]{x}\) and \(y=\sqrt[3]{-x}\) (i.e., the options \(y =-\sqrt[3]{x}\) and \(y=\sqrt[3]{-x}\)). In the given options, we should select the checkboxes for \(y =-\sqrt[3]{x}\) and \(y=\sqrt[3]{-x}\).