Question

the graph of h(x) is a translation of f(x) = sqrt3{x}. which equation represents h(x)? \
\bigcirc \\ h(x) = \sqrt3{x - 2} \
\bigcirc \\ h(x) = \sqrt3{x + 2} \
\bigcirc \\ h(x) = \sqrt3{x} - 2 \
\bigcirc \\ h(x) = \sqrt3{x} + 2

Explanation:

Step1: Recall parent function

The parent function is \( f(x) = \sqrt[3]{x} \). Its key point is \((0, 0)\) (since \( \sqrt[3]{0} = 0 \)).

Step2: Analyze the translated point

The graph of \( h(x) \) passes through \((-1, 1)\). Let's test each option by plugging \( x = -1 \) and seeing if \( y = 1 \).

• For \( h(x) = \sqrt[3]{x - 2} \): Plug \( x = -1 \), we get \( \sqrt[3]{-1 - 2} = \sqrt[3]{-3}

eq 1 \).

• For \( h(x) = \sqrt[3]{x + 2} \): Plug \( x = -1 \), we get \( \sqrt[3]{-1 + 2} = \sqrt[3]{1} = 1 \). This matches the point \((-1, 1)\).
• For \( h(x) = \sqrt[3]{x} - 2 \): Plug \( x = -1 \), we get \( \sqrt[3]{-1} - 2 = -1 - 2 = -3

eq 1 \).

• For \( h(x) = \sqrt[3]{x} + 2 \): Plug \( x = -1 \), we get \( \sqrt[3]{-1} + 2 = -1 + 2 = 1 \). Wait, but let's check another point. The parent function \( f(x) = \sqrt[3]{x} \) has a point \((-8, -2)\) (since \( \sqrt[3]{-8} = -2 \)). For \( h(x) = \sqrt[3]{x} + 2 \), at \( x = -8 \), \( h(-8) = \sqrt[3]{-8} + 2 = -2 + 2 = 0 \), but the graph shows a point \((-3, -1)\). Let's check \( h(x) = \sqrt[3]{x + 2} \) at \( x = -3 \): \( \sqrt[3]{-3 + 2} = \sqrt[3]{-1} = -1 \), which matches \((-3, -1)\). For \( h(x) = \sqrt[3]{x} + 2 \) at \( x = -3 \): \( \sqrt[3]{-3} + 2

eq -1 \). So the correct one is \( h(x) = \sqrt[3]{x + 2} \).

Answer:

\( h(x) = \sqrt[3]{x + 2} \) (the second option: \( h(x) = \sqrt[3]{x + 2} \))