Cora is working on graphing h(x)= ^h square root -x-3. She starts by graphing f(x)=^3 square root x as the parent function. Next she takes her graph of f(x) and reflects it across the y-axis and calls the results g(x)

Question

In order to transform the graph g(x) to make the graph h(x) Cora should translate g(x) 3 units to the right
True or False

Answer 1

To determine if Cora's statement about translating the graph \( g(x) \) to obtain \( h(x) \) is true, we need to analyze the functions involved.

1. The parent function is \( f(x) = \sqrt[3]{x} \), which is a cube root function.

2. To reflect it across the y-axis, we obtain \( g(x) = \sqrt[3]{-x} \).

3. The function \( h(x) \) is given as \( h(x) = \sqrt{-x - 3} \). We can rewrite \( h(x) \) to understand the transformations better: \[ h(x) = \sqrt{-(x + 3)}. \]

   This suggests that it involves a horizontal translation and a reflection.

4. To get from \( g(x) = \sqrt[3]{-x} \) to \( h(x) \), we see that there are not only horizontal transformations involved but also a change from a cube root to a square root, which is a more significant change from a different type of function.

Now, examining the translation specifically:

• \( g(x) = \sqrt[3]{-x} \) reflects across the y-axis, and if we translate this 3 units to the right, we get: \[ g(x - 3) = \sqrt[3]{-(x - 3)} = \sqrt[3]{-x + 3}. \]

This does not yield the function \( h(x) = \sqrt{-(x + 3)} \).

Thus, the assertion that Cora should translate the graph \( g(x) \) 3 units to the right in order to obtain the graph of \( h(x) \) is False.

In summary, Cora's translation claim is false because the transformations required to reach \( h(x) \) from \( g(x) \) are not as simple as just translating \( g(x) \) 3 units to the right.