Mrs. Garcia assigns her student the cube root function \begin{align*}y = -\sqrt[3]{(x + 1)}\end{align*}

Alendro says that because of the negative sign, all *y* values are negative. Therefore his graph is only in the third and fourth quadrants quadrant.

Dako says that his graph is in the third and fourth quadrants as well but it is also in the second quadrant.

Marisha says they are both wrong and that her graph of the function is in all four quadrants.

Which one of them is correct?

### Graphing Cubed Root Functions

A cubed root function is different from that of a square root. Their general forms look very similar, \begin{align*}y=a \sqrt[3]{x-h}+k\end{align*}

x |
y |
---|---|

-27 | -3 |

-8 | -2 |

-1 | -1 |

0 | 0 |

1 | 1 |

8 | 2 |

27 | 3 |

For \begin{align*}y= \sqrt[3]{x}\end{align*}**point of inflection** (see the *Analyzing the Graph of a Polynomial Function* concept).

#### Graph the cube root functions

Describe how to obtain the graph of \begin{align*}y= \sqrt[3]{x}+5\end{align*}

From the previous concept, we know that the +5 indicates a vertical shift of 5 units up. Therefore, this graph will look exactly the same as the parent graph, shifted up five units.

Graph \begin{align*}f(x)=- \sqrt[3]{x+2}-3\end{align*}

From the previous example, we know that from the parent graph, this function is going to shift to the left two units and down three units. The negative sign will result in a reflection.

**Alternate Method:** If you want to use a table (like in the previous concept), that will also work. Here is a table, then plot the points. \begin{align*}(h, k)\end{align*}

x |
y |
---|---|

6 | -5 |

-1 | -4 |

-2 | -3 |

-3 | -2 |

-10 | -1 |

Graph \begin{align*}f(x)= \frac{1}{2} \sqrt[3]{x-4}\end{align*}

The -4 tells us that, from the parent graph, the function will shift to the right four units. The \begin{align*}\frac{1}{2}\end{align*}

**Using the graphing calculator:** If you wanted to graph this function using the TI-83 or 84, press \begin{align*}Y=\end{align*}**MATH** and scroll down to **4**: \begin{align*}\sqrt[3]{\;\;}\end{align*}**ENTER**. Then, type in the rest of the function, so that \begin{align*}Y= \left(\frac{1}{2}\right) \sqrt[3]{\;\;}(X-4)\end{align*}. Press **GRAPH** and adjust the window.

**Important Note:** The domain and range of all cubed root functions are both all real numbers.

### Examples

#### Example 1

Earlier, you were asked who was correct.

If you graph the function \begin{align*}y = -\sqrt[3]{(x + 1)}\end{align*}, you see that the domain is all real numbers, which makes all quadrants possible. However, for all positive values of *x*, *y* is negative because of the negative sign in front of the cube root. That rules out the first quadrant. Therefore, Dako is correct.

#### Example 2

Evaluate \begin{align*}y= \sqrt[3]{x+4}-11\end{align*} when \begin{align*}x=-12\end{align*}.

Plug in \begin{align*}x=-12\end{align*} and solve for \begin{align*}y\end{align*}.

\begin{align*}y= \sqrt[3]{-12+4}-11= \sqrt[3]{-8}+4=-2+4=2\end{align*}

#### Example 3

Describe how to obtain the graph of \begin{align*}y= \sqrt[3]{x+4}-11\end{align*} from \begin{align*}y= \sqrt[3]{x}\end{align*}.

Graph the following cubed root functions. Check your graphs on the graphing calculator.

Starting with \begin{align*}y= \sqrt[3]{x}\end{align*}, you would obtain \begin{align*}y= \sqrt[3]{x+4}-11\end{align*} by shifting the function to the left four units and down 11 units.

#### Example 4

\begin{align*}y= \sqrt[3]{x-2}-4\end{align*}

This function is a horizontal shift to the right two units and down four units.

#### Example 5

\begin{align*}f(x)=-3 \sqrt{x}-1\end{align*}

This function is a reflection of \begin{align*}y= \sqrt[3]{x}\end{align*} and stretched to be three times as large. Lastly, it is shifted down one unit.

### Review

Evaluate \begin{align*}f(x)=\sqrt[3]{2x-1}\end{align*} for the following values of *x*.

- \begin{align*}f(14)\end{align*}
- \begin{align*}f(-62)\end{align*}
- \begin{align*}f(20)\end{align*}

Graph the following cubed root functions. Use your calculator to check your answers.

- \begin{align*}y=\sqrt[3]{x}+4\end{align*}
- \begin{align*}y=\sqrt[3]{x-3}\end{align*}
- \begin{align*}f(x)=\sqrt[3]{x+2}-1\end{align*}
- \begin{align*}g(x)=- \sqrt[3]{x}-6\end{align*}
- \begin{align*}f(x)=2 \sqrt[3]{x+1}\end{align*}
- \begin{align*}h(x)=-3 \sqrt[3]{x}+5\end{align*}
- \begin{align*}y=\frac{1}{2} \sqrt[3]{1-x}\end{align*}
- \begin{align*}y=2 \sqrt[3]{x+4}-3\end{align*}
- \begin{align*}y=- \frac{1}{3} \sqrt[3]{x-5}+2\end{align*}
- \begin{align*}g(x)=\sqrt[3]{6-x}+7\end{align*}
- \begin{align*}f(x)=-5 \sqrt[3]{x-1}+3\end{align*}
- \begin{align*}y=4 \sqrt[3]{7-x}-8\end{align*}

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 7.5.