Suppose you were given a function rule like \begin{align*}f(x)=2x^2+1\end{align*}, how could you graph the function? After completing this Concept, you'll be able to create a table of values to graph functions in the coordinate plane.

### Watch This

CK-12 Foundation: 0113S Graph a Function from a Rule

### Try This

Once you’ve had some practice graphing functions by hand, you may want to use a graphing calculator to make graphing easier. If you don’t have one, you can also use the applet at http://rechneronline.de/function-graphs/. Just type a function in the blank and press Enter. You can use the options under Display Properties to zoom in or pan around to different parts of the graph.

### Guidance

Of course, we can always make a graph from a function rule by substituting values in for the variable and getting the corresponding output value.

#### Example A

*Graph the following function: \begin{align*}f(x) = |x - 2|\end{align*}*

**Solution**

Make a table of values. Pick a variety of negative and positive values for \begin{align*}x\end{align*}. Use the function rule to find the value of \begin{align*}y\end{align*} for each value of \begin{align*}x\end{align*}. Then, graph each of the coordinate points.

\begin{align*}x\end{align*} | \begin{align*}y = f(x) = \mid x - 2 \mid\end{align*} |
---|---|

-4 | \begin{align*}\mid -4 - 2 \mid = \mid -6 \mid = 6\end{align*} |

-3 | \begin{align*}\mid -3 - 2 \mid = \mid -5 \mid = 5\end{align*} |

-2 | \begin{align*}\mid -2 - 2 \mid = \mid -4 \mid = 4\end{align*} |

-1 | \begin{align*}\mid -1 - 2 \mid = \mid -3 \mid = 3\end{align*} |

0 | \begin{align*}\mid 0 - 2 \mid = \mid -2 \mid = 2 \end{align*} |

1 | \begin{align*}\mid 1 - 2 \mid = \mid -1 \mid = 1\end{align*} |

2 | \begin{align*}\mid 2 - 2 \mid = \mid 0 \mid = 0\end{align*} |

3 | \begin{align*}\mid 3 - 2 \mid = \mid 1 \mid = 1\end{align*} |

4 | \begin{align*}\mid 4 - 2 \mid = \mid 2 \mid = 2\end{align*} |

5 | \begin{align*}\mid 5 - 2 \mid = \mid 3 \mid = 3\end{align*} |

6 | \begin{align*}\mid 6 - 2 \mid = \mid 4 \mid = 4\end{align*} |

7 | \begin{align*}\mid 7 - 2 \mid = \mid 5 \mid = 5\end{align*} |

8 | \begin{align*}\mid 8 - 2 \mid = \mid 6 \mid = 6\end{align*} |

It is wise to work with a lot of values when you begin graphing. As you learn about different types of functions, you will start to only need a few points in the table of values to create an accurate graph.

#### Example B

*Graph the following function: \begin{align*}f(x)=\sqrt{x}\end{align*}*

**Solution**

Make a table of values. We know \begin{align*}x\end{align*} can’t be negative because we can't take the square root of a negative number. The domain is all positive real numbers, so we pick a variety of positive integer values for \begin{align*}x\end{align*}. Use the function rule to find the value of \begin{align*}y\end{align*} for each value of \begin{align*}x\end{align*}.

\begin{align*}x\end{align*} | \begin{align*}y = f(x) = \sqrt{x}\end{align*} |
---|---|

0 | \begin{align*}\sqrt{0} = 0\end{align*} |

1 | \begin{align*}\sqrt{1} = 1\end{align*} |

2 | \begin{align*}\sqrt{2} \approx 1.41\end{align*} |

3 | \begin{align*}\sqrt{3} \approx 1.73\end{align*} |

4 | \begin{align*}\sqrt{4} = 2\end{align*} |

5 | \begin{align*}\sqrt{5} \approx 2.24\end{align*} |

6 | \begin{align*}\sqrt{6} \approx 2.45\end{align*} |

7 | \begin{align*}\sqrt{7} \approx 2.65\end{align*} |

8 | \begin{align*}\sqrt{8} \approx 2.83\end{align*} |

9 | \begin{align*}\sqrt{9} = 3\end{align*} |

Note that the range is all positive real numbers.

#### Example C

*The post office charges 41 cents to send a letter that is one ounce or less and an extra 17 cents for each additional ounce or fraction of an ounce. This rate applies to letters up to 3.5 ounces.*

**Solution**

Make a table of values. We can’t use negative numbers for \begin{align*}x\end{align*} because it doesn’t make sense to have negative weight. We pick a variety of positive values for \begin{align*}x\end{align*}, making sure to include some decimal values because prices can be decimals too. Then we use the function rule to find the value of \begin{align*}y\end{align*} for each value of \begin{align*}x\end{align*}.

\begin{align*}& x \quad 0 \quad 0.2 \quad 0.5 \quad 0.8 \quad 1 \quad 1.2 \quad 1.5 \quad 1.8 \quad 2 \quad 2.2 \quad 2.5 \quad 2.8 \quad 3 \quad 3.2 \quad 3.5\\ & y \quad 0 \quad 41 \quad \ 41 \quad \ 41 \quad 41 \quad 58 \quad 58 \quad \ 58 \quad 58 \quad 75 \quad 75 \quad \ 75 \quad 75 \quad 92 \quad 92\end{align*}

Watch this video for help with the Examples above.

CK-12 Foundation: Graph a Function from a Rule

### Guided Practice

*Graph the following function: \begin{align*}f(x)=\sqrt{x^2}\end{align*}*

**Solution**

Make a table of values. Even though \begin{align*}x\end{align*} can’t be negative inside the square root, because we are squaring \begin{align*}x\end{align*} first, the domain is all real numbers. So we integer values for \begin{align*}x\end{align*} which are on either side of zero. Use the function rule to find the value of \begin{align*}y\end{align*} for each value of \begin{align*}x\end{align*}.

\begin{align*}x\end{align*} | \begin{align*}y = f(x) = \sqrt{x^2}\end{align*} |
---|---|

-2 | \begin{align*}\sqrt{(-2)^2} = 2\end{align*} |

-1 | \begin{align*}\sqrt{(-1)^2} = 1\end{align*} |

0 | \begin{align*}\sqrt{0^2} = 0\end{align*} |

1 | \begin{align*}\sqrt{1^2} = 1\end{align*} |

2 | \begin{align*}\sqrt{2^2}=2\end{align*} |

Note that the range is all positive real numbers, and that this looks like an absolute value function.

### Explore More

Graph the following functions.

- Vanson spends $20 a month on his cat.
- Brandon is a member of a movie club. He pays a $50 annual membership and $8 per movie.
- \begin{align*}f(x) = (x - 2)^2\end{align*}
- \begin{align*}f(x) = 3.2^x\end{align*}
- \begin{align*}f(t) = 27t-t^2\end{align*}
- \begin{align*}f(w) = \frac{w}{4}+5\end{align*}
- \begin{align*}f(x) = t+2t^2+3t^3\end{align*}
- \begin{align*}f(x) = (x-1)(x+3)\end{align*}
- \begin{align*}f(x) = \frac{x}{3}+\frac{x^2}{5}\end{align*}
- \begin{align*}f(x) = \sqrt{2x}\end{align*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 1.13.