We choose several convenient values for \(x\) and evaluate the function to find the corresponding \(f(x)\)-values. For this function we cannot choose \(x\)-values less than \(-4\text{,}\) because the square root of a negative number is not a real number.

\begin{equation*}
f(\alert{-4}) =\sqrt{\alert{-4} + 4}=\sqrt{0}= 0
\end{equation*}

\begin{equation*}
f(\alert{-3}) =\sqrt{\alert{-3} + 4}=\sqrt{1}= 1
\end{equation*}

\begin{equation*}
f(\alert{0}) =\sqrt{\alert{0} + 4}=\sqrt{4}=2
\end{equation*}

\begin{equation*}
f(\alert{2}) =\sqrt{\alert{2} + 4}=\sqrt{6}\approx 2.45
\end{equation*}

\begin{equation*}
f(\alert{5}) =\sqrt{\alert{5} + 4}=\sqrt{9}=3
\end{equation*}

The results are shown in the table.

\(x\) |
\(f(x)\) |

\(-4\) |
\(0\) |

\(-3\) |
\(1\) |

\(0\) |
\(2\) |

\(2\) |
\(\sqrt{6}\) |

\(5\) |
\(3\) |

Points on the graph have coordinates \((x, f(x))\text{,}\) so the vertical coordinate of each popint is given by the value of \(f(x)\text{.}\) We plot the points and connect them with a smooth curve, as shown in the figure. Notice that no points on the graph have \(x\)-coordinates less than \(-4\text{.}\)