Sure, let's complete the function table step-by-step and plot the points on the graph. We'll find the value of the function [tex]\( f(x) = -x^2 + 2x + 5 \)[/tex] for each [tex]\( x \)[/tex] in the given domain: [tex]\( -1, 0, 1, 2, 3 \)[/tex].
1. For [tex]\( x = -1 \)[/tex]:
[tex]\[
f(-1) = -(-1)^2 + 2(-1) + 5 = -1 - 2 + 5 = 2
\][/tex]
So, [tex]\( f(-1) = 2 \)[/tex].
2. For [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = -(0)^2 + 2(0) + 5 = 0 + 0 + 5 = 5
\][/tex]
So, [tex]\( f(0) = 5 \)[/tex].
3. For [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = -(1)^2 + 2(1) + 5 = -1 + 2 + 5 = 6
\][/tex]
So, [tex]\( f(1) = 6 \)[/tex].
4. For [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = -(2)^2 + 2(2) + 5 = -4 + 4 + 5 = 5
\][/tex]
So, [tex]\( f(2) = 5 \)[/tex].
5. For [tex]\( x = 3 \)[/tex]:
[tex]\[
f(3) = -(3)^2 + 2(3) + 5 = -9 + 6 + 5 = 2
\][/tex]
So, [tex]\( f(3) = 2 \)[/tex].
Now let's fill in the function table:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & -1 & 0 & 1 & 2 & 3 \\
\hline
$f(x)$ & 2 & 5 & 6 & 5 & 2 \\
\hline
\end{tabular}
\][/tex]
Next, we plot the points [tex]\((-1, 2)\)[/tex], [tex]\((0, 5)\)[/tex], [tex]\((1, 6)\)[/tex], [tex]\((2, 5)\)[/tex], and [tex]\((3, 2)\)[/tex] on the graph.
1. Point [tex]\((-1, 2)\)[/tex] is [tex]\( 1 \)[/tex] unit to the left of the y-axis (negative direction) and [tex]\( 2 \)[/tex] units up.
2. Point [tex]\((0, 5)\)[/tex] is on the y-axis itself and [tex]\( 5 \)[/tex] units up.
3. Point [tex]\((1, 6)\)[/tex] is [tex]\( 1 \)[/tex] unit to the right of the y-axis (positive direction) and [tex]\( 6 \)[/tex] units up.
4. Point [tex]\((2, 5)\)[/tex] is [tex]\( 2 \)[/tex] units to the right of the y-axis (positive direction) and [tex]\( 5 \)[/tex] units up.
5. Point [tex]\((3, 2)\)[/tex] is [tex]\( 3 \)[/tex] units to the right of the y-axis (positive direction) and [tex]\( 2 \)[/tex] units up.
Make sure to properly label the axes and the points as you plot them on the graph. This will help visualize the parabolic curve described by the function [tex]\( f(x) = -x^2 + 2x + 5 \)[/tex].
