Certainly! Let's solve the problem step-by-step and create a table to help us graph the quadratic function [tex]\(y = \frac{1}{2} x^2\)[/tex] using the given [tex]\(x\)[/tex]-values [tex]\(-4, -2, 0, 2, 4\)[/tex].
### Step-by-Step Solution:
1. Identify the given [tex]\(x\)[/tex]-values: [tex]\(-4, -2, 0, 2, 4\)[/tex].
2. Substitute each [tex]\(x\)[/tex]-value into the equation [tex]\(y = \frac{1}{2} x^2\)[/tex] and calculate the corresponding [tex]\(y\)[/tex]-values:
Let's compute each one:
- For [tex]\(x = -4\)[/tex]:
[tex]\[
y = \frac{1}{2} (-4)^2 = \frac{1}{2} \times 16 = 8.0
\][/tex]
- For [tex]\(x = -2\)[/tex]:
[tex]\[
y = \frac{1}{2} (-2)^2 = \frac{1}{2} \times 4 = 2.0
\][/tex]
- For [tex]\(x = 0\)[/tex]:
[tex]\[
y = \frac{1}{2} (0)^2 = 0.0
\][/tex]
- For [tex]\(x = 2\)[/tex]:
[tex]\[
y = \frac{1}{2} (2)^2 = \frac{1}{2} \times 4 = 2.0
\][/tex]
- For [tex]\(x = 4\)[/tex]:
[tex]\[
y = \frac{1}{2} (4)^2 = \frac{1}{2} \times 16 = 8.0
\][/tex]
### Table of Values:
Now, let's put these values into a table format:
[tex]\[
\begin{array}{c|c}
x & y = \frac{1}{2} x^2 \\
\hline
-4 & 8.0 \\
-2 & 2.0 \\
0 & 0.0 \\
2 & 2.0 \\
4 & 8.0 \\
\end{array}
\][/tex]
### Graphing the Points:
To graph the quadratic function, plot the points [tex]\((-4, 8.0)\)[/tex], [tex]\((-2, 2.0)\)[/tex], [tex]\((0, 0.0)\)[/tex], [tex]\((2, 2.0)\)[/tex], and [tex]\((4, 8.0)\)[/tex] on a coordinate plane. Connect these points smoothly to form a parabola.
### Final Graph:
The graph of the function [tex]\(y = \frac{1}{2} x^2\)[/tex] will be a parabola opening upwards, with its vertex at the origin (0,0). The points we plotted fall symmetrically on either side of the y-axis, showing the characteristic shape of a quadratic function.
This completes our step-by-step solution and allows us to visualize the quadratic function through the table and graphing process!
