To find 5 solutions to the equation [tex]\( y = 0.5x + 1 \)[/tex], let's carefully evaluate the equation for specific values of [tex]\( x \)[/tex]. We'll substitute these [tex]\( x \)[/tex] values into the equation and solve for [tex]\( y \)[/tex]. The values of [tex]\( x \)[/tex] we'll use are: -2, -1, 0, 1, and 2.
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = 0.5(-2) + 1 = -1 + 1 = 0
\][/tex]
So the first point is (-2, 0.0).
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = 0.5(-1) + 1 = -0.5 + 1 = 0.5
\][/tex]
So the second point is (-1, 0.5).
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
y = 0.5(0) + 1 = 0 + 1 = 1.0
\][/tex]
So the third point is (0, 1.0).
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 0.5(1) + 1 = 0.5 + 1 = 1.5
\][/tex]
So the fourth point is (1, 1.5).
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = 0.5(2) + 1 = 1 + 1 = 2.0
\][/tex]
So the fifth point is (2, 2.0).
Summarizing the solutions, we get the coordinates:
- (-2, 0.0)
- (-1, 0.5)
- (0, 1.0)
- (1, 1.5)
- (2, 2.0)
Now, let's complete the table with these solutions:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
-2 & 0.0 \\
\hline
-1 & 0.5 \\
\hline
0 & 1.0 \\
\hline
1 & 1.5 \\
\hline
2 & 2.0 \\
\hline
\end{tabular}
\][/tex]
Next, we plot the points and graph the function.
The points to be plotted are:
- (-2, 0.0)
- (-1, 0.5)
- (0, 1.0)
- (1, 1.5)
- (2, 2.0)
The graph of the equation [tex]\( y = 0.5x + 1 \)[/tex] is a straight line.
To draw the graph:
- Plot the points on a coordinate plane.
- Draw a straight line through these points since the equation represents a linear function.
Here is a visualization of the plotted points and the straight line representing [tex]\( y = 0.5x + 1 \)[/tex]:
```
Graph of y = 0.5x + 1
Y
^
| (2,2)
| /
| /
| /
| /
| /
| /
| /
| /
-+----------------------------------------> X
-2 -1 0 1 2
(-2, 0) (-1, 0.5) (0, 1) (1, 1.5) (2, 2)
```
The line passes through all the plotted points, which verifies that the solutions are correct.
