Answer :
Certainly! Let's break this down step by step:
### Part (a) Complete the table of values
We are given the linear equation \( y = 2x + 3 \) and a set of \( x \)-values: \(-2, -1, 0, 1, 2, 3\). We need to calculate the corresponding \( y \)-values.
1. When \( x = -2 \):
[tex]\[ y = 2(-2) + 3 = -4 + 3 = -1 \][/tex]
2. When \( x = -1 \):
[tex]\[ y = 2(-1) + 3 = -2 + 3 = 1 \][/tex]
3. When \( x = 0 \):
[tex]\[ y = 2(0) + 3 = 0 + 3 = 3 \][/tex]
4. When \( x = 1 \):
[tex]\[ y = 2(1) + 3 = 2 + 3 = 5 \][/tex]
5. When \( x = 2 \):
[tex]\[ y = 2(2) + 3 = 4 + 3 = 7 \][/tex]
6. When \( x = 3 \):
[tex]\[ y = 2(3) + 3 = 6 + 3 = 9 \][/tex]
Now, we can complete the table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline [tex]$x$[/tex] & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$y$[/tex] & -1 & 1 & 3 & 5 & 7 & 9 \\
\hline
\end{tabular}
\][/tex]
### Part (b) Draw the graph of \( y = 2x + 3 \)
1. Draw the Axes: First, draw the \( x \)-axis and \( y \)-axis on the grid. Label each axis with appropriate values.
2. Plot the Points: Based on the completed table, plot the points \((-2, -1)\), \((-1, 1)\), \( (0, 3) \), \( (1, 5) \), \( (2, 7) \), and \( (3, 9) \).
3. Draw the Line: Since \( y = 2x + 3 \) is a linear equation, the points should form a straight line. After plotting the points, draw a straight line through them. Make sure the line extends across the entire grid, indicating that it continues indefinitely in both directions.
4. Label the Line: Optionally, label the line with the equation \( y = 2x + 3 \).
And there you have it! Remember, the graph should be a straight line intersecting the [tex]\( y \)[/tex]-axis at [tex]\( y = 3 \)[/tex] (the y-intercept) and having a slope of 2 (indicating that for each unit increase in [tex]\( x \)[/tex], [tex]\( y \)[/tex] increases by 2 units).
### Part (a) Complete the table of values
We are given the linear equation \( y = 2x + 3 \) and a set of \( x \)-values: \(-2, -1, 0, 1, 2, 3\). We need to calculate the corresponding \( y \)-values.
1. When \( x = -2 \):
[tex]\[ y = 2(-2) + 3 = -4 + 3 = -1 \][/tex]
2. When \( x = -1 \):
[tex]\[ y = 2(-1) + 3 = -2 + 3 = 1 \][/tex]
3. When \( x = 0 \):
[tex]\[ y = 2(0) + 3 = 0 + 3 = 3 \][/tex]
4. When \( x = 1 \):
[tex]\[ y = 2(1) + 3 = 2 + 3 = 5 \][/tex]
5. When \( x = 2 \):
[tex]\[ y = 2(2) + 3 = 4 + 3 = 7 \][/tex]
6. When \( x = 3 \):
[tex]\[ y = 2(3) + 3 = 6 + 3 = 9 \][/tex]
Now, we can complete the table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline [tex]$x$[/tex] & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$y$[/tex] & -1 & 1 & 3 & 5 & 7 & 9 \\
\hline
\end{tabular}
\][/tex]
### Part (b) Draw the graph of \( y = 2x + 3 \)
1. Draw the Axes: First, draw the \( x \)-axis and \( y \)-axis on the grid. Label each axis with appropriate values.
2. Plot the Points: Based on the completed table, plot the points \((-2, -1)\), \((-1, 1)\), \( (0, 3) \), \( (1, 5) \), \( (2, 7) \), and \( (3, 9) \).
3. Draw the Line: Since \( y = 2x + 3 \) is a linear equation, the points should form a straight line. After plotting the points, draw a straight line through them. Make sure the line extends across the entire grid, indicating that it continues indefinitely in both directions.
4. Label the Line: Optionally, label the line with the equation \( y = 2x + 3 \).
And there you have it! Remember, the graph should be a straight line intersecting the [tex]\( y \)[/tex]-axis at [tex]\( y = 3 \)[/tex] (the y-intercept) and having a slope of 2 (indicating that for each unit increase in [tex]\( x \)[/tex], [tex]\( y \)[/tex] increases by 2 units).
