Answer :
Certainly! In order to solve this problem, we first need to understand and plot the functions [tex]\( y = 6 + 2x - x^2 \)[/tex] and [tex]\( y = x + 2 \)[/tex] on the same set of axes. Once we have the graphs plotted, we can use them to determine the points of intersection which are the solutions to the simultaneous equations. Let's go step by step.
### Step 1: Define the Functions
1. Function 1: [tex]\( y = 6 + 2x - x^2 \)[/tex]
2. Function 2: [tex]\( y = x + 2 \)[/tex]
### Step 2: Setting up the Axes
- For the x-axis, we use a scale of 2 cm to 1 unit.
- For the y-axis, we use a scale of 2 cm to 2 units.
### Step 3: Plot the Range for [tex]\( x \)[/tex]
- We will plot the graphs for [tex]\( -3 \leq x \leq 5 \)[/tex].
### Step 4: Calculate and Plot Points for Each Function
#### Function 1: [tex]\( y = 6 + 2x - x^2 \)[/tex]
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = 6 + 2(-3) - (-3)^2 = 6 - 6 - 9 = -9 \][/tex]
2. For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = 6 + 2(-2) - (-2)^2 = 6 - 4 - 4 = -2 \][/tex]
3. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 6 + 2(-1) - (-1)^2 = 6 - 2 - 1 = 3 \][/tex]
4. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 6 + 2(0) - 0^2 = 6 \][/tex]
5. For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 6 + 2(1) - 1^2 = 6 + 2 - 1 = 7 \][/tex]
6. For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 6 + 2(2) - 2^2 = 6 + 4 - 4 = 6 \][/tex]
7. For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 6 + 2(3) - 3^2 = 6 + 6 - 9 = 3 \][/tex]
8. For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 6 + 2(4) - 4^2 = 6 + 8 - 16 = -2 \][/tex]
9. For [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 6 + 2(5) - 5^2 = 6 + 10 - 25 = -9 \][/tex]
Plot these points and draw a smooth curve through them.
#### Function 2: [tex]\( y = x + 2 \)[/tex]
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = -3 + 2 = -1 \][/tex]
2. For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -2 + 2 = 0 \][/tex]
3. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -1 + 2 = 1 \][/tex]
4. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 0 + 2 = 2 \][/tex]
5. For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 1 + 2 = 3 \][/tex]
6. For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2 + 2 = 4 \][/tex]
7. For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 3 + 2 = 5 \][/tex]
8. For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 4 + 2 = 6 \][/tex]
9. For [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 5 + 2 = 7 \][/tex]
Plot these points and draw a straight line through them.
### Step 5: Finding the Intersection Points
To find the intersection points, use the graphs:
- Look for points where the parabola [tex]\( y = 6 + 2x - x^2 \)[/tex] intersects the line [tex]\( y = x + 2 \)[/tex].
- From observing the graph, identify the [tex]\( x \)[/tex]-coordinates of these intersection points.
### Step 6: Solve Algebraically (Confirm the Graphical Solution)
To confirm, solve the equation algebraically:
[tex]\[ x + 2 = 6 + 2x - x^2 \][/tex]
[tex]\[ x^2 - x - 4 = 0 \][/tex]
Solve using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], [tex]\( c = -4 \)[/tex]:
[tex]\[ x = \frac{1 \pm \sqrt{1 + 16}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{17}}{2} \][/tex]
Approximately:
[tex]\[ x \approx 2.6 \text{ and } x \approx -1.6 \][/tex]
So, the solutions are:
- [tex]\( x \approx 2.6 \)[/tex]
- [tex]\( x \approx -1.6 \)[/tex]
These [tex]\( x \)[/tex]-values correspond to the intersections observed.
Finally, let's solve the remaining equation:
[tex]\[ 2 - x = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
Thus, the solutions to the equations [tex]\( y = x + 2 \)[/tex] and [tex]\( y = 6 + 2x - x^2 \)[/tex] correct to one decimal place are:
[tex]\[ x \approx 2.6 \text{ and } x \approx -1.6 \][/tex]
And the solution for [tex]\( 2 - x = 0 \)[/tex] is:
[tex]\[ x = 2 \][/tex]
I hope this step-by-step solution helps! If anything is unclear, please let me know.
### Step 1: Define the Functions
1. Function 1: [tex]\( y = 6 + 2x - x^2 \)[/tex]
2. Function 2: [tex]\( y = x + 2 \)[/tex]
### Step 2: Setting up the Axes
- For the x-axis, we use a scale of 2 cm to 1 unit.
- For the y-axis, we use a scale of 2 cm to 2 units.
### Step 3: Plot the Range for [tex]\( x \)[/tex]
- We will plot the graphs for [tex]\( -3 \leq x \leq 5 \)[/tex].
### Step 4: Calculate and Plot Points for Each Function
#### Function 1: [tex]\( y = 6 + 2x - x^2 \)[/tex]
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = 6 + 2(-3) - (-3)^2 = 6 - 6 - 9 = -9 \][/tex]
2. For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = 6 + 2(-2) - (-2)^2 = 6 - 4 - 4 = -2 \][/tex]
3. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 6 + 2(-1) - (-1)^2 = 6 - 2 - 1 = 3 \][/tex]
4. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 6 + 2(0) - 0^2 = 6 \][/tex]
5. For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 6 + 2(1) - 1^2 = 6 + 2 - 1 = 7 \][/tex]
6. For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 6 + 2(2) - 2^2 = 6 + 4 - 4 = 6 \][/tex]
7. For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 6 + 2(3) - 3^2 = 6 + 6 - 9 = 3 \][/tex]
8. For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 6 + 2(4) - 4^2 = 6 + 8 - 16 = -2 \][/tex]
9. For [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 6 + 2(5) - 5^2 = 6 + 10 - 25 = -9 \][/tex]
Plot these points and draw a smooth curve through them.
#### Function 2: [tex]\( y = x + 2 \)[/tex]
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = -3 + 2 = -1 \][/tex]
2. For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -2 + 2 = 0 \][/tex]
3. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -1 + 2 = 1 \][/tex]
4. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 0 + 2 = 2 \][/tex]
5. For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 1 + 2 = 3 \][/tex]
6. For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2 + 2 = 4 \][/tex]
7. For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 3 + 2 = 5 \][/tex]
8. For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 4 + 2 = 6 \][/tex]
9. For [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 5 + 2 = 7 \][/tex]
Plot these points and draw a straight line through them.
### Step 5: Finding the Intersection Points
To find the intersection points, use the graphs:
- Look for points where the parabola [tex]\( y = 6 + 2x - x^2 \)[/tex] intersects the line [tex]\( y = x + 2 \)[/tex].
- From observing the graph, identify the [tex]\( x \)[/tex]-coordinates of these intersection points.
### Step 6: Solve Algebraically (Confirm the Graphical Solution)
To confirm, solve the equation algebraically:
[tex]\[ x + 2 = 6 + 2x - x^2 \][/tex]
[tex]\[ x^2 - x - 4 = 0 \][/tex]
Solve using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], [tex]\( c = -4 \)[/tex]:
[tex]\[ x = \frac{1 \pm \sqrt{1 + 16}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{17}}{2} \][/tex]
Approximately:
[tex]\[ x \approx 2.6 \text{ and } x \approx -1.6 \][/tex]
So, the solutions are:
- [tex]\( x \approx 2.6 \)[/tex]
- [tex]\( x \approx -1.6 \)[/tex]
These [tex]\( x \)[/tex]-values correspond to the intersections observed.
Finally, let's solve the remaining equation:
[tex]\[ 2 - x = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
Thus, the solutions to the equations [tex]\( y = x + 2 \)[/tex] and [tex]\( y = 6 + 2x - x^2 \)[/tex] correct to one decimal place are:
[tex]\[ x \approx 2.6 \text{ and } x \approx -1.6 \][/tex]
And the solution for [tex]\( 2 - x = 0 \)[/tex] is:
[tex]\[ x = 2 \][/tex]
I hope this step-by-step solution helps! If anything is unclear, please let me know.
