First, let's focus on graphing each equation step by step:
### Step 1: Graph the Linear Equation [tex]\( y = x - 8 \)[/tex]
1. Find the Y-intercept (b):
The equation [tex]\( y = x - 8 \)[/tex] has a y-intercept of -8. This means it crosses the y-axis at the point (0, -8).
2. Find the Slope (m):
The slope of the line is 1, indicating that for every unit increase in [tex]\( x \)[/tex], [tex]\( y \)[/tex] increases by 1.
3. Plot Points Using the Slope:
Start at the point (0, -8). From this point, move 1 unit to the right and 1 unit up to get to (1, -7). Repeat this to get multiple points.
4. Draw the Line:
Connect the points with a straight line.
### Step 2: Graph the Quadratic Equation [tex]\( y = x^2 - 2x - 3 \)[/tex]
1. Find the Y-intercept:
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[
y = 0^2 - 2(0) - 3 = -3
\][/tex]
This means it crosses the y-axis at the point (0, -3).
2. Find the Vertex:
The vertex form of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] can be determined by using the formula [tex]\( x = -\frac{b}{2a} \)[/tex]. Here, [tex]\( a = 1 \)[/tex] and [tex]\( b = -2 \)[/tex]:
[tex]\[
x = -\frac{-2}{2 \cdot 1} = 1
\][/tex]
Substitute [tex]\( x = 1 \)[/tex] back into the equation to find the y-coordinate of the vertex:
[tex]\[
y = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4
\][/tex]
So, the vertex is at (1, -4).
3. Plot Additional Points:
Plot the points around the vertex. For example:
- At [tex]\( x = -1 \)[/tex]:
[tex]\[
y = (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0
\][/tex]
- At [tex]\( x = 2 \)[/tex]:
[tex]\[
y = (2)^2 - 2(2) - 3 = 4 - 4 - 3 = -3
\][/tex]
- At [tex]\( x = 3 \)[/tex]:
[tex]\[
y = (3)^2 - 2(3) - 3 = 9 - 6 - 3 = 0
\][/tex]
4. Draw the Parabola:
Connect these points to form a parabola. Ensure it passes through the vertex (1, -4) and other calculated points.
### Step 3: Identify Intersections
After graphing both equations on the same coordinate plane, observe the points where the two graphs intersect. These intersection points represent the solutions to the system of equations.
### Conclusion:
Based on the given equations and their graphs:
- Intersection Points: After graphing and analyzing, you should see that the line [tex]\( y = x - 8 \)[/tex] and the parabola [tex]\( y = x^2 - 2x - 3 \)[/tex] intersect at the points where both [tex]\( y \)[/tex]- values are equal for the same [tex]\( x \)[/tex]- values.
To identify the solutions accurately, solve the system algebraically:
Set the equations equal to each other:
[tex]\[
x - 8 = x^2 - 2x - 3
\][/tex]
Rearrange and solve for [tex]\( x \)[/tex]:
[tex]\[
x^2 - 3x + 5 = 0
\][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[
a = 1, b = -3, c = 5
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{9 - 20}}{2}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{-11}}{2}
\][/tex]
Since the discriminant ([tex]\( -11 \)[/tex]) is negative, there are no real solutions.
This means that the graphs of the line and the parabola do not intersect at any point in the real plane.
### Matched Answer:
- Answer Choice: A. No solution
