Question
on the set of aces below solve the following system of equations graphically and steate the coordinates of all points in the soloution set
y=x^2+10x+24
6x+3y=-24
where do you put the graph
give me the right answer
graph the parabola and the line
y=x^2+10x+24
6x+3y=-24
where do you put the graph
give me the right answer
graph the parabola and the line
Answers
GPT-4o mini
To solve the system of equations graphically, we first need to rewrite the equations so they can be plotted.
### Equations:
1. **Parabola:** \( y = x^2 + 10x + 24 \)
2. **Line:** Rearranging \( 6x + 3y = -24 \) to solve for \( y \):
\[
3y = -6x - 24 \implies y = -2x - 8
\]
### Steps for Plotting:
1. **Plot the Parabola \( y = x^2 + 10x + 24 \)**:
- This is a downward-opening parabola. You can calculate a few points to plot it.
- Vertex: Use the vertex formula \( x = -\frac{b}{2a} \) where \( a = 1 \) and \( b = 10 \):
\[
x = -\frac{10}{2 \cdot 1} = -5
\]
- Calculate \( y \) at \( x = -5 \):
\[
y = (-5)^2 + 10(-5) + 24 = 25 - 50 + 24 = -1
\]
- So the vertex is at \( (-5, -1) \).
- Additional points:
- For \( x = -6 \): \( y = 36 - 60 + 24 = 0 \) (point: (-6, 0))
- For \( x = -4 \): \( y = 16 - 40 + 24 = 0 \) (point: (-4, 0))
- For \( x = -10 \): \( y = 100 - 100 + 24 = 24 \) (point: (-10, 24))
- For \( x = 0 \): \( y = 0 + 0 + 24 = 24 \) (point: (0, 24))
2. **Plot the Line \( y = -2x - 8 \)**:
- You can plot some points as well.
- For \( x = 0 \): \( y = -8 \) (point: (0, -8))
- For \( x = -5 \): \( y = -2(-5) - 8 = 10 - 8 = 2 \) (point: (-5, 2))
- For \( x = -4 \): \( y = -2(-4) - 8 = 8 - 8 = 0 \) (point: (-4, 0))
- For \( x = -6 \): \( y = -2(-6) - 8 = 12 - 8 = 4 \) (point: (-6, 4))
### Finding Points of Intersection:
To find the intersection points of \( y = x^2 + 10x + 24 \) and \( y = -2x - 8 \), we can set them equal to each other:
\[
x^2 + 10x + 24 = -2x - 8
\]
Rearranging gives:
\[
x^2 + 12x + 32 = 0
\]
### Solve the Quadratic Equation:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 12 \), and \( c = 32 \):
\[
x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1}
\]
\[
x = \frac{-12 \pm \sqrt{144 - 128}}{2}
\]
\[
x = \frac{-12 \pm \sqrt{16}}{2}
\]
\[
x = \frac{-12 \pm 4}{2}
\]
So we obtain:
1. \( x = \frac{-12 + 4}{2} = -4 \)
2. \( x = \frac{-12 - 4}{2} = -8 \)
### Finding corresponding y-values:
- For \( x = -4 \):
\[
y = -2(-4) - 8 = 8 - 8 = 0 \quad \text{(point: (-4, 0))}
\]
- For \( x = -8 \):
\[
y = -2(-8) - 8 = 16 - 8 = 8 \quad \text{(point: (-8, 8))}
\]
### Solution Set:
The points of intersection (the solution set) are:
- \( (-4, 0) \)
- \( (-8, 8) \)
### Graphing:
You can use graphing software or graphing tools to plot the equations. The parabola will open upward, and the line will intersect it at the points mentioned above.
### Summary:
The solutions to the given system of equations graphically intersect at the coordinates:
- \( (-4, 0) \)
- \( (-8, 8) \)
### Equations:
1. **Parabola:** \( y = x^2 + 10x + 24 \)
2. **Line:** Rearranging \( 6x + 3y = -24 \) to solve for \( y \):
\[
3y = -6x - 24 \implies y = -2x - 8
\]
### Steps for Plotting:
1. **Plot the Parabola \( y = x^2 + 10x + 24 \)**:
- This is a downward-opening parabola. You can calculate a few points to plot it.
- Vertex: Use the vertex formula \( x = -\frac{b}{2a} \) where \( a = 1 \) and \( b = 10 \):
\[
x = -\frac{10}{2 \cdot 1} = -5
\]
- Calculate \( y \) at \( x = -5 \):
\[
y = (-5)^2 + 10(-5) + 24 = 25 - 50 + 24 = -1
\]
- So the vertex is at \( (-5, -1) \).
- Additional points:
- For \( x = -6 \): \( y = 36 - 60 + 24 = 0 \) (point: (-6, 0))
- For \( x = -4 \): \( y = 16 - 40 + 24 = 0 \) (point: (-4, 0))
- For \( x = -10 \): \( y = 100 - 100 + 24 = 24 \) (point: (-10, 24))
- For \( x = 0 \): \( y = 0 + 0 + 24 = 24 \) (point: (0, 24))
2. **Plot the Line \( y = -2x - 8 \)**:
- You can plot some points as well.
- For \( x = 0 \): \( y = -8 \) (point: (0, -8))
- For \( x = -5 \): \( y = -2(-5) - 8 = 10 - 8 = 2 \) (point: (-5, 2))
- For \( x = -4 \): \( y = -2(-4) - 8 = 8 - 8 = 0 \) (point: (-4, 0))
- For \( x = -6 \): \( y = -2(-6) - 8 = 12 - 8 = 4 \) (point: (-6, 4))
### Finding Points of Intersection:
To find the intersection points of \( y = x^2 + 10x + 24 \) and \( y = -2x - 8 \), we can set them equal to each other:
\[
x^2 + 10x + 24 = -2x - 8
\]
Rearranging gives:
\[
x^2 + 12x + 32 = 0
\]
### Solve the Quadratic Equation:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 12 \), and \( c = 32 \):
\[
x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1}
\]
\[
x = \frac{-12 \pm \sqrt{144 - 128}}{2}
\]
\[
x = \frac{-12 \pm \sqrt{16}}{2}
\]
\[
x = \frac{-12 \pm 4}{2}
\]
So we obtain:
1. \( x = \frac{-12 + 4}{2} = -4 \)
2. \( x = \frac{-12 - 4}{2} = -8 \)
### Finding corresponding y-values:
- For \( x = -4 \):
\[
y = -2(-4) - 8 = 8 - 8 = 0 \quad \text{(point: (-4, 0))}
\]
- For \( x = -8 \):
\[
y = -2(-8) - 8 = 16 - 8 = 8 \quad \text{(point: (-8, 8))}
\]
### Solution Set:
The points of intersection (the solution set) are:
- \( (-4, 0) \)
- \( (-8, 8) \)
### Graphing:
You can use graphing software or graphing tools to plot the equations. The parabola will open upward, and the line will intersect it at the points mentioned above.
### Summary:
The solutions to the given system of equations graphically intersect at the coordinates:
- \( (-4, 0) \)
- \( (-8, 8) \)