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
y=x^2+10x+24
6x+3y=-24
where do you put the graph
Answers
GPT-4o mini
To solve the system of equations graphically, we will plot the two equations and find their intersection points.
1. **Equation 1**: \( y = x^2 + 10x + 24 \)
This is a quadratic equation, and it represents a parabola that opens upwards.
2. **Equation 2**: \( 6x + 3y = -24 \)
We can rearrange this into slope-intercept form (y = mx + b):
\[
3y = -6x - 24 \implies y = -2x - 8
\]
This represents a straight line with a slope of -2 and a y-intercept at -8.
### Step 1: Plotting the Equations
To plot these equations, we can find a few key points for both curves.
**For the quadratic equation:**
To find some points for \( y = x^2 + 10x + 24 \):
- When \( x = -12 \):
\( y = (-12)^2 + 10(-12) + 24 = 144 - 120 + 24 = 48 \) -> Point (-12, 48)
- When \( x = -8 \):
\( y = (-8)^2 + 10(-8) + 24 = 64 - 80 + 24 = 8 \) -> Point (-8, 8)
- When \( x = -4 \):
\( y = (-4)^2 + 10(-4) + 24 = 16 - 40 + 24 = 0 \) -> Point (-4, 0)
- When \( x = 0 \):
\( y = (0)^2 + 10(0) + 24 = 24 \) -> Point (0, 24)
**For the linear equation:**
To find some points for \( y = -2x - 8 \):
- When \( x = -8 \):
\( y = -2(-8) - 8 = 16 - 8 = 8 \) -> Point (-8, 8)
- When \( x = 0 \):
\( y = -2(0) - 8 = -8 \) -> Point (0, -8)
- When \( x = 4 \):
\( y = -2(4) - 8 = -8 - 8 = -16 \) -> Point (4, -16)
### Step 2: Plotting the Graph
You can use graphing software, a graphing calculator, or graph paper to plot these points.
- The parabolic curve \( y = x^2 + 10x + 24 \) will be a U-shaped graph.
- The line \( y = -2x - 8 \) will be a straight line descending from left to right.
### Step 3: Finding Intersection Points
To find where the curves intersect, we can set the equations equal to each other:
\[
x^2 + 10x + 24 = -2x - 8
\]
Rearranging gives us:
\[
x^2 + 10x + 2x + 24 + 8 = 0 \implies x^2 + 12x + 32 = 0
\]
Now we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} = \frac{-12 \pm \sqrt{144 - 128}}{2} = \frac{-12 \pm \sqrt{16}}{2} = \frac{-12 \pm 4}{2}
\]
Calculating the two possible solutions:
1. \( x = \frac{-12 + 4}{2} = \frac{-8}{2} = -4 \)
2. \( x = \frac{-12 - 4}{2} = \frac{-16}{2} = -8 \)
### Step 4: Finding Corresponding \( y \)-values
Now we can find the corresponding y-values for these x-values using either equation. Let's use the linear equation \( y = -2x - 8 \):
- For \( x = -4 \):
\[
y = -2(-4) - 8 = 8 - 8 = 0 \implies (-4, 0)
\]
- For \( x = -8 \):
\[
y = -2(-8) - 8 = 16 - 8 = 8 \implies (-8, 8)
\]
### Solution Set
The coordinates of the points in the solution set are:
1. \( (-4, 0) \)
2. \( (-8, 8) \)
These two points are where the parabola and the line intersect.
1. **Equation 1**: \( y = x^2 + 10x + 24 \)
This is a quadratic equation, and it represents a parabola that opens upwards.
2. **Equation 2**: \( 6x + 3y = -24 \)
We can rearrange this into slope-intercept form (y = mx + b):
\[
3y = -6x - 24 \implies y = -2x - 8
\]
This represents a straight line with a slope of -2 and a y-intercept at -8.
### Step 1: Plotting the Equations
To plot these equations, we can find a few key points for both curves.
**For the quadratic equation:**
To find some points for \( y = x^2 + 10x + 24 \):
- When \( x = -12 \):
\( y = (-12)^2 + 10(-12) + 24 = 144 - 120 + 24 = 48 \) -> Point (-12, 48)
- When \( x = -8 \):
\( y = (-8)^2 + 10(-8) + 24 = 64 - 80 + 24 = 8 \) -> Point (-8, 8)
- When \( x = -4 \):
\( y = (-4)^2 + 10(-4) + 24 = 16 - 40 + 24 = 0 \) -> Point (-4, 0)
- When \( x = 0 \):
\( y = (0)^2 + 10(0) + 24 = 24 \) -> Point (0, 24)
**For the linear equation:**
To find some points for \( y = -2x - 8 \):
- When \( x = -8 \):
\( y = -2(-8) - 8 = 16 - 8 = 8 \) -> Point (-8, 8)
- When \( x = 0 \):
\( y = -2(0) - 8 = -8 \) -> Point (0, -8)
- When \( x = 4 \):
\( y = -2(4) - 8 = -8 - 8 = -16 \) -> Point (4, -16)
### Step 2: Plotting the Graph
You can use graphing software, a graphing calculator, or graph paper to plot these points.
- The parabolic curve \( y = x^2 + 10x + 24 \) will be a U-shaped graph.
- The line \( y = -2x - 8 \) will be a straight line descending from left to right.
### Step 3: Finding Intersection Points
To find where the curves intersect, we can set the equations equal to each other:
\[
x^2 + 10x + 24 = -2x - 8
\]
Rearranging gives us:
\[
x^2 + 10x + 2x + 24 + 8 = 0 \implies x^2 + 12x + 32 = 0
\]
Now we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} = \frac{-12 \pm \sqrt{144 - 128}}{2} = \frac{-12 \pm \sqrt{16}}{2} = \frac{-12 \pm 4}{2}
\]
Calculating the two possible solutions:
1. \( x = \frac{-12 + 4}{2} = \frac{-8}{2} = -4 \)
2. \( x = \frac{-12 - 4}{2} = \frac{-16}{2} = -8 \)
### Step 4: Finding Corresponding \( y \)-values
Now we can find the corresponding y-values for these x-values using either equation. Let's use the linear equation \( y = -2x - 8 \):
- For \( x = -4 \):
\[
y = -2(-4) - 8 = 8 - 8 = 0 \implies (-4, 0)
\]
- For \( x = -8 \):
\[
y = -2(-8) - 8 = 16 - 8 = 8 \implies (-8, 8)
\]
### Solution Set
The coordinates of the points in the solution set are:
1. \( (-4, 0) \)
2. \( (-8, 8) \)
These two points are where the parabola and the line intersect.