Answer :
Certainly! Let's break down the problem step-by-step to find the required solutions and coordinates.
### Step 1: Solutions of the Equation [tex]\(2x^2 = -x^2 - 5x - 1\)[/tex]
First, let's solve the equation [tex]\(2x^2 = -x^2 - 5x - 1\)[/tex]. Rearranging all terms to one side gives us:
[tex]\[ 2x^2 + x^2 + 5x + 1 = 0 \][/tex]
[tex]\[ 3x^2 + 5x + 1 = 0 \][/tex]
We solve the quadratic equation [tex]\(3x^2 + 5x + 1 = 0\)[/tex]. The solutions for [tex]\(x\)[/tex] are:
[tex]\[ x = -\frac{5}{6} - \frac{\sqrt{13}}{6} \][/tex]
[tex]\[ x = -\frac{5}{6} + \frac{\sqrt{13}}{6} \][/tex]
### Step 2: [tex]\(y\)[/tex]-Coordinates of the Intersection Points
Next, we need the [tex]\(y\)[/tex]-coordinates of the intersection points of the graphs [tex]\(y = 2x^2\)[/tex] and [tex]\(y = -x^2 - 5x - 1\)[/tex]. We substitute the [tex]\(x\)[/tex]-values found from the solutions into [tex]\(y = 2x^2\)[/tex]:
For [tex]\(x = -\frac{5}{6} - \frac{\sqrt{13}}{6}\)[/tex]:
[tex]\[ y = 2 \left( -\frac{5}{6} - \frac{\sqrt{13}}{6} \right)^2 \][/tex]
For [tex]\(x = -\frac{5}{6} + \frac{\sqrt{13}}{6} \)[/tex]:
[tex]\[ y = 2 \left( -\frac{5}{6} + \frac{\sqrt{13}}{6} \right)^2 \][/tex]
### Step 3: [tex]\(x\)[/tex]-Coordinates of the [tex]\(x\)[/tex]-Intercepts
Now, let's find the [tex]\(x\)[/tex]-coordinates of the [tex]\(x\)[/tex]-intercepts for the graphs [tex]\(y = 2x^2\)[/tex] and [tex]\(y = -x^2 - 5x - 1\)[/tex]:
For [tex]\(y = 2x^2\)[/tex]:
[tex]\[ 2x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
For [tex]\(y = -x^2 - 5x - 1\)[/tex], we solve:
[tex]\[ -x^2 - 5x - 1 = 0 \][/tex]
The solutions for [tex]\(x\)[/tex] are:
[tex]\[ x = -\frac{5}{2} - \frac{\sqrt{21}}{2} \][/tex]
[tex]\[ x = -\frac{5}{2} + \frac{\sqrt{21}}{2} \][/tex]
### Step 4: [tex]\(y\)[/tex]-Coordinates of the [tex]\(y\)[/tex]-Intercepts
Finally, we find the [tex]\(y\)[/tex]-coordinates of the [tex]\(y\)[/tex]-intercepts for the graphs [tex]\(y = 2x^2\)[/tex] and [tex]\(y = -x^2 - 5x - 1\)[/tex]:
For [tex]\(y = 2x^2\)[/tex], setting [tex]\(x = 0\)[/tex]:
[tex]\[ y = 2 \cdot 0^2 = 0 \][/tex]
For [tex]\(y = -x^2 - 5x - 1\)[/tex], setting [tex]\(x = 0\)[/tex]:
[tex]\[ y = -0^2 - 5\cdot0 - 1 = -1 \][/tex]
### Summary of Results
1. Solutions of the equation [tex]\(2x^2 = -x^2 - 5x - 1\)[/tex]:
[tex]\[ x = -\frac{5}{6} - \frac{\sqrt{13}}{6}, \quad x = -\frac{5}{6} + \frac{\sqrt{13}}{6} \][/tex]
2. [tex]\(y\)[/tex]-coordinates of the intersections points of [tex]\(y=2x^2\)[/tex] and [tex]\(y=-x^2-5x-1\)[/tex]:
[tex]\[ y = 2 \left( -\frac{5}{6} - \frac{\sqrt{13}}{6} \right)^2, \quad y = 2 \left( -\frac{5}{6} + \frac{\sqrt{13}}{6} \right)^2 \][/tex]
3. [tex]\(x\)[/tex]-coordinates of the [tex]\(x\)[/tex]-intercepts of [tex]\(y=2x^2\)[/tex] and [tex]\(y=-x^2-5x-1\)[/tex]:
[tex]\[ x = 0, \quad x = -\frac{5}{2} - \frac{\sqrt{21}}{2}, \quad x = -\frac{5}{2} + \frac{\sqrt{21}}{2} \][/tex]
4. [tex]\(y\)[/tex]-coordinates of the [tex]\(y\)[/tex]-intercepts of [tex]\(y=2x^2\)[/tex] and [tex]\(y=-x^2-5x-1\)[/tex]:
[tex]\[ y = 0, \quad y = -1 \][/tex]
### Step 1: Solutions of the Equation [tex]\(2x^2 = -x^2 - 5x - 1\)[/tex]
First, let's solve the equation [tex]\(2x^2 = -x^2 - 5x - 1\)[/tex]. Rearranging all terms to one side gives us:
[tex]\[ 2x^2 + x^2 + 5x + 1 = 0 \][/tex]
[tex]\[ 3x^2 + 5x + 1 = 0 \][/tex]
We solve the quadratic equation [tex]\(3x^2 + 5x + 1 = 0\)[/tex]. The solutions for [tex]\(x\)[/tex] are:
[tex]\[ x = -\frac{5}{6} - \frac{\sqrt{13}}{6} \][/tex]
[tex]\[ x = -\frac{5}{6} + \frac{\sqrt{13}}{6} \][/tex]
### Step 2: [tex]\(y\)[/tex]-Coordinates of the Intersection Points
Next, we need the [tex]\(y\)[/tex]-coordinates of the intersection points of the graphs [tex]\(y = 2x^2\)[/tex] and [tex]\(y = -x^2 - 5x - 1\)[/tex]. We substitute the [tex]\(x\)[/tex]-values found from the solutions into [tex]\(y = 2x^2\)[/tex]:
For [tex]\(x = -\frac{5}{6} - \frac{\sqrt{13}}{6}\)[/tex]:
[tex]\[ y = 2 \left( -\frac{5}{6} - \frac{\sqrt{13}}{6} \right)^2 \][/tex]
For [tex]\(x = -\frac{5}{6} + \frac{\sqrt{13}}{6} \)[/tex]:
[tex]\[ y = 2 \left( -\frac{5}{6} + \frac{\sqrt{13}}{6} \right)^2 \][/tex]
### Step 3: [tex]\(x\)[/tex]-Coordinates of the [tex]\(x\)[/tex]-Intercepts
Now, let's find the [tex]\(x\)[/tex]-coordinates of the [tex]\(x\)[/tex]-intercepts for the graphs [tex]\(y = 2x^2\)[/tex] and [tex]\(y = -x^2 - 5x - 1\)[/tex]:
For [tex]\(y = 2x^2\)[/tex]:
[tex]\[ 2x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
For [tex]\(y = -x^2 - 5x - 1\)[/tex], we solve:
[tex]\[ -x^2 - 5x - 1 = 0 \][/tex]
The solutions for [tex]\(x\)[/tex] are:
[tex]\[ x = -\frac{5}{2} - \frac{\sqrt{21}}{2} \][/tex]
[tex]\[ x = -\frac{5}{2} + \frac{\sqrt{21}}{2} \][/tex]
### Step 4: [tex]\(y\)[/tex]-Coordinates of the [tex]\(y\)[/tex]-Intercepts
Finally, we find the [tex]\(y\)[/tex]-coordinates of the [tex]\(y\)[/tex]-intercepts for the graphs [tex]\(y = 2x^2\)[/tex] and [tex]\(y = -x^2 - 5x - 1\)[/tex]:
For [tex]\(y = 2x^2\)[/tex], setting [tex]\(x = 0\)[/tex]:
[tex]\[ y = 2 \cdot 0^2 = 0 \][/tex]
For [tex]\(y = -x^2 - 5x - 1\)[/tex], setting [tex]\(x = 0\)[/tex]:
[tex]\[ y = -0^2 - 5\cdot0 - 1 = -1 \][/tex]
### Summary of Results
1. Solutions of the equation [tex]\(2x^2 = -x^2 - 5x - 1\)[/tex]:
[tex]\[ x = -\frac{5}{6} - \frac{\sqrt{13}}{6}, \quad x = -\frac{5}{6} + \frac{\sqrt{13}}{6} \][/tex]
2. [tex]\(y\)[/tex]-coordinates of the intersections points of [tex]\(y=2x^2\)[/tex] and [tex]\(y=-x^2-5x-1\)[/tex]:
[tex]\[ y = 2 \left( -\frac{5}{6} - \frac{\sqrt{13}}{6} \right)^2, \quad y = 2 \left( -\frac{5}{6} + \frac{\sqrt{13}}{6} \right)^2 \][/tex]
3. [tex]\(x\)[/tex]-coordinates of the [tex]\(x\)[/tex]-intercepts of [tex]\(y=2x^2\)[/tex] and [tex]\(y=-x^2-5x-1\)[/tex]:
[tex]\[ x = 0, \quad x = -\frac{5}{2} - \frac{\sqrt{21}}{2}, \quad x = -\frac{5}{2} + \frac{\sqrt{21}}{2} \][/tex]
4. [tex]\(y\)[/tex]-coordinates of the [tex]\(y\)[/tex]-intercepts of [tex]\(y=2x^2\)[/tex] and [tex]\(y=-x^2-5x-1\)[/tex]:
[tex]\[ y = 0, \quad y = -1 \][/tex]
