Answer :
Sure, let's go through each part of the problem step-by-step.
### 1. Solutions of [tex]\( x^2 - 2x - 4 = -3x + 9 \)[/tex]
To solve the equation [tex]\( x^2 - 2x - 4 = -3x + 9 \)[/tex], we start by setting the equations equal to each other and solving for [tex]\( x \)[/tex].
[tex]\[ x^2 - 2x - 4 = -3x + 9 \][/tex]
We bring all terms to one side to form a standard quadratic equation:
[tex]\[ x^2 - 2x - 4 + 3x - 9 = 0 \][/tex]
[tex]\[ x^2 + x - 13 = 0 \][/tex]
Solving this quadratic equation, we find the solutions:
[tex]\[ x = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
Thus, the solutions for the intersection points are:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
### 2. [tex]\( y \)[/tex]-Coordinates of the [tex]\( y \)[/tex]-Intercepts
The [tex]\( y \)[/tex]-intercept is found by setting [tex]\( x = 0 \)[/tex] in the equation:
For [tex]\( y = x^2 - 2x - 4 \)[/tex]:
[tex]\[ y = 0^2 - 2 \cdot 0 - 4 \][/tex]
[tex]\[ y = -4 \][/tex]
For [tex]\( y = -3x + 9 \)[/tex]:
[tex]\[ y = -3 \cdot 0 + 9 \][/tex]
[tex]\[ y = 9 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercepts are:
[tex]\[ y_{\text{intercept of } y = x^2 - 2x - 4} = -4 \][/tex]
[tex]\[ y_{\text{intercept of } y = -3x + 9} = 9 \][/tex]
### 3. [tex]\( x \)[/tex]-Coordinates of the [tex]\( x \)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercept is found by setting [tex]\( y = 0 \)[/tex] in the equation:
For [tex]\( y = x^2 - 2x - 4 \)[/tex]:
[tex]\[ x^2 - 2x - 4 = 0 \][/tex]
Solving the above quadratic equation, we get:
[tex]\[ x = 1 - \sqrt{5} \][/tex]
[tex]\[ x = 1 + \sqrt{5} \][/tex]
For [tex]\( y = -3x + 9 \)[/tex]:
[tex]\[ 0 = -3x + 9 \][/tex]
[tex]\[ 3x = 9 \][/tex]
[tex]\[ x = 3 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ x_{\text{intercepts of } y = x^2 - 2x - 4} = 1 - \sqrt{5}, \; 1 + \sqrt{5} \][/tex]
[tex]\[ x_{\text{intercept of } y = -3x + 9} = 3 \][/tex]
### 4. [tex]\( y \)[/tex]-Coordinates of the Intersection Points
We have already found the [tex]\( x \)[/tex]-coordinates of the intersection points as:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
To find the [tex]\( y \)[/tex]-coordinates of these intersection points, we substitute these [tex]\( x \)[/tex]-values back into either of the original equations (since they are equal at the intersection points):
For [tex]\( x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \)[/tex]:
[tex]\[ y_1 = \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 - 2 \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right) - 4 \][/tex]
For [tex]\( x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \)[/tex]:
[tex]\[ y_2 = \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 - 2 \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right) - 4 \][/tex]
So, the [tex]\( y \)[/tex]-coordinates of the intersection points are:
[tex]\[ y_1 = -\sqrt{53} - 3 + \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 \][/tex]
[tex]\[ y_2 = -3 + \sqrt{53} + \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 \][/tex]
### 5. [tex]\( x \)[/tex]-Coordinates of the Intersection Points
The solutions are found in step 1:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
Putting it all together, the results are as follows:
- [tex]\( x \)[/tex]-Coordinates of intersection points: [tex]\( x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \)[/tex], [tex]\( x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \)[/tex]
- [tex]\( y \)[/tex]-Intercepts: [tex]\( y_{\text{intercept of } y = x^2 - 2x - 4} = -4 \)[/tex], [tex]\( y_{\text{intercept of } y = -3x + 9} = 9 \)[/tex]
- [tex]\( x \)[/tex]-Intercepts: [tex]\( x_{\text{intercepts of } y = x^2 - 2x - 4} = 1 - \sqrt{5}, 1 + \sqrt{5} \)[/tex], [tex]\( x_{\text{intercept of } y = -3x + 9} = 3 \)[/tex]
- [tex]\( y \)[/tex]-Coordinates of intersection points: [tex]\( y_1 = -\sqrt{53} - 3 + \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 \)[/tex], [tex]\( y_2 = -3 + \sqrt{53} + \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 \)[/tex]
### 1. Solutions of [tex]\( x^2 - 2x - 4 = -3x + 9 \)[/tex]
To solve the equation [tex]\( x^2 - 2x - 4 = -3x + 9 \)[/tex], we start by setting the equations equal to each other and solving for [tex]\( x \)[/tex].
[tex]\[ x^2 - 2x - 4 = -3x + 9 \][/tex]
We bring all terms to one side to form a standard quadratic equation:
[tex]\[ x^2 - 2x - 4 + 3x - 9 = 0 \][/tex]
[tex]\[ x^2 + x - 13 = 0 \][/tex]
Solving this quadratic equation, we find the solutions:
[tex]\[ x = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
Thus, the solutions for the intersection points are:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
### 2. [tex]\( y \)[/tex]-Coordinates of the [tex]\( y \)[/tex]-Intercepts
The [tex]\( y \)[/tex]-intercept is found by setting [tex]\( x = 0 \)[/tex] in the equation:
For [tex]\( y = x^2 - 2x - 4 \)[/tex]:
[tex]\[ y = 0^2 - 2 \cdot 0 - 4 \][/tex]
[tex]\[ y = -4 \][/tex]
For [tex]\( y = -3x + 9 \)[/tex]:
[tex]\[ y = -3 \cdot 0 + 9 \][/tex]
[tex]\[ y = 9 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercepts are:
[tex]\[ y_{\text{intercept of } y = x^2 - 2x - 4} = -4 \][/tex]
[tex]\[ y_{\text{intercept of } y = -3x + 9} = 9 \][/tex]
### 3. [tex]\( x \)[/tex]-Coordinates of the [tex]\( x \)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercept is found by setting [tex]\( y = 0 \)[/tex] in the equation:
For [tex]\( y = x^2 - 2x - 4 \)[/tex]:
[tex]\[ x^2 - 2x - 4 = 0 \][/tex]
Solving the above quadratic equation, we get:
[tex]\[ x = 1 - \sqrt{5} \][/tex]
[tex]\[ x = 1 + \sqrt{5} \][/tex]
For [tex]\( y = -3x + 9 \)[/tex]:
[tex]\[ 0 = -3x + 9 \][/tex]
[tex]\[ 3x = 9 \][/tex]
[tex]\[ x = 3 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ x_{\text{intercepts of } y = x^2 - 2x - 4} = 1 - \sqrt{5}, \; 1 + \sqrt{5} \][/tex]
[tex]\[ x_{\text{intercept of } y = -3x + 9} = 3 \][/tex]
### 4. [tex]\( y \)[/tex]-Coordinates of the Intersection Points
We have already found the [tex]\( x \)[/tex]-coordinates of the intersection points as:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
To find the [tex]\( y \)[/tex]-coordinates of these intersection points, we substitute these [tex]\( x \)[/tex]-values back into either of the original equations (since they are equal at the intersection points):
For [tex]\( x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \)[/tex]:
[tex]\[ y_1 = \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 - 2 \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right) - 4 \][/tex]
For [tex]\( x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \)[/tex]:
[tex]\[ y_2 = \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 - 2 \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right) - 4 \][/tex]
So, the [tex]\( y \)[/tex]-coordinates of the intersection points are:
[tex]\[ y_1 = -\sqrt{53} - 3 + \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 \][/tex]
[tex]\[ y_2 = -3 + \sqrt{53} + \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 \][/tex]
### 5. [tex]\( x \)[/tex]-Coordinates of the Intersection Points
The solutions are found in step 1:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
Putting it all together, the results are as follows:
- [tex]\( x \)[/tex]-Coordinates of intersection points: [tex]\( x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \)[/tex], [tex]\( x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \)[/tex]
- [tex]\( y \)[/tex]-Intercepts: [tex]\( y_{\text{intercept of } y = x^2 - 2x - 4} = -4 \)[/tex], [tex]\( y_{\text{intercept of } y = -3x + 9} = 9 \)[/tex]
- [tex]\( x \)[/tex]-Intercepts: [tex]\( x_{\text{intercepts of } y = x^2 - 2x - 4} = 1 - \sqrt{5}, 1 + \sqrt{5} \)[/tex], [tex]\( x_{\text{intercept of } y = -3x + 9} = 3 \)[/tex]
- [tex]\( y \)[/tex]-Coordinates of intersection points: [tex]\( y_1 = -\sqrt{53} - 3 + \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 \)[/tex], [tex]\( y_2 = -3 + \sqrt{53} + \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 \)[/tex]
