Answer :
To solve this system of equations:
[tex]\[ \left\{\begin{array}{l} -2x + y = -2 \\ y = -\frac{1}{2}x^2 + 3x + 2 \end{array}\right. \][/tex]
we first rearrange and solve the linear equation:
1. Rearrange the Linear Equation:
[tex]\[ -2x + y = -2 \quad \implies \quad y = 2x - 2 \][/tex]
2. Intercept Points:
These are the points where the linear equation intersects the quadratic equation.
Given the solution [tex]\([-2.0, 4.0]\)[/tex], we have the x-coordinates where the quadratic and linear equations intersect.
3. Plotting the Linear Equation on the Graph:
- Identify two points that satisfy the equation [tex]\(y = 2x - 2\)[/tex]. Let's take [tex]\(x = 0\)[/tex] and [tex]\(x = 1\)[/tex] for simplicity:
- For [tex]\(x = 0\)[/tex]:
[tex]\[ y = 2(0) - 2 = -2 \][/tex]
So, we have point [tex]\( (0, -2) \)[/tex].
- For [tex]\(x = 1\)[/tex]:
[tex]\[ y = 2(1) - 2 = 0 \][/tex]
So, we have point [tex]\( (1, 0) \)[/tex].
- Now draw a line through the points [tex]\((0, -2)\)[/tex] and [tex]\((1, 0)\)[/tex].
4. Mark Points of Intersection:
- With [tex]\(x = -2\)[/tex]:
Substitute [tex]\(x = -2\)[/tex] back into either equation to find [tex]\(y\)[/tex]:
For the quadratic equation:
[tex]\[ y = -\frac{1}{2}(-2)^2 + 3(-2) + 2 \implies y = -2 - 6 + 2 = -6 \][/tex]
So, point [tex]\((-2, -6)\)[/tex].
- With [tex]\(x = 4\)[/tex]:
Substitute [tex]\(x = 4\)[/tex] back into either equation to find [tex]\(y\)[/tex]:
For the quadratic equation:
[tex]\[ y = -\frac{1}{2}(4)^2 + 3(4) + 2 \implies y = -8/2 + 12 + 2 = 12/2 + 2 = 6 + 2 = 8 \][/tex]
So, point [tex]\(( 4, 8 )\)[/tex].
5. Marking Points of Intersection on the Graph:
Mark the points [tex]\((-2, -6)\)[/tex] and [tex]\((4, 8)\)[/tex] on the graph using the "Mark Feature".
To help visualize, the steps in summary:
- Draw the line [tex]\(y = 2x - 2\)[/tex].
- Plot the points [tex]\((0, -2)\)[/tex] and [tex]\((1, 0)\)[/tex] and draw a straight line through them.
- Mark the intersection points [tex]\((-2, -6)\)[/tex] and [tex]\((4, 8)\)[/tex] on the graph.
These marked points represent the solutions to the system of equations where both the linear and quadratic equations intersect.
[tex]\[ \left\{\begin{array}{l} -2x + y = -2 \\ y = -\frac{1}{2}x^2 + 3x + 2 \end{array}\right. \][/tex]
we first rearrange and solve the linear equation:
1. Rearrange the Linear Equation:
[tex]\[ -2x + y = -2 \quad \implies \quad y = 2x - 2 \][/tex]
2. Intercept Points:
These are the points where the linear equation intersects the quadratic equation.
Given the solution [tex]\([-2.0, 4.0]\)[/tex], we have the x-coordinates where the quadratic and linear equations intersect.
3. Plotting the Linear Equation on the Graph:
- Identify two points that satisfy the equation [tex]\(y = 2x - 2\)[/tex]. Let's take [tex]\(x = 0\)[/tex] and [tex]\(x = 1\)[/tex] for simplicity:
- For [tex]\(x = 0\)[/tex]:
[tex]\[ y = 2(0) - 2 = -2 \][/tex]
So, we have point [tex]\( (0, -2) \)[/tex].
- For [tex]\(x = 1\)[/tex]:
[tex]\[ y = 2(1) - 2 = 0 \][/tex]
So, we have point [tex]\( (1, 0) \)[/tex].
- Now draw a line through the points [tex]\((0, -2)\)[/tex] and [tex]\((1, 0)\)[/tex].
4. Mark Points of Intersection:
- With [tex]\(x = -2\)[/tex]:
Substitute [tex]\(x = -2\)[/tex] back into either equation to find [tex]\(y\)[/tex]:
For the quadratic equation:
[tex]\[ y = -\frac{1}{2}(-2)^2 + 3(-2) + 2 \implies y = -2 - 6 + 2 = -6 \][/tex]
So, point [tex]\((-2, -6)\)[/tex].
- With [tex]\(x = 4\)[/tex]:
Substitute [tex]\(x = 4\)[/tex] back into either equation to find [tex]\(y\)[/tex]:
For the quadratic equation:
[tex]\[ y = -\frac{1}{2}(4)^2 + 3(4) + 2 \implies y = -8/2 + 12 + 2 = 12/2 + 2 = 6 + 2 = 8 \][/tex]
So, point [tex]\(( 4, 8 )\)[/tex].
5. Marking Points of Intersection on the Graph:
Mark the points [tex]\((-2, -6)\)[/tex] and [tex]\((4, 8)\)[/tex] on the graph using the "Mark Feature".
To help visualize, the steps in summary:
- Draw the line [tex]\(y = 2x - 2\)[/tex].
- Plot the points [tex]\((0, -2)\)[/tex] and [tex]\((1, 0)\)[/tex] and draw a straight line through them.
- Mark the intersection points [tex]\((-2, -6)\)[/tex] and [tex]\((4, 8)\)[/tex] on the graph.
These marked points represent the solutions to the system of equations where both the linear and quadratic equations intersect.
