Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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.
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.