Which graph shows a system of equations with a solution at (–1, 1)?

A. a Negative parabola with a slope of 2/1, vertex of (0,1) and a line with a slope of 1/2 and y-intercept of -3
B. a Negative parabola with a slope of 2/1. vertex of (0,1) and a line with a slope of 2/1 and a y-intercept of 3
C. a parabola with x-intercepts of 0.5 and 2.5, and a line with a slope of 1, and a y-intercept of -3
D. a Negative parabola with a slope of 2/1, vertex of (0,3), and a line with a negative slope of 1, and y-intercept of -1

To find which graph represents a system of equations with a solution at \((-1, 1)\), we need to check if the point \((-1, 1)\) lies on both the parabola and the line in each option.

Let's analyze each option:

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**A. Negative parabola with a slope of \(2/1\), vertex at \((0,1)\), and a line with a slope of \(1/2\) and y-intercept of \(-3\)**:
- The parabola: Since the vertex is at \((0, 1)\), the parabola opens downwards (as it's a "negative" parabola). A parabola’s equation with vertex form can be written as:
  \[
  y = -2(x - 0)^2 + 1 = -2x^2 + 1
  \]
  Substituting \((-1, 1)\) into this equation:
  \[
  1 = -2(-1)^2 + 1 \implies 1 = -2 + 1 \implies 1 \neq -1
  \]
  \(\Rightarrow\) The point \((-1, 1)\) does **not** lie on the parabola.

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**B. Negative parabola with a slope of \(2/1\), vertex at \((0,1)\), and a line with a slope of \(2/1\) and y-intercept of \(3\)**:
- The parabola: Same as before, the equation is:
  \[
  y = -2x^2 + 1
  \]
  Substituting \((-1, 1)\):
  \[
  1 = -2(-1)^2 + 1 \implies 1 = -2 + 1 \implies 1 \neq -1
  \]
  \(\Rightarrow\) The point \((-1, 1)\) does **not** lie on the parabola.

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**C. Parabola with x-intercepts of \(0.5\) and \(2.5\), and a line with a slope of \(1\) and y-intercept of \(-3\)**:
- The parabola has roots at \(x = 0.5\) and \(x = 2.5\), so its equation is in factored form:
  \[
  y = a(x - 0.5)(x - 2.5)
  \]
  We don't know the value of \(a\), but we need to substitute \((-1, 1)\) into the equation and check if it's a solution. Without \(a\), it is unclear whether \((-1, 1)\) lies on the parabola, so we'll move on.
- The line: The equation is \(y = x - 3\). Substituting \((-1, 1)\):
  \[
  1 = -1 - 3 \implies 1 \neq -4
  \]
  \(\Rightarrow\) The point \((-1, 1)\) does **not** lie on the line.

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**D. Negative parabola with a slope of \(2/1\), vertex at \((0, 3)\), and a line with a negative slope of \(-1\) and y-intercept of \(-1\)**:
- The parabola: The vertex is at \((0, 3)\), and the equation in vertex form is:
  \[
  y = -2x^2 + 3
  \]
  Substituting \((-1, 1)\) into this equation:
  \[
  1 = -2(-1)^2 + 3 \implies 1 = -2 + 3 \implies 1 = 1
  \]
  \(\Rightarrow\) The point \((-1, 1)\) **lies on the parabola**.
  
- The line: The equation of the line is \(y = -x - 1\). Substituting \((-1, 1)\):
  \[
  1 = -(-1) - 1 \implies 1 = 1
  \]
  \(\Rightarrow\) The point \((-1, 1)\) **lies on the line**.

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**Conclusion:**
The graph described in **Option D** shows a system of equations with a solution at \((-1, 1)\). 

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Would you like further details on how the equations were formed or on another related topic?

Here are 5 related questions for further exploration:
1. How do you find the equation of a parabola given its vertex?
2. What is the general form of a quadratic function, and how can it be transformed?
3. How can you determine the solution to a system of equations graphically?
4. What is the significance of the vertex in the graph of a quadratic function?
5. How does changing the slope of a line affect the solution of a system?

**Tip:** When solving systems graphically, always check if the solution satisfies both equations by substituting the coordinates back into each equation.

Identify which graph shows a system of equations with a solution at (-1, 1). This problem covers systems of quadratic and linear equations, with detailed analysis of the vertex form for parabolas and the slope-intercept form for lines. Suitable for students in Grades 9-10.

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