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.