Solve Graphed System Of Equations Y=-x^2+x+2 And Y=-x+3

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In mathematics, solving systems of equations is a fundamental concept with applications spanning various fields. One method of solving these systems is through graphing, which provides a visual representation of the solutions. This article delves into the process of identifying solutions to graphed systems of equations, focusing on the intersection points of the graphs. Specifically, we will address the question: Which represents the solution(s) of the graphed system of equations, y **= -x2 + x + 2 and y **= -x + 3?

Understanding Systems of Equations

A system of equations is a set of two or more equations containing the same variables. The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously. Graphically, each equation in the system represents a curve or line on a coordinate plane. The solution(s) to the system are the points where these curves or lines intersect. These intersection points represent the values of the variables that make all equations true at the same time.

To fully grasp the concept of solving systems of equations, it's essential to understand the different types of equations we might encounter. In our given problem, we have two equations:

  1. y **= -x2 + x + 2: This is a quadratic equation, which, when graphed, forms a parabola. The negative coefficient of the x2 term indicates that the parabola opens downwards.
  2. y **= -x + 3: This is a linear equation, which, when graphed, forms a straight line. The slope of this line is -1, and the y-intercept is 3.

The solutions to this system will be the points where the parabola and the line intersect on the coordinate plane. These points of intersection are crucial as they represent the (x, y) pairs that satisfy both equations simultaneously.

Methods for Solving Systems of Equations

While graphing is a visual method, there are other algebraic techniques to solve systems of equations, including:

  • Substitution: Solve one equation for one variable and substitute that expression into the other equation.
  • Elimination: Add or subtract the equations to eliminate one variable.
  • Matrix Methods: Use matrices to represent and solve the system, particularly useful for larger systems.

However, for the purpose of this article, we will primarily focus on the graphical method, as it directly relates to the problem at hand. Graphing allows us to visualize the equations and identify the intersection points, which are the solutions we seek. Understanding these methods provides a comprehensive toolkit for solving various types of systems of equations.

Graphing the Equations

To solve the system graphically, we need to plot both equations on the same coordinate plane. Let's start with the quadratic equation, y **= -x2 + x + 2. This is a parabola, and to graph it accurately, we need to identify key features such as the vertex, axis of symmetry, and a few additional points.

Graphing the Parabola y **= -**x2 + x + 2

The vertex of a parabola in the form y = ax2 + bx + c can be found using the formula x = -b/(2a). In our case, a = -1 and b = 1, so the x-coordinate of the vertex is:

x = -1 / (2 * -1) = 1/2

Now, we can find the y-coordinate of the vertex by substituting x = 1/2 into the equation:

y = -(1/2)2 + (1/2) + 2 = -1/4 + 1/2 + 2 = 9/4

So, the vertex of the parabola is (1/2, 9/4), or (0.5, 2.25).

The axis of symmetry is a vertical line that passes through the vertex, which in this case is x = 1/2. This line divides the parabola into two symmetrical halves.

To plot the parabola, we can find a few additional points by substituting different values of x into the equation. For example:

  • If x = -1, y = -(-1)2 + (-1) + 2 = -1 - 1 + 2 = 0. So, the point is (-1, 0).
  • If x = 0, y = -(0)2 + (0) + 2 = 2. So, the point is (0, 2).
  • If x = 1, y = -(1)2 + (1) + 2 = -1 + 1 + 2 = 2. So, the point is (1, 2).
  • If x = 2, y = -(2)2 + (2) + 2 = -4 + 2 + 2 = 0. So, the point is (2, 0).

Plotting these points and the vertex, we can sketch the parabola. Graphing the parabola accurately is crucial as it sets the stage for finding the intersection points with the linear equation.

Graphing the Line y **= -**x + 3

Graphing the linear equation y **= -x + 3 is simpler. We can identify the slope and y-intercept directly from the equation.

The slope (m) is -1, and the y-intercept (b) is 3. This means the line crosses the y-axis at (0, 3) and goes down 1 unit for every 1 unit it moves to the right.

To plot the line, we need at least two points. We already have the y-intercept (0, 3). Let's find another point:

  • If x = 1, y = -1 + 3 = 2. So, the point is (1, 2).

Plotting these points (0, 3) and (1, 2), we can draw the line. The line's negative slope indicates that it slopes downwards from left to right. Accurately graphing the line is as important as graphing the parabola, as the intersection points of these two graphs will provide the solutions to the system of equations.

Identifying the Solutions

Once both equations are graphed on the same coordinate plane, the solutions to the system are the points where the parabola and the line intersect. By visually inspecting the graph, we can identify these points. In this case, the parabola y **= -x2 + x + 2 and the line y **= -x + 3 intersect at two points: (-1, 4) and (1, 2).

Verifying the Solutions

To ensure accuracy, it's always a good practice to verify the solutions by substituting the x and y values back into the original equations:

  1. For the point (-1, 4):
    • In y **= -x2 + x + 2: 4 = -(-1)2 + (-1) + 2 = -1 - 1 + 2 = 0 (This is incorrect)
    • In y **= -x + 3: 4 = -(-1) + 3 = 1 + 3 = 4 (This is correct)

It appears there's an error in our initial visual inspection. Let’s re-evaluate the intersection points more closely. Upon closer examination, the intersection points are actually (-1, 4) and (1, 2).

Let’s re-verify the point (-1, 4):

  • In y **= -x2 + x + 2: 4 = -(-1)2 + (-1) + 2 = -1 - 1 + 2 = 0 (Incorrect)
  • In y **= -x + 3: 4 = -(-1) + 3 = 1 + 3 = 4 (Correct)

There seems to be a mistake in the problem or the initial identification of the intersection points. The point (-1, 4) does not satisfy the first equation. Let’s verify the other potential solution, (1, 2).

  1. For the point (1, 2):
    • In y **= -x2 + x + 2: 2 = -(1)2 + (1) + 2 = -1 + 1 + 2 = 2 (Correct)
    • In y **= -x + 3: 2 = -(1) + 3 = -1 + 3 = 2 (Correct)

The point (1, 2) satisfies both equations. Thus, (1, 2) is a solution to the system.

To find the other solution, we need to carefully re-examine the graph or use an algebraic method such as substitution or elimination to find the exact intersection points.

Algebraic Verification: Substitution Method

Let's use the substitution method to find the correct solutions algebraically. We have:

  1. y **= -**x2 + x + 2
  2. y **= -x + 3

Since both equations are equal to y, we can set them equal to each other:

-x2 + x + 2 = -x + 3

Rearrange the equation to form a quadratic equation:

0 = x2 - 2x + 1

This is a perfect square trinomial, which can be factored as:

0 = (x - 1)2

Solving for x:

x = 1

Now, substitute x = 1 into either equation to find y. Using the second equation:

y = -1 + 3 = 2

So, the only solution we found algebraically is (1, 2). This indicates that the line is tangent to the parabola at the point (1, 2), meaning they only intersect at one point. The initial graphical inspection suggesting two intersection points was likely an oversight.

Conclusion

In summary, the solution(s) to the graphed system of equations y **= -x2 + x + 2 and y **= -x + 3 is (1, 2). This point represents the intersection of the parabola and the line on the coordinate plane. Solving systems of equations graphically involves plotting the equations and identifying the intersection points. Verifying the solutions algebraically, as we did with the substitution method, is crucial to ensure accuracy.

Understanding how to solve systems of equations graphically is an essential skill in mathematics, offering a visual method to find solutions. This approach, combined with algebraic techniques, provides a robust toolkit for tackling various mathematical problems. The graphical method not only aids in finding solutions but also enhances the understanding of the relationships between different equations. By carefully graphing the equations and verifying the intersection points, we can confidently determine the solutions to the system.

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