Question

Solve the system by graphing. Check your solution(s). (See Example 1.) \(y=x+2\) \(y=0.5(x+2)^2\)

Short answer

The solution to the system of equations is given by the intersection points of the graphs of the two equations.

Step-by-step solution

Rewrite equations

The equations are given in a convenient form to graph them, so rewriting is not necessary. The standard form, \(y = f(x)\), is already provided. The first equation is a linear one, it represents a straight line with a slope equals to 1. The second equation is a quadratic, representing a parabola with a vertex (lowest point) that we can find by setting \(x+2=0\) to \(x=-2\), so the vertex is \(-2,0.5(-2+2)^2\) which simplifies to \(-2,0\).

Graph the equations

Sketch the graphs of the two equations on the same set of axes. For \(y = x + 2\), choose a few x values, say -3, -2, -1, 0, and 1, and calculate the corresponding y values. For \(y = 0.5(x+2)^2\), you can use the same x values and calculate the y values. Plot the points and join them.

Find intersection points

After drawing the graphs, find the points where the two graphs intersect. These points are the solutions to the system of equations. Check these points by substituting into the given equations and confirming that both equations hold true.

Key concepts

Graphing Equations

Graphing equations is a powerful method to visually solve systems of equations. When we graph equations, we represent each one as a line or curve on a coordinate plane. This helps us to see where they might intersect and thus find their solutions. To graph an equation:

• First identify its type: linear, quadratic, etc.
• Determine a few key points to draw the curve.
• Plot these points on graph paper or a digital graphing tool.
• Connect the points smoothly for quadratic curves or use a ruler for straight lines.

For instance, if you have both a line and a parabola, you will look for points where the line and the parabola intersect. In our example, graphing the equations provides a visual method to find the solution of the system by directly observing the x and y coordinates where the curves meet. This is a practical and often simpler solution method compared to algebraic techniques when you're dealing with straightforward equations in a system.

Linear Equations

Linear equations are equations of the first degree, meaning they graph as straight lines. They take the form of \( y = mx + b \), where:

• \( m \) is the slope, indicating how steep the line is.
• \( b \) is the y-intercept, the point where the line crosses the y-axis.

In the context of our exercise, the linear equation is \( y = x + 2 \). Here, the slope \( m \) is 1, suggesting the line increases one unit in the y direction for each unit increase in the x direction. The y-intercept \( b \) is 2, so the line crosses the y-axis at (0,2).To graph this equation, you can choose several values for \( x \), calculate the corresponding \( y \) values, plot these points, and draw a straight line through them. This line represents all possible solutions for \( y \) given any \( x \) value according to this equation. It's important to note that every point on the line satisfies the linear equation and showcases its simplicity when graphed.

Quadratic Equations

Quadratic equations introduce a second-degree variable, resulting in a parabola when graphed. The standard form is \( y = ax^2 + bx + c \) and for our exercise, it was given as \( y = 0.5(x+2)^2 \). This form is often easier to work with for graphing, since it highlights the vertex form, \[(x-h)^2+k\], where:

• \( h \) is the x-coordinate of the vertex.
• \( k \) is the y-coordinate of the vertex.

Here, the equation \( y = 0.5(x+2)^2 \) features a parabola opening upwards. The expression \((x+2)^2\) suggests a horizontal shift left by 2 units, moving the vertex to (-2,0). Since 0.5 is the leading coefficient, the parabola is broader than one with a coefficient of 1.To effectively graph a quadratic equation:

• Start by finding the vertex of the parabola.
• Use symmetry to find additional points on either side of the vertex.
• Plot these points and draw the curve.

This graphical process reveals the parabolic shape and helps visualize solutions where it intersects with other graphs, such as our linear equation in the exercise. Quadratic equations, due to their non-linear nature, provide a unique graphical representation that is vital in solving systems visually.