Problem 13 - Algebra 2 and Trigonometry

Question

In \(9-17,\) graph each system and determine the common solution from the graph. $$ \begin{array}{l}{y=2 x^{2}+2 x+3} \\ {y-x=3}\end{array} $$

Step-by-Step Solution

Answer

The common solutions are (0, 3) and (-0.5, 2.5).

Step 1: Understand the Equations

We are given a system of equations: \( y = 2x^2 + 2x + 3 \) and \( y - x = 3 \). The first equation is a quadratic equation representing a parabola, and the second is a linear equation representing a straight line.

Step 2: Graph the Quadratic Function

The quadratic equation \( y = 2x^2 + 2x + 3 \) is a parabola that opens upwards because the coefficient of \( x^2 \) is positive. Identify the vertex. The vertex form is \( y = 2(x + 0.5)^2 + 2.5 \). Plot the vertex at (-0.5, 2.5) and other points symmetric about the vertex.

Step 3: Graph the Linear Function

Transform the linear equation \( y - x = 3 \) into slope-intercept form, which is \( y = x + 3 \). The graph of this equation is a line with a slope of 1 and a y-intercept of 3. Plot the y-intercept and use the slope to find another point, then draw the line.

Step 4: Determine the Intersection Points

The intersection points of the graphs represent the solutions for the system of equations where both equations are true simultaneously. Calculate the intersections by substituting \( y = x + 3 \) into \( y = 2x^2 + 2x + 3 \). Simplify to get \( 2x^2 + 2x + 3 = x + 3 \).

Step 5: Solve the System Algebraically

Simplify the equation to find the x-values of the intersection. Calculate: \( 2x^2 + x = 0 \), which simplifies to \( x(2x + 1) = 0 \). Hence, \( x = 0 \) or \( x = -0.5 \).

Step 6: Find Corresponding y-values

Substitute \( x = 0 \) into \( y = x + 3 \) to get \( y = 3 \). Substitute \( x = -0.5 \) into \( y = x + 3 \) to get \( y = 2.5 \). Thus, intersection points are (0, 3) and (-0.5, 2.5).

Step 7: Verify the Solution from the Graph

Plot both equations on the graph. Check the points of intersection which visually confirm that the equations intersect at (0, 3) and (-0.5, 2.5).

Key Concepts

Graphing Quadratic Functions

When we talk about graphing quadratic functions, we're dealing with a special type of curve called a parabola. For the equation \(y = 2x^2 + 2x + 3\), the parabola opens upwards because the coefficient of \(x^2\) is positive.
Understanding the structure is key:

• The **vertex** is the turning point: Here, it is at \((-0.5, 2.5)\).
• The **axis of symmetry** helps find points on either side: It's the vertical line through the vertex.
• **Direction**: This parabola points upwards, so it expands as it moves away from the vertex.

To plot, start by marking the vertex and a few other points on either side according to their symmetry. Connecting them smoothly forms the parabola's curve.

Graphing Linear Equations

Linear equations like \(y - x = 3\) simplify to \(y = x + 3\), representing lines with consistent angles and measures. A straight line requires two main components:

• The **slope**: A measure of steepness. Here, the slope is 1, meaning 45 degrees up to the right.
• The **y-intercept**: Where the line crosses the y-axis. At 3 for this equation.

Start by plotting the y-intercept, then use the slope to find another point. Keep rising one unit up for every one unit you move right, to draw the line. Connect these points seamlessly to complete the graph.

Intersection Points

The intersection points between two graphs tell us where the two equations share common solutions. They show where the lines cross the parabola.
For the equations at hand:

• We solved \(2x^2 + x = 0\) by factoring to find \(x = 0\) and \(x = -0.5\).
• Using these x-values in \(y = x + 3\) gives \(y = 3\) and \(y = 2.5\).
• Thus, the intersection points are \((0, 3)\) and \((-0.5, 2.5)\).

These points represent where both equations are true at the same time on the graph.

Quadratic and Linear Systems

Quadratic and linear systems involve equations of both types together, typically resulting in multiple solutions.

• A **system** means solving for points where both conditions—the parabola from the quadratic and the line from the linear—are satisfied.
• Solutions are graphed visually as intersection points on the plot. In this exercise, we found two such solutions.

Understanding these systems is crucial as they illustrate scenarios where different mathematical worlds—the curvy nature of quadratics and straight-lined behavior of linear equations—interact. These interactions tell us important information about dynamics like motion and force in physics via graphs.