Problem 40 - Algebra and Trigonometry

Question

Use the graphical method to find all solutions of the system of equations, correct to two decimal places. \(\left\\{\begin{array}{l}{y=x^{2}-4 x} \\ {2 x-y=2}\end{array}\right.\)

Step-by-Step Solution

Answer

Solutions are approximately (5.24, 8.48) and (0.76, -0.48).

Step 1: Interpret the System of Equations

The given system of equations is: 1. \( y = x^2 - 4x \) which represents a parabola.2. \( 2x - y = 2 \) can be rewritten as \( y = 2x - 2 \), representing a straight line.

Step 2: Plot the Parabola

Sketch the graph of the parabola \( y = x^2 - 4x \). Determine key points by finding the vertex and intercepts.- Vertex using \( x = -\frac{b}{2a} = 2 \) gives coordinates (2, -4).- Intercepts: At \( x=0 \), \( y=0 \). - Additionally, find points for specific \( x \) values like \( x=1, 3, 4 \).

Step 3: Plot the Line

Graph the linear equation \( y = 2x - 2 \). Determine intercepts:- X-intercept by setting \( y = 0 \), gives \( x = 1 \).- Y-intercept by setting \( x = 0 \), gives \( y = -2 \).

Step 4: Find Intersection Points

Look for the points where the parabola and the line intersect. These intersections are the solutions to the system. The approximate graphical solutions appear around the points where the curves cross.

Step 5: Verify Intersections Algebraically

Set the equations equal to verify intersection points: \( x^2 - 4x = 2x - 2 \) simplifies to \( x^2 - 6x + 2 = 0 \).- Solve the quadratic equation using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -6, c = 2 \).- Calculate \( x = 5.24 \) and \( x = 0.76 \).

Step 6: Calculate Corresponding Y-values

Substitute the values of \( x \) back into the linear equation \( y = 2x - 2 \) to find corresponding \( y \) values.- For \( x = 5.24 \), \( y = 8.48 \).- For \( x = 0.76 \), \( y = -0.48 \).

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of degree two. It usually takes the form \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In our problem, we are working with the quadratic equation \( y = x^2 - 4x \). This equation describes a curve known as a parabola.

Quadratic equations have distinctive properties, such as a vertex, axis of symmetry, and can open upwards or downwards depending on the coefficient \( a \). Here are some key features of the parabola from our equation:

• **Vertex**: The vertex is the point where the parabola has its maximum or minimum value. For the equation \( y = x^2 - 4x \), the vertex is at \((2, -4)\). You can find the x-coordinate of the vertex using the formula \( x = -\frac{b}{2a} \).
• **Intercepts**: These are the points where the parabola crosses the axes. To find the y-intercept, set \( x = 0 \). For the x-intercepts (roots), set \( y = 0 \) and solve the equation for \( x \).

Understanding these properties helps in sketching the parabola and predicting its path on a graph.

Graphical Method

The graphical method involves sketching the graphs of equations on the same coordinate plane to find solutions visually. By plotting the equations, points of intersection indicate possible solutions to the system.

Here's a step-by-step approach to using this method:

• **Sketch Each Function**: Begin by plotting each equation individually. For a quadratic like \( y = x^2 - 4x \), sketch the parabola. For a linear equation like \( y = 2x - 2 \), draw the straight line.
• **Identify Key Points**: Important features such as intercepts and vertices should be clearly marked. These points guide you in tracing the graph accurately.
• **Look for Intersections**: Observe where the graphs intersect. Each intersection represents a potential solution where both equations satisfy the x and y values.

This visual method helps clarify what solutions look like graphically, making abstract algebraic solutions more tangible.

Intersection Points

Intersection points occur where two graphs cross each other. These are critical in solving systems of equations graphically as they represent the solutions.

In the context of our problem, the intersection points between the parabola \( y = x^2 - 4x \) and the line \( y = 2x - 2 \) are the solutions we seek.

• **Significance**: Each intersection point gives a pair \((x, y)\) that satisfies both equations simultaneously.
• **Finding Solutions**: In our example, we found intersections at approximately \((5.24, 8.48)\) and \((0.76, -0.48)\). These points were verified algebraically by setting the equations equal and solving for \( x \) then \( y \).

Once you identify the intersections visually, it's often beneficial to verify them algebraically for precision.

Linear Equations

Linear equations are of the form \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. These equations describe straight lines on the graph.

In this exercise, one of the equations was \( y = 2x - 2 \), translating to a line with a slope of 2 and a y-intercept of -2.

• **Slope**: The slope \( m \) dictates the steepness and direction of the line. A positive slope means the line increases from left to right. In our example, the slope is 2, indicating a relatively steep ascent.
• **Intercepts**: Set \( y = 0 \) to find the x-intercept, and set \( x = 0 \) to find the y-intercept. For our linear equation, the x-intercept is \( x = 1 \). These intercepts help anchor the line on the graph.

Understanding these elements is fundamental when plotting a linear equation, ensuring accuracy in conjunction with other graphs like parabolas.