Chapter 12: Problem 12
Graph each equation of the system. Then solve the system to find the points of intersection. $$ \left\\{\begin{array}{l} y=x-1 \\ y=x^{2}-6 x+9 \end{array}\right. $$
Short Answer
Expert verified
The points of intersection are (2, 1) and (5, 4).
Step by step solution
01
Identify and Understand the Equations
The given system of equations consists of two equations: 1. Linear Equation: \(y = x - 1\) 2. Quadratic Equation: \(y = x^2 - 6x + 9\).
02
Graph the Linear Equation
To graph the linear equation \(y = x - 1\), find two points on this line. For example, if \(x = 0\), then \(y = -1\) (point (0, -1)); and if \(x = 2\), then \(y = 1\) (point (2, 1)). Plot these points and draw the line through them.
03
Graph the Quadratic Equation
To graph the quadratic equation \(y = x^2 - 6x + 9\), find key points such as the vertex and the x-intercepts (if any). The vertex form of this quadratic is \(y = (x - 3)^2\), indicating the vertex is at (3, 0). Calculate the values of \(y\) for some \(x\)-values around the vertex, for example: - If \(x = 2\), then \(y = 2^2 - 6\cdot 2 + 9 = 1\) (point (2, 1)). - If \(x = 4\), then \(y = 4^2 - 6\cdot 4 + 9 = 1\) (point (4, 1)). Plot these points and draw the parabola.
04
Find Points of Intersection
To find the points of intersection algebraically, set the equations equal to each other: \(x - 1 = x^2 - 6x + 9\). Rearrange this to get a quadratic equation: \(x^2 - 7x + 10 = 0\). Solve this by factoring: \((x - 2)(x - 5) = 0\) Thus, \(x = 2\) or \(x = 5\).
05
Find Corresponding y-values
Substitute \(x = 2\) back into the linear equation: \(y = 2 - 1 = 1\). So, the point (2, 1) is one intersection. Substitute \(x = 5\) into the linear equation: \(y = 5 - 1 = 4\). So, the point (5, 4) is the other intersection.
06
Verify Intersection Points on the Graph
Plot the points (2, 1) and (5, 4) to confirm they lie on both the linear and quadratic graphs.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing
Graphing is a crucial method for visually representing mathematical equations.
It allows you to see where lines or curves intersect, making it particularly useful for solving systems of equations.
There are a few key points to remember when graphing:
Additionally, practice graphing different types of equations to become more familiar with various graphical features.
This will help you in recognizing patterns and intersections quickly.
It allows you to see where lines or curves intersect, making it particularly useful for solving systems of equations.
There are a few key points to remember when graphing:
- Always find a few key points for each equation to ensure accuracy.
- For linear equations, you need just two points because the graph is a straight line.
- For quadratic equations, identify important features like the vertex and possible x-intercepts.
- Plot each point accurately, then draw the line or curve through them.
Additionally, practice graphing different types of equations to become more familiar with various graphical features.
This will help you in recognizing patterns and intersections quickly.
Linear Equations
Linear equations are equations of the first degree, meaning they graph as straight lines.
The general form is: \[\begin{equation} y = mx + b otag \text{, where } m \text{ is the slope and } b \text{ is the y-intercept.} otag onumber \tag{1} onumber onumber \end{equation}\],
In our exercise, the linear equation given is: \[\begin{equation} y = x - 1 otag onumber \tag{2} onumber onumber \end{equation}\]
To graph this, we can easily determine two points by assigning values to x:
By plotting these points and drawing a straight line through them, we get the graph of the linear equation.
Remember that the slope (m) determines the steepness of the line, and the y-intercept (b) is where it crosses the y-axis.
The general form is: \[\begin{equation} y = mx + b otag \text{, where } m \text{ is the slope and } b \text{ is the y-intercept.} otag onumber \tag{1} onumber onumber \end{equation}\],
In our exercise, the linear equation given is: \[\begin{equation} y = x - 1 otag onumber \tag{2} onumber onumber \end{equation}\]
To graph this, we can easily determine two points by assigning values to x:
- If x = 0, then y = -1 (Point: (0, -1)).
- If x = 2, then y = 1 (Point: (2, 1)).
By plotting these points and drawing a straight line through them, we get the graph of the linear equation.
Remember that the slope (m) determines the steepness of the line, and the y-intercept (b) is where it crosses the y-axis.
Quadratic Equations
Quadratic equations are second-degree equations, meaning the variable has an exponent of 2.
These equations produce a parabolic curve when graphed.
The general form is: \[\begin{equation} y = ax^2 + bx + c otag onumber \tag{3} onumber \end{equation}\], where a determines the direction of the parabola (up for positive, down for negative), b affects the position of the vertex, and c is the y-intercept.
In our exercise, the quadratic equation is:
To graph this, identify the vertex and other key points:
Plotting these helps in drawing the parabolic curve accurately.
These equations produce a parabolic curve when graphed.
The general form is: \[\begin{equation} y = ax^2 + bx + c otag onumber \tag{3} onumber \end{equation}\], where a determines the direction of the parabola (up for positive, down for negative), b affects the position of the vertex, and c is the y-intercept.
In our exercise, the quadratic equation is:
- \[\begin{equation} y = x^2 - 6x + 9 otag onumber \tag{4} onumber \end{equation}\].
To graph this, identify the vertex and other key points:
- The vertex can be found by completing the square or using the vertex form. Here, the vertex is at (3, 0).
- Calculate y-values for x-values around the vertex, e.g., x = 2 and x = 4, both giving y = 1 (Points: (2, 1) and (4, 1)).
Plotting these helps in drawing the parabolic curve accurately.
Intersection Points
Intersection points are where two or more graphs meet.
These points satisfy all equations simultaneously.
To find intersection points, you can equate the equations and solve for the common x-values.
In our exercise, we set the linear and quadratic equations equal to each other:
Simplify to form a quadratic equation:
Factor this equation:
Yielding the solutions:
Substituting these x-values back into the linear equation gives the y-values:
These points (2, 1) and (5, 4) are the intersection points, confirmed by plotting them on the graph.
These points satisfy all equations simultaneously.
To find intersection points, you can equate the equations and solve for the common x-values.
In our exercise, we set the linear and quadratic equations equal to each other:
- \[\begin{equation} x - 1 = x^2 - 6x + 9 otag onumber \tag{5} onumber \end{equation}\].
Simplify to form a quadratic equation:
\[\begin{equation} x^2 - 7x + 10 = 0 otag onumber \tag{6} onumber \end{equation}\].
Factor this equation:
- \[\begin{equation} (x - 2)(x - 5) = 0 otag onumber \tag{7} onumber \end{equation}\].
Yielding the solutions:
- x = 2
- x = 5
Substituting these x-values back into the linear equation gives the y-values:
- If x = 2, then y = 1 (Point: (2, 1)).
- If x = 5, then y = 4 (Point: (5, 4)).
These points (2, 1) and (5, 4) are the intersection points, confirmed by plotting them on the graph.