Question

Solve the system graphically. $$\left\\{\begin{aligned} x-y+3 &=0 \\ x^{2}-4 x+7 &=y \end{aligned}\right.$$

Short answer

After plotting the functions and identifying the intersection points, solutions to the system are the coordinates of these intersection points. Precise coordinates will depend on the accuracy of the graphical representation, or can be found by solving the equation \(x + 3 = x^{2} - 4x + 7\) algebraically.

Step-by-step solution

Rearrange Equations

First, express both equations as functions of x. From the first equation \(x - y + 3 = 0\), we can express y in terms of x as \(y = x + 3\). From the second equation \(x^{2} - 4x + 7 = y\), y is already expressed in terms of x as \(y = x^{2} - 4x + 7\)

Graph the Functions

Draw a graph and plot the two functions \(y = x + 3\) and \(y = x^{2} - 4x + 7\). The first function is a straight line with a slope of 1 and y-intercept of 3. The second function is a parabola opening upwards with the vertex at \((2,3)\) which can be found by applying vertex form of parabola \(f(x)=a(x-h)^{2}+k\) where (h,k) is the vertex of parabola.

Identify Intersection Points

Identify the points of intersection of the two graphs. These points are the solutions to the system of equations. Depending on the graphics you have, you may need to solve the equation \(x + 3 = x^{2} - 4x + 7\) mathematically to find precise coordinates for the intersection points.

Key concepts

Graphing Linear Equations

Graphing linear equations is one of the fundamental skills in algebra. A linear equation represents a straight line and is of the form \(y = mx + b\) where \(m\) is the slope and \(b\) is the y-intercept, the point where the line crosses the y-axis. The process of graphing a linear equation involves identifying the y-intercept first and plotting it on the graph. Then, we use the slope, which represents the rise over run, to determine the direction and steepness of the line.

For instance, given the linear equation \(y = x + 3\), we can see that the slope (\(m\)) is 1 and the y-intercept (\(b\)) is 3. This tells us to start at the point (0, 3) on the graph and move one unit up for every one unit we move to the right to draw the line. This method gives students a clear path to follow and ensures they understand the relationship between the equation and the graph they create.

Graphing Quadratic Equations

Quadratic equations form parabolas when graphed and are represented by an equation of the form \(y = ax^2 + bx + c\). Unlike straight lines, parabolas can open upward or downward depending on the sign of \(a\). When \(a\) is positive, the parabola opens upward; when \(a\) is negative, it opens downward.

A crucial point on a parabola is its vertex, the highest or lowest point depending on the direction it opens. The vertex form of a parabola's equation, \(y = a(x - h)^2 + k\), provides an easy way to find the vertex at point \((h, k)\). Following the given example equation, \(y = x^2 - 4x + 7\), we can complete the square to express it in the vertex form and find the vertex of the parabola. In this case, the parabola has a vertex at \((2, 3)\), telling us that our parabola opens upwards from that point. Understanding the shape and orientation of the parabola is essential for accurately graphing the quadratic equation.

Intersection Points of Graphs

The intersection points of graphs occur where two or more functions have the same value for \(x\) and \(y\). These points are significant because they represent the solutions to a system of equations. When we graph the functions on the same coordinate plane, the intersection points can be visually identified as the points where the graphs meet.

In the context of our example, to find the intersection points between the linear equation \(y = x + 3\) and the quadratic equation \(y = x^2 - 4x + 7\), we set them equal to each other and solve for \(x\). This gives us the exact value(s) of \(x\) where the intersections occur. However, when graphing by hand or using technology, we can often estimate the intersection points by closely examining the points where the line intersects with the parabola. For precise solutions, algebraic methods or graphing calculators can be employed to find the exact coordinates. The intersection points are crucial as they provide the answer to which set of coordinates satisfy both equations simultaneously.