Solve Systems of Linear and Quadratic Equations Graphically
Algebra-1
1. Fundamental Concepts
System of Linear and Quadratic Equations: A system consisting of a linear equation (in the form of $y = mx + b$ , whose graph is a straight line) and a quadratic equation (in the form of $y = ax^2+bx + c$ , where $a\neq0$ , and its graph is a parabola).
Geometric Meaning of Solutions: The solutions of the system are the coordinates $(x,y)$ of the intersection points of the straight line and the parabola. The number of solutions corresponds to the number of intersection points, and there are three possibilities:
2 solutions: The straight line intersects the parabola at two points.
1 solution: The straight line is tangent to the parabola (only one intersection point).
0 solutions: The straight line and the parabola have no intersection points.
2. Key Concepts
Key Points for Graph Plotting:
Parabola: It is necessary to determine the vertex, the direction of opening (determined by the quadratic coefficient a; if a > 0, it opens upwards; if a<0, it opens downwards), and the intersection points with the coordinate axes and other key features.
Straight line: Determine the inclination and the intersection point with the y-axis according to the slope m and the intercept b, and determine the position of the straight line through two points.
Reading of Solutions: After plotting the parabola and the straight line in the same coordinate system, the coordinates $(x,y)$ of the intersection points are the solutions of the system of equations, and attention should be paid to the accuracy of the coordinates.
3. Examples
1. System of equations: $y=x^2 - 5x + 7$ and $y = 2x + 1$
Steps: Plot the parabola $y=x^2 - 5x + 7$ (opens upwards; the vertex can be calculated by the formula $x=-\frac{b}{2a}=\frac{5}{2}$ , and substituting it in gives the vertex coordinates $(2.5,0.75)$ ) and the straight line $y = 2x + 1$ (with a slope of 2 and a y-intercept of 1). Find the intersection points of the two, and the solutions are $x=1,y=3$ and $x=6,y=13$ .
2. System of equations: $y = 2x^2 + 4x + 1$ and $y=2x + 1$
Steps: Plot the parabola $y = 2x^2 + 4x + 1$ (opens upwards; the vertex is at $x=-\frac{4}{2\times2}=-1$ , and substituting it in gives the vertex coordinates $(-1,-1)$ ) and the straight line $y=2x + 1$ (with a slope of 2 and a y-intercept of 1). The intersection points are $(0,1)$ and $(-1,-1)$ .
3. System of equations: $y = 2x^2 + 4x + 1$ and $y=2x - 4$
Steps: Plot the parabola $y = 2x^2 + 4x + 1$ (with the same features as above) and the straight line $y=2x - 4$ (with a slope of 2 and a y-intercept of -4). Observing the graph, it can be seen that there is no intersection point between them, meaning the system of equations has no solution.
4. Problem-Solving Techniques
Preparation: Identify the quadratic equation and the linear equation in the system, and rearrange them into the form of $y=$ an expression in terms of x to facilitate graph plotting.
Graph Plotting:
For the parabola, first determine the vertex coordinates and the direction of opening, then find several symmetric points (such as the intersection points with the x-axis, y-axis, or points near the vertex), and connect them smoothly to form the parabola.
For the straight line, determine two points according to the slope and intercept, and connect them to form the straight line.
Determine Solutions: Carefully observe the graph to find the intersection points of the straight line and the parabola. The coordinates of the intersection points are the solutions of the system of equations. If there is no intersection point, the system of equations has no solution; if there is only one intersection point, it is the unique solution.
Verification: Substitute the obtained solutions into the two equations of the original system to check whether the left and right sides are equal, so as to ensure the correctness of the solutions.