solve the system by graphing. y=2×^2+3×+1 y=-2×+1
To solve the system by graphing, we can plot the graphs of the two equations and find their point of intersection.
First, let's graph the equation y = 2x^2 + 3x + 1:
To start graphing the quadratic equation, we can find the vertex and the y-intercept.
The equation is in the form y = ax^2 + bx + c, where a = 2, b = 3, and c = 1.
The x-coordinate of the vertex can be found using the formula x = -b / (2a), which becomes:
x = -(3) / (2 * 2) = -3/4
Then, substitute this x-value to find the corresponding y-value:
y = 2 * (-3/4)^2 + 3(-3/4) + 1
y = 2 * 9/16 - 9/4 + 1
y = 9/8 - 9/4 + 1
y = 9/8 - 18/8 + 8/8
y = 1/8
So, the vertex is (-3/4, 1/8).
Next, let's find the y-intercept by setting x = 0:
y = 2(0)^2 + 3(0) + 1
y = 1
Therefore, the y-intercept is (0, 1).
Now, let's graph the equation y = -2x + 1:
To graph a linear equation, we can use the y-intercept (0, 1) and the slope (-2).
With a slope of -2, we can find another point on the line by using the rise over run. For every one unit to the right, the line goes two units down.
Using the y-intercept point (0, 1), we can plot another point using the slope:
(0 + 1, 1 - 2) = (1, -1)
Now, we can plot these two equations on a graph:
On the graph, we can see that the two lines intersect at the point (-3/4, 1/8). This is the solution to the system of equations.
To solve the system by graphing, we'll plot the graphs of both equations on the same coordinate plane and find the point where they intersect. This point will represent the solution to the system.
Let's start with the first equation, y = 2x^2 + 3x + 1. To graph this equation, we'll choose some x-values, plug them into the equation, and then plot the corresponding y-values. Repeat this process to get multiple points, and then connect them to form a smooth curve.
For example, if we choose x = -2, we can calculate y as follows:
y = 2(-2)^2 + 3(-2) + 1
y = 8 - 6 + 1
y = 3
So, one point on the graph is (-2, 3). We can repeat this process with other values of x to get more points.
Next, let's graph the second equation, y = -2x + 1. Similarly, choose various x-values, calculate the corresponding y-values, and plot the points to form a line.
For x = -2, y = -2(-2) + 1 = 5. So, one point on the graph is (-2, 5). Plot more points using different x-values.
Once both equations are graphed, identify the point where the two graphs intersect. This point represents the solution to the system of equations.
In this case, the graphs intersect at the point (-1, 3). Hence, the solution to the system is x = -1 and y = 3.
Remember, graphing is just one method to solve a system of equations. Other methods like substitution or elimination may also be used depending on the complexity of the equations.
To solve the system of equations by graphing, we need to plot the graphs of both equations on the same coordinate plane and find their point of intersection(s).
First, let's graph the equation y = 2x^2 + 3x + 1:
- Choose some x-values and calculate the corresponding y-values to create a table of values.
For example, let's use x-values of -2, -1, 0, 1, and 2:
When x = -2: y = 2(-2)^2 + 3(-2) + 1 = 9
When x = -1: y = 2(-1)^2 + 3(-1) + 1 = 0
When x = 0: y = 2(0)^2 + 3(0) + 1 = 1
When x = 1: y = 2(1)^2 + 3(1) + 1 = 6
When x = 2: y = 2(2)^2 + 3(2) + 1 = 15
Now, plot these points on the graph.
Next, let's graph the equation y = -2x + 1:
- Choose some x-values and calculate the corresponding y-values to create a table of values.
For example, let's use x-values of -2, -1, 0, 1, and 2:
When x = -2: y = -2(-2) + 1 = 5
When x = -1: y = -2(-1) + 1 = 3
When x = 0: y = -2(0) + 1 = 1
When x = 1: y = -2(1) + 1 = -1
When x = 2: y = -2(2) + 1 = -3
Now, plot these points on the graph.
After plotting both equations on the same graph, look for the point(s) where the two graphs intersect. These intersections represent the solutions to the system of equations.
Based on the graph, it appears that the two equations intersect at one point, which is approximately at (0.5, 0.5).
Therefore, the solution to the system of equations is:
x = 0.5
y = 0.5