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.
solve the system by graphing. y=2×^2+3×+1 y=-2×+1
1 answer