Solve the following system graphically. Verify the solution(s) using substitution.
• y=2x^2+3x+1
• y=x+13
I can verify, I just need help graphing, please
irdk
there are lots of handy graphing web sites, such as
desmos.com
wolframalpha.com
To solve the given system graphically, we will plot the two equations on a graph and find the point(s) where they intersect. These points will represent the solutions to the system.
Let's start by graphing the first equation, y = 2x^2 + 3x + 1.
To plot this equation, we can use a table of values or find the vertex and a few additional points. Let's find the vertex using the formula x = -b/2a, where a, b, and c are the coefficients from the standard form equation.
For the equation y = 2x^2 + 3x + 1:
a = 2, b = 3, c = 1
Using the formula x = -b/2a, we get:
x = -3 / (2*2) = -3/4
Substituting this value of x back into the equation, we can find the y-value:
y = 2 * (-3/4)^2 + 3 * (-3/4) + 1
y = 9/8 - 9/4 + 1
y = 9/8 - 18/8 + 8/8
y = -1/8
So the vertex of the parabolic curve is (-3/4, -1/8).
Now let's find a few more points to plot. We can choose x-values around the vertex, such as x = -2, -1, 0, 1, and 2.
For x = -2:
y = 2 * (-2)^2 + 3 * (-2) + 1
y = 8 - 6 + 1
y = 3
For x = -1:
y = 2 * (-1)^2 + 3 * (-1) + 1
y = 2 - 3 + 1
y = 0
For x = 0:
y = 2 * 0^2 + 3 * 0 + 1
y = 1
For x = 1:
y = 2 * 1^2 + 3 * 1 + 1
y = 2 + 3 + 1
y = 6
For x = 2:
y = 2 * 2^2 + 3 * 2 + 1
y = 8 + 6 + 1
y = 15
Now we have a few points to plot: (-2, 3), (-1, 0), (0, 1), (1, 6), and (2, 15). We can plot these on a graph.
Next, let's graph the second equation, y = x + 13. This is a linear equation, so we can easily find two points to plot.
For x = -2:
y = -2 + 13
y = 11
For x = 2:
y = 2 + 13
y = 15
Now we have the points (-2, 11) and (2, 15) to plot.
Plotting both equations on the same graph and connecting the points, we get a parabola for the first equation (y = 2x^2 + 3x + 1) and a straight line for the second equation (y = x + 13).
To visually verify the solution to the system, we identify the point(s) where the parabola and line intersect. These point(s) represent the solution(s) to the system.
Finally, to verify the solution(s) algebraically, we substitute the x and y values of the intersection point(s) obtained from the graph back into the original equations. If the equations hold true, the solution(s) are valid.
This method allows us to solve the system graphically and verify the solution(s) using substitution.