y=2x^+3 use integer values of x from -3 to 3 to graph the equation.
Someone please tell me how to graph this
evaluate the y for the given x's, then plot the points.
Join them with a smooth curve.
You will get a parabola
I'm only in seventh grade and I don't understand any of this.what does a parabola look like
Actually I just noticed your equation is not quite clear.
is the exponent 3 ?
then you would have written it as y = 2x^3
or
did you forget to type the exponent ?
It would be highly unusual for a text to put the + sign in front of the exponent like you typed it.
y= -2x+3.the x has a 2 above it
To graph the equation y = 2x^2 + 3 using integer values of x from -3 to 3, you can follow these steps:
1. Start by determining the values of y for each x-coordinate. Plug in each x-value from -3 to 3 into the equation and calculate the corresponding y-value.
For x = -3:
y = 2(-3)^2 + 3 = 2(9) + 3 = 18 + 3 = 21
For x = -2:
y = 2(-2)^2 + 3 = 2(4) + 3 = 8 + 3 = 11
For x = -1:
y = 2(-1)^2 + 3 = 2(1) + 3 = 2 + 3 = 5
For x = 0:
y = 2(0)^2 + 3 = 2(0) + 3 = 0 + 3 = 3
For x = 1:
y = 2(1)^2 + 3 = 2(1) + 3 = 2 + 3 = 5
For x = 2:
y = 2(2)^2 + 3 = 2(4) + 3 = 8 + 3 = 11
For x = 3:
y = 2(3)^2 + 3 = 2(9) + 3 = 18 + 3 = 21
2. Now, you can plot these points on a graph. Create a coordinate plane with x and y axes. Mark the x-coordinates -3, -2, -1, 0, 1, 2, 3 on the x-axis, and plot the corresponding y-values above each x-coordinate.
(-3, 21), (-2, 11), (-1, 5), (0, 3), (1, 5), (2, 11), (3, 21)
3. Once all the points are plotted, you can connect them to form a smooth curve. In this case, since the equation is y = 2x^2 + 3, the graph will be a U-shaped curve opening upwards.
The graph of the equation y = 2x^2 + 3 using integer values of x from -3 to 3 will look like a U-shaped curve, passing through the points (-3, 21), (-2, 11), (-1, 5), (0, 3), (1, 5), (2, 11), and (3, 21).