Solve graphically.
x-y=0
y=x²
The reason I don't understand the question is the second equation with the x². How do I graph that? I'm not sure if I ever graphed with an exponent, and if I did, it was a long time ago.
I'm reposting this because I didn't understand Damon's answer (I don't know what a parabola is) and I doubt anyone would look down again that far. Could you explain it in more simple terms?
Thanks!
For the second equation just graph the points I gave you
(-3 9)
(-2,4)
(-1,1)
(0,0)
(1,1)
(2,4)
(3,9)
Then you will see what a parabola is.
Just so I can know in the future, how exactly did you figure those points out?
Thanks!
I just put in some values of x and calculated y. Of course I had the advantage of knowing what the curve would look like.
No worries! I'll explain the process in simple terms for you.
To graph the equation y = x², you need to understand what a parabola is. A parabola is a U-shaped curve that can open upwards or downwards. In this case, since the equation is y = x², the parabola will open upwards.
To graph this equation, you can start by making a table of values. Choose some x-values and plug them into the equation to find the corresponding y-values. For example, if you choose x = -2, -1, 0, 1, and 2, you can calculate the corresponding y-values as follows:
For x = -2: y = (-2)² = 4
For x = -1: y = (-1)² = 1
For x = 0: y = (0)² = 0
For x = 1: y = (1)² = 1
For x = 2: y = (2)² = 4
Now that you have a set of x and y-values, plot these points on a graph. Connect the points with a smooth curve, and you'll have a graph of the equation y = x².
For the first equation, x - y = 0, you can rearrange it to y = x. This equation represents a straight line that passes through the origin (0, 0) and has a slope of 1.
Now, to find the solution to the system of equations graphically, you need to look for the points where the graph of the parabola (y = x²) intersects with the line (y = x).
When you plot both the line and the parabola on the same graph, the points where they intersect will give you the solutions to the system of equations.
I hope this explanation helps! Let me know if you have any further questions.