The delta x and delta y (dx and dy) are the changes in x and in y. When you take the derivative you are provided with a ratio (rise over run) of these changes.
Take f(x)=3x^2 + 2x
Derivative: (dy/dx)=6x + 2
From this, you can solve for dx and dy algebraically.
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if you are looking at a graph the y intercept is when the graph crosses the y axis and the x intercept is when the graph crosses the x axis. if you have a formula... plug zero in for x to find the y intercept, and plug zero in for y to find the x intercept
If the point (x,y) is on the graph of the even function y = f(x) then so is (-x,y)
Select a set of x values and find the value of y or f(x) - depending on how the parabola is defined. These are the values that you need to graph.
The graph of the function y = (sin x)^2 has the same amplitude 1, and the same period 2pi, as the graph of the function y = sin x. The only difference between them is that the part of the graph of y = sin x which is below the x-axis is reflected above x axis. In order to graph the function y = (sin x)^2, we need to find the values of (x, y) for the five key points, where 0 ≤ x ≤ 2pi. Values of (x, y) on y = (sin x)^2: x = 0, y = 0 x = pi/2, y = 1 x = pi, y = 0 x = 3pi/2, y = 1 x = 2pi, y = 0 Plot these five key points and connect them with a smooth curve and graph one complete cycle of the given function.
A graph has both an x and y axis.