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I assume you are asking about using differentiation. In this case, where dy-by-dx=0, there is a stationary point on the graph i.e. where the gradient is equal to 0.

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What is the definition of unsquaring a number?

unsquaring a number is to chage it back


(03.04 LC) Use the graph below to answer the following question graph of parabola going through 0 4 2 negative 4 and 4 4 What is the average rate of change from x 0 to x 4?

To find the average rate of change of the function from ( x = 0 ) to ( x = 4 ), you can use the formula: [ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} ] Here, ( f(0) = 4 ) and ( f(4) = 4 ). Thus, the average rate of change is: [ \frac{4 - 4}{4 - 0} = \frac{0}{4} = 0 ] Therefore, the average rate of change from ( x = 0 ) to ( x = 4 ) is 0.


A graph that is a line has a rate of change?

with y=mx+b dy/dx=m d^2.y/dx^2=0 The rate of change is 0


Can a rate change and the slope of the line be different quantities?

The instantaneous rate change of the variable y with respect to x must be the slope of the line at the point represented by that instant. However, the rate of change of x, with respect to y will be different [it will be the x/y slope, not the y/x slope]. It will be the reciprocal of the slope of the line. Also, if you have a time-distance graph the slope is the rate of chage of distance, ie speed. But, there is also the rate of change of speed - the acceleration - which is not DIRECTLY related to the slope. It is the rate at which the slope changes! So the answer, in normal circumstances, is no: they are the same. But you can define situations where they can be different.


What is the rate of change in the equation x equal 9?

0. Differentiation of a constant gives f'(x)=0.