does the rate of change of y with respect to x equal m?

Question

anonymous · Answer

in the equation y=mx+b

anonymous · Answer

yes

nincompoop · Answer

yes

ankit042 · Answer

yes dy/dx = m

anonymous · Answer

can you explain? just so I know why

nincompoop · Answer

why m? no body knows. but that's just a placeholder - meaning it can be any letter or symbol assignment so long as it is indicated in the definition that it is the "slope" or the steepness of a line.

ankit042 · Answer

The equation says|dw:1375403071127:dw| 
The equation says y=mx +b

anonymous · Answer

@paigeRG $m$ is our slope and (non-vertical) lines are special because their explicit equations have constant slope

nincompoop · Answer

slope is important because it tells us (describes) the influence of the independent variable to the on the dependent variable.

anonymous · Answer

the importance of slope is that it characterizes the local behavior of a function

ankit042 · Answer

y2 = mx2+c
y1 = mx1+c
y2-y1=m(x2-x1)
y2-y1/x2-x1=m 
y2-y1/x2-x1 = rate of change of y wrt to x

anonymous · Answer

I know m is the slope and I know what slope is but I just would like to know what the question means...

nincompoop · Answer

meaning, for example that if we have a large value, may it be negative or positive, it indicates that the independent variable has a greater influence on the dependent variable.

ankit042 · Answer

@paigeRG i did a proof have a look

anonymous · Answer

$$y=mx+b$$So say we have two points $x,x+\Delta x$ and $mx+b,m(x+\Delta x)+b$. our rate of change is given:$$\frac{m(x+\Delta x)+b-(mx+b)}{x+\Delta x-x}=\frac{mx+m\Delta x+b-mx-b}{\Delta x}=\frac{m\Delta x}{\Delta x}=m$$

nincompoop · Answer

the question is essentially asking you to identify a part of the equation.

anonymous · Answer

Oookay thanks guys! is there a possible way to award all of you with a medal cus thatd be great

nincompoop · Answer

I don't need an award.

ankit042 · Answer

give it to @oldrin.bataku he has the best explanation :)

anonymous · Answer

nah, @ankit042 had it first :-p

anonymous · Answer

. . . .  well then

ankit042 · Answer

@paigeRG in dilemma :D

anonymous · Answer

That I am... okay I have com up with a solution.... Smiles for everyone!! 
:)    :D   :)    :D    :)   :D

anonymous · Answer

*come

ankit042 · Answer

:) good solution