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Determine the slope of the secant line for the curve defined by the equation: (attached below)

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The slope of the secant line is basically the average rate of change from \(\bf x_0\) to \(\bf x_1\). This is given by:\[\bf m_{secant}=A.R.C.H|_{x_0}^{x_1}=\frac{ f(x_1)-f(x_0) }{ x_1-x_0 }\]
@mathcalculus Can you do that?

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Other answers:

is that like the differentiation but in other terms?
is it = to 1?
do we substitute just for it?
can we solve this together? @genius12
im stuck.
Ok I will solve step by step. Firstly, you should realise that this is essentially the slope formula which is:\[\bf m=\frac{ y_2-y_1 }{ x_2-x_1 }\]You remember that formula? @mathstudent55
that's the formula to find the slope .
are you there?
Well that what this essentially is! It's the slope, of the secant line, and the secant line is the line that connects the two points on the graph when x = -5 to when x = -6. I'll draw the graph and show to you:|dw:1375078386992:dw|
well i used that formula you gave me and i got 1/1
the y2 and y1 are such the y values at the -5 and -6. So plug in x = -5 and x = -6 in to f(x) = -2x^2 - 1 and find the y-values of each. Now just calculate the slope with the:\[\bf m=\frac{y_2-y_1}{x_2-x_1}=\frac{ f(-6)-f(-5) }{ -6-(-5) }\]
\[\bf f(-6)=-2(-6)^2-1=-73\]\[\bf f(-5)=-2(-5)^2-1=-51\]Plugging these values in we get:\[\bf slope_{secant}=\frac{ f(-6)-f(-5) }{ -6-(-5) }=\frac{ -73-(-51) }{ -6-(-5) }=\frac{ -22 }{ -1 }=22\]
@mathcalculus That's what you are supposed to do. How did you get 1?
so it;s 22?
Yes. How did you get 1?
im not sure i did it the other way around.
hey, thank you so much , i'll be here tomorrow. need to sleep, thanks again. @genius12
Well I'll list out the steps for you. Find the y-value at both x-values. Here we were finding the slope of the secant line (the line that connects the two y-values at x = -5 and x = -6). We first find the y-value at x = -6 by plugging -6 for x in to f(x). Then we find y-value at x = -5 by plugging in -5 for x. When we have these 2 y-values, you just use the slope formula.
lol thank you ^_^

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