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mathcalculus
Determine the slope of the secant line for the curve defined by the equation: (attached below)
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?
is that like the differentiation but in other terms?
do we substitute just for it?
can we solve this together? @genius12
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 .
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?
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.