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mathcalculus Group Title

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

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    • one year ago
  2. genius12 Group Title
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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:\[\bf m_{secant}=A.R.C.H|_{x_0}^{x_1}=\frac{ f(x_1)-f(x_0) }{ x_1-x_0 }\]

    • one year ago
  3. genius12 Group Title
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    @mathcalculus Can you do that?

    • one year ago
  4. mathcalculus Group Title
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    is that like the differentiation but in other terms?

    • one year ago
  5. mathcalculus Group Title
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    @genius12

    • one year ago
  6. mathcalculus Group Title
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    is it = to 1?

    • one year ago
  7. mathcalculus Group Title
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    do we substitute just for it?

    • one year ago
  8. mathcalculus Group Title
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    can we solve this together? @genius12

    • one year ago
  9. mathcalculus Group Title
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    im stuck.

    • one year ago
  10. genius12 Group Title
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    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

    • one year ago
  11. mathcalculus Group Title
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    yes

    • one year ago
  12. mathcalculus Group Title
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    that's the formula to find the slope .

    • one year ago
  13. mathcalculus Group Title
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    are you there?

    • one year ago
  14. genius12 Group Title
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    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|

    • one year ago
  15. mathcalculus Group Title
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    well i used that formula you gave me and i got 1/1

    • one year ago
  16. genius12 Group Title
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    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:\[\bf m=\frac{y_2-y_1}{x_2-x_1}=\frac{ f(-6)-f(-5) }{ -6-(-5) }\]

    • one year ago
  17. genius12 Group Title
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    \[\bf f(-6)=-2(-6)^2-1=-73\]\[\bf f(-5)=-2(-5)^2-1=-51\]Plugging these values in we get:\[\bf slope_{secant}=\frac{ f(-6)-f(-5) }{ -6-(-5) }=\frac{ -73-(-51) }{ -6-(-5) }=\frac{ -22 }{ -1 }=22\]

    • one year ago
  18. genius12 Group Title
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    @mathcalculus That's what you are supposed to do. How did you get 1?

    • one year ago
  19. mathcalculus Group Title
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    so it;s 22?

    • one year ago
  20. genius12 Group Title
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    Yes. How did you get 1?

    • one year ago
  21. mathcalculus Group Title
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    im not sure i did it the other way around.

    • one year ago
  22. mathcalculus Group Title
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    hey, thank you so much , i'll be here tomorrow. need to sleep, thanks again. @genius12

    • one year ago
  23. genius12 Group Title
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    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.

    • one year ago
  24. mathcalculus Group Title
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    lol thank you ^_^

    • one year ago
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