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pottersheep Group Title

Intro to Rates of Change - Grade 12. I don't really get what my teachertaught? Can someone explain this step please?? He wrote "Find the equation of the tangent to y = x^3 - 3x^2 - x +2 at x = 2. y1 = 3x^2 - 6x -1 at x = 2. Steps: at x=2. m1 = 3(2)^2 - 6(2) - 1 = -1 (What??? m?? Where did that come from??) y = 2^3 - 3(2)^2 - 2 + 3 = -3 y-y1 m(x-x1) therefore y+3 = -1(x-2) y = -x-1 Can someone please explain?

  • one year ago
  • one year ago

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  1. campbell_st Group Title
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    ok. y1 is the 1st derivative of the function \[y = x^3 -3x^2 -x + 2\] if you differentiate you get \[\frac{dy}{dx} = 3x^2 - 6x - 1\] the 1st derivative is also the equation for the slope of the tangent at any point on the curve. in your question you need to find the equation of the tangent, which will be a straight line, st the point where x = 2. so substitute x = 2 into the 1st derivative. but your teacher has made a shortcut and said the slope of the tangent m, is \[\frac{dy}{dx} = m = 3(2)^2 - 6(2) - 1 = -1\] does this make sense...?

    • one year ago
  2. pottersheep Group Title
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    Did he divide the two equations together? :s sorry im not sure

    • one year ago
  3. pottersheep Group Title
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    this is our very first lesson on rtes of change ever

    • one year ago
  4. campbell_st Group Title
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    once you have the slope of the tangent at x = 2 you need a point (x, y) so you can use the point slope formula to find the equation of the tangent. so if x = 2 substitute it into the original equation to find y \[y = (2)^3 - 3(2)^2 - 2 + 2 = -4\] so now you have a point (2, -4) and a slope of m = -1 I'm not sure if the original equation has a constant of +2 or +3 so you point may be (2, -3) with m = -1 then use \[y - y_{1} = m(x - x_{1})\] and you'll get the equation of the tangent

    • one year ago
  5. campbell_st Group Title
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    he has differentiated.... I'd presume that you know how to differentiate...

    • one year ago
  6. pottersheep Group Title
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    No...he didnt teach us that...ill look over my notes. seems like hes teaching himself sometimes

    • one year ago
  7. pottersheep Group Title
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    Nope he taught us power rules though, does that mean anything? And thank you so much for your explanation

    • one year ago
  8. pottersheep Group Title
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    It'll take a while but I'lltry toprocess what you said haha

    • one year ago
  9. pottersheep Group Title
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    should I google differentiate and teach myself?

    • one year ago
  10. campbell_st Group Title
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    well he used the power rule to find the 1st derivative... if you look at the original equation and use the power rule on each term you will get y1 = 3x^2 - 6x - 1

    • one year ago
  11. pottersheep Group Title
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    You said "the 1st derivative is also the equation for the slope of the tangent at any point on the curve". is that why m = 3(2)^2 - 6(2) - 1 = -1? I didnt know what rule....or is there another wway to know?

    • one year ago
  12. campbell_st Group Title
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    thats correct... once you have the 1st derivative... then you can find the slope of a tangent at a specific point on the curve. here is all the information graphed. Hope this helps

    • one year ago
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  13. precal Group Title
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    rates of change is the slope at that point

    • one year ago
  14. precal Group Title
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    http://www.calculus-help.com/tutorials these are good videos for an intro of calculus

    • one year ago
  15. precal Group Title
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    btw the power rule is one of many ways to take the derivative of a function.

    • one year ago
  16. pottersheep Group Title
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    Thank you very much. @campbell_st Thanks for the graph!

    • one year ago
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