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pottersheep

  • 2 years ago

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?

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  1. campbell_st
    • 2 years ago
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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...?

  2. pottersheep
    • 2 years ago
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    Did he divide the two equations together? :s sorry im not sure

  3. pottersheep
    • 2 years ago
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    this is our very first lesson on rtes of change ever

  4. campbell_st
    • 2 years ago
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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

  5. campbell_st
    • 2 years ago
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    he has differentiated.... I'd presume that you know how to differentiate...

  6. pottersheep
    • 2 years ago
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    No...he didnt teach us that...ill look over my notes. seems like hes teaching himself sometimes

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

  8. pottersheep
    • 2 years ago
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    It'll take a while but I'lltry toprocess what you said haha

  9. pottersheep
    • 2 years ago
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    should I google differentiate and teach myself?

  10. campbell_st
    • 2 years ago
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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

  11. pottersheep
    • 2 years ago
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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?

  12. campbell_st
    • 2 years ago
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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

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  13. precal
    • 2 years ago
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    rates of change is the slope at that point

  14. precal
    • 2 years ago
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    http://www.calculus-help.com/tutorials these are good videos for an intro of calculus

  15. precal
    • 2 years ago
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    btw the power rule is one of many ways to take the derivative of a function.

  16. pottersheep
    • 2 years ago
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    Thank you very much. @campbell_st Thanks for the graph!

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