pottersheep
  • pottersheep
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?
Mathematics
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pottersheep
  • pottersheep
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?
Mathematics
chestercat
  • chestercat
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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campbell_st
  • campbell_st
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...?
pottersheep
  • pottersheep
Did he divide the two equations together? :s sorry im not sure
pottersheep
  • pottersheep
this is our very first lesson on rtes of change ever

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campbell_st
  • campbell_st
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
campbell_st
  • campbell_st
he has differentiated.... I'd presume that you know how to differentiate...
pottersheep
  • pottersheep
No...he didnt teach us that...ill look over my notes. seems like hes teaching himself sometimes
pottersheep
  • pottersheep
Nope he taught us power rules though, does that mean anything? And thank you so much for your explanation
pottersheep
  • pottersheep
It'll take a while but I'lltry toprocess what you said haha
pottersheep
  • pottersheep
should I google differentiate and teach myself?
campbell_st
  • campbell_st
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
pottersheep
  • pottersheep
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?
campbell_st
  • campbell_st
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
1 Attachment
precal
  • precal
rates of change is the slope at that point
precal
  • precal
http://www.calculus-help.com/tutorials these are good videos for an intro of calculus
precal
  • precal
btw the power rule is one of many ways to take the derivative of a function.
pottersheep
  • pottersheep
Thank you very much. @campbell_st Thanks for the graph!

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