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pottersheep
 4 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
yy1 m(xx1)
therefore y+3 = 1(x2)
y = x1
Can someone please explain?
pottersheep
 4 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 yy1 m(xx1) therefore y+3 = 1(x2) y = x1 Can someone please explain?

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campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.3ok. 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
 4 years ago
Best ResponseYou've already chosen the best response.0Did he divide the two equations together? :s sorry im not sure

pottersheep
 4 years ago
Best ResponseYou've already chosen the best response.0this is our very first lesson on rtes of change ever

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.3once 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
 4 years ago
Best ResponseYou've already chosen the best response.3he has differentiated.... I'd presume that you know how to differentiate...

pottersheep
 4 years ago
Best ResponseYou've already chosen the best response.0No...he didnt teach us that...ill look over my notes. seems like hes teaching himself sometimes

pottersheep
 4 years ago
Best ResponseYou've already chosen the best response.0Nope he taught us power rules though, does that mean anything? And thank you so much for your explanation

pottersheep
 4 years ago
Best ResponseYou've already chosen the best response.0It'll take a while but I'lltry toprocess what you said haha

pottersheep
 4 years ago
Best ResponseYou've already chosen the best response.0should I google differentiate and teach myself?

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.3well 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
 4 years ago
Best ResponseYou've already chosen the best response.0You 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
 4 years ago
Best ResponseYou've already chosen the best response.3thats 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

precal
 4 years ago
Best ResponseYou've already chosen the best response.0rates of change is the slope at that point

precal
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.calculushelp.com/tutorials these are good videos for an intro of calculus

precal
 4 years ago
Best ResponseYou've already chosen the best response.0btw the power rule is one of many ways to take the derivative of a function.

pottersheep
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you very much. @campbell_st Thanks for the graph!
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