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?

Question

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?

campbell_st · Answer

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 · Answer

Did he divide the two equations together? :s sorry im not sure

pottersheep · Answer

this is our very first lesson on rtes of change ever

campbell_st · Answer

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 · Answer

he has differentiated.... I'd presume that you know how to differentiate...

pottersheep · Answer

No...he didnt teach us that...ill look over my notes. seems like hes teaching himself sometimes

pottersheep · Answer

Nope he taught us power rules though, does that mean anything? And thank you so much for your explanation

pottersheep · Answer

It'll take a while but I'lltry toprocess what you said haha

pottersheep · Answer

should I google differentiate and teach myself?

campbell_st · Answer

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 · Answer

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 · Answer

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

precal · Answer

rates of change is the slope at that point

precal · Answer

http://www.calculus-help.com/tutorials these are good videos for an intro of calculus

precal · Answer

btw the power rule is one of many ways to take the derivative of a function.

pottersheep · Answer

Thank you very much. @campbell_st Thanks for the graph!