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Mathematics 7 Online
OpenStudy (anonymous):

(a) Find the equation of the tangent line to f(x) = x3 − 3x2 + 2x − 2 at x = 2.

OpenStudy (anonymous):

what does y= ?

OpenStudy (anonymous):

im confused on what im supposed to do? i understand to find the tangent line i must find the derivative. right?

OpenStudy (gorv):

find derivative

OpenStudy (gorv):

that will give you slope

OpenStudy (gorv):

f ' (x)=3x^2-6x+2

OpenStudy (anonymous):

\[f \prime (x) = 3x ^{2}-6x+2\]

OpenStudy (gorv):

at x=2 find its value ???

OpenStudy (gorv):

f ' (2) =???? replace x by 2 here

OpenStudy (anonymous):

2

OpenStudy (anonymous):

\[f \prime (2) = 2\]

OpenStudy (anonymous):

\[f \prime (2) = 3(2)^{2}-6(2)+2 = 12-12+2 = 2\]

OpenStudy (anonymous):

right?

OpenStudy (gorv):

now this is slope of the tangent

OpenStudy (anonymous):

so what is Y= ?

OpenStudy (anonymous):

does* y= ?

OpenStudy (gorv):

now we have x=2 we need valur of y so that we can find the point from which it will pass

OpenStudy (anonymous):

find f(2)

OpenStudy (gorv):

that is m=slope

OpenStudy (gorv):

y=mx+c

OpenStudy (anonymous):

so y-y1 = m(x-x1) ?

OpenStudy (anonymous):

or y - f(a)=m(x-a)

OpenStudy (gorv):

so to find y put x=2 in f(x) equation which is in start

OpenStudy (gorv):

yeah but we need y1

OpenStudy (anonymous):

yea but you need you actual function value at 2 so just plug 2 into the original

OpenStudy (anonymous):

oh yeah

OpenStudy (gorv):

which we will get when we calculate f(2)

OpenStudy (gorv):

you got it ??

OpenStudy (anonymous):

but we dont have m do we? oh yeah m=2, and x=2

OpenStudy (anonymous):

yea m = f'(x) derivative is your slope

OpenStudy (anonymous):

y=-2

OpenStudy (anonymous):

then plug in my factors and get y?

OpenStudy (anonymous):

so using the format of y - y1 = m(x - x1) you get?

OpenStudy (anonymous):

\[y-(-2)= 2(x-2)\]

OpenStudy (anonymous):

yes that is you tangent line equation

OpenStudy (anonymous):

y=2x-6

OpenStudy (gorv):

f(x) = x3 − 3x2 + 2x − 2 find f(2) here

OpenStudy (gorv):

f(x)=x3-3x2=2x-2 find f(2) from here

OpenStudy (anonymous):

she has that it is -2

OpenStudy (anonymous):

so my steps are= 1. find m by finding the derivative 2. find y by plugging x into f(x) 3. set my equation y-y1=m(x-x1) and plug in my factors

OpenStudy (anonymous):

correct

OpenStudy (gorv):

yeppp

OpenStudy (gorv):

you got the concept

OpenStudy (anonymous):

to find M i plug x=2 into f1(x) right?

OpenStudy (anonymous):

that will give me the slope at x=2 ?

OpenStudy (anonymous):

yes you plut your x-value given into your derivative formula

OpenStudy (anonymous):

plug*

OpenStudy (gorv):

no no no slope we calculated by derivative

OpenStudy (anonymous):

yyaaayyy i think i'm finally getting the hang of this lol

OpenStudy (gorv):

we have to find value of y corresponding to x=2

OpenStudy (anonymous):

yea F(prime)X = derivative

OpenStudy (gorv):

f(x) give value of y corrresponding to x

OpenStudy (anonymous):

basic format of these problems if your given that x equals a value 1. find f'(x) 2. find f'(a) 3.find f(a) 4. plug values into y - y1 = m(x - x1)

OpenStudy (gorv):

y=f(2)=2^3-3*2^2+2*2-2

OpenStudy (gorv):

y=f(2)=8-12+4-2=-2 so tangent pass through (2,-2)

OpenStudy (gorv):

and slope = 2

OpenStudy (gorv):

y-(-2)=2(x-2)

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

A solution and plot is attached. The calculations and plot are from Mathematica 9.

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