Find an equation of the tangent line to the curve at the given point

[afcwestwarrior]

y=1+2x-xcubed, (1,2)

 

[afcwestwarrior]

first i set up the problem, i put 1+2x-xcubed-1/x-1
i got stuck because of the xcubed, i forgot how to factor out polynomials that hae xcubed in them, i know this part doesn't have to do with calculus, but i forgot when i did polynomials like this

 

[afcwestwarrior]

hey i set up the problem help me out mane

 

[afcwestwarrior]

the answer isy=-x+3, but i need to figure out the slope

 

[afcwestwarrior]

i figured out my problem, except i don't know how to factor out xcubed polynomials, or put them into parenthesis

 

[tony873004]

the answer isy=-x+3, but i need to figure out the slope

You've got the slope in this formula. y=-x+3 is the same as saying y=-1x+3. And since y=mx+b, what do you suppose m is?

I got the answer without factoring anything.

I'm not sure how you got the slope (even though you don't realize you got it), but what I did was take the derivative of 1+2x-x³ Then I plugged my x-value into that formula.

 

[Bitter]

Typically speaking, when a professor wants a tangent line to a curve, he probably wants you to find the derivative. so i guess that would be 2x-3x^2

So since you are given points...

 

[afcwestwarrior]

well i know what the slope is but how do i find it, obviously it's -1, but the equation is confusing

 

[ultimateguy]

y = mx + b
y = -1x +3

b = 3, so m = ...

 

[afcwestwarrior]

i have to find the slope from y=1+2x-x3

 

[afcwestwarrior]

i factor it out to x(x+1)(x+1) but I'm wrong because i plug in 1 into the x, and i get 4

 

[afcwestwarrior]

here is what the slope form looks like 1+2x-x3-1/x-1 which is f(x)-a/x-a
the only part i am having trouble with is the 1+2x-x3, so what do i do with this

 

[ultimateguy]

1+2x-x^3

Derivative is 2-3x^2

You know that the slope is -1 and the point, so use the equation

y-y_0=m(x-x_0)

where m = -1, x_0=1 and y_0=2

 

[afcwestwarrior]

how did u get 2-3x2

 

[ultimateguy]

I took the derivative of the very first equation in my post. Since the derivative is equal to the slope of the tangent, then plug in x=1

2-3(1)^2=-1

 

[afcwestwarrior]

so this equals negative -1, ahh makes sense, so what do you do about the x-1 in the denominator

 

[ultimateguy]

Ah, you're doing it from first principles. Give me a second to see if I can work it out.

 

[ultimateguy]

Ok, from first principles, I get

\lim_{x\rightarrow a} \frac{-x^3+2x-1}{x-1}

When I plug in x=1 into the equation, the nominator and denominator both go to zero. So I need to use Hopital's rule and take the derivative of the top and the bottom.

\lim_{x\rightarrow a} \frac{-3x^2+2}{1}

Plug in x=1 and you get -1.

Have you learned Hopital's rule yet?

 

[afcwestwarrior]

nope i checked my notes, actually my teacher may have tought us but i probably missed that day or something

 

[afcwestwarrior]

what is this rule, if you don't feel like explaining it, i'll look it up on the internet

 

[ultimateguy]

It's basically this:

When you have a fraction inside of a limit, and the numerator and denominator both go to zero or infinity, then you need to differentiate the numerator and the denominator. Just treat them separately though, don't go use the quotient rule.

I'm sure your textbook has a better explanation.

 

[HallsofIvy]

Yes, that was the whole idea: the derivative of a function, at a given x, is the slope of the tangent line to the graph at that point! If you do not know that you cannot possibly do the problem. If do know that, then it is easy.

 

[afcwestwarrior]

i know how u got it

 

[MrSparky]

dy/dx = gradient of tangent = 2-3x squared
Then sub in point (1,2) and the gradient is -1
y=mx+b
we know m = -1
y= -x + b
Because the point (1,2) satisfy both the tangent and curve you sub it in
2= -1 + b
b= 3
therefore the equation of the tangent is y= -x + 3