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

1. Feb 11, 2007

### afcwestwarrior

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

2. Feb 11, 2007

### 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

3. Feb 11, 2007

### afcwestwarrior

hey i set up the problem help me out mane

4. Feb 11, 2007

### afcwestwarrior

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

5. Feb 11, 2007

### afcwestwarrior

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

6. Feb 12, 2007

### tony873004

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.

7. Feb 12, 2007

### 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....

8. Feb 12, 2007

### afcwestwarrior

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

9. Feb 12, 2007

### ultimateguy

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

b = 3, so m = ...

10. Feb 12, 2007

### afcwestwarrior

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

11. Feb 12, 2007

### 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

12. Feb 12, 2007

### 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

13. Feb 12, 2007

### 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 $$y_0=2$$

14. Feb 12, 2007

### afcwestwarrior

how did u get 2-3x2

Last edited: Feb 12, 2007
15. Feb 12, 2007

### 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

$$2-3(1)^2=-1$$

16. Feb 12, 2007

### afcwestwarrior

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

17. Feb 12, 2007

### ultimateguy

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

18. Feb 12, 2007

### 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?

19. Feb 12, 2007

### afcwestwarrior

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

20. Feb 12, 2007

### afcwestwarrior

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