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

In summary, the conversation revolves around finding the slope of the tangent line to the curve y=1+2x-x^3 at the point (1,2). The slope is found using the derivative, which is 2-3x^2. The slope is then used to find the equation of the tangent line, y=-x+3. Various methods, such as using the derivative, Hopital's rule, and the equation of a line, are discussed to find the slope.
  • #1
afcwestwarrior
457
0
y=1+2x-xcubed, (1,2)
 
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  • #2
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
hey i set up the problem help me out mane
 
  • #4
the answer isy=-x+3, but i need to figure out the slope
 
  • #5
i figured out my problem, except i don't know how to factor out xcubed polynomials, or put them into parenthesis
 
  • #6
afcwestwarrior said:
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.
 
  • #7
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
well i know what the slope is but how do i find it, obviously it's -1, but the equation is confusing
 
  • #9
y = mx + b
y = -1x +3

b = 3, so m = ...
 
  • #10
i have to find the slope from y=1+2x-x3
 
  • #11
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
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
[tex]1+2x-x^3[/tex]

Derivative is [tex]2-3x^2[/tex]

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

[tex]y-y_0=m(x-x_0)[/tex]

where m = -1, [tex]x_0=1[/tex] and [tex]y_0=2[/tex]
 
  • #14
how did u get 2-3x2
 
Last edited:
  • #15
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

[tex]2-3(1)^2=-1[/tex]
 
  • #16
so this equals negative -1, ahh makes sense, so what do you do about the x-1 in the denominator
 
  • #17
Ah, you're doing it from first principles. Give me a second to see if I can work it out.
 
  • #18
Ok, from first principles, I get

[tex]\lim_{x\rightarrow a} \frac{-x^3+2x-1}{x-1}[/tex]

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.

[tex]\lim_{x\rightarrow a} \frac{-3x^2+2}{1}[/tex]

Plug in x=1 and you get -1.

Have you learned Hopital's rule yet?
 
  • #19
nope i checked my notes, actually my teacher may have tought us but i probably missed that day or something
 
  • #20
what is this rule, if you don't feel like explaining it, i'll look it up on the internet
 
  • #21
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.
 
  • #22
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.
 
  • #23
i know how u got it
 
  • #24
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
 

1. What does it mean to find an equation of the tangent line to a curve?

When finding the equation of the tangent line to a curve, we are essentially determining the line that touches the curve at a specific point. This line represents the instantaneous rate of change, or slope, of the curve at that point.

2. How do I find the equation of the tangent line to a curve?

To find the equation of the tangent line, we use the derivative of the curve at the given point. This derivative represents the slope of the tangent line. Then, we use the point-slope form of a line to plug in the slope and the given point to find the equation.

3. Can I find the tangent line at any point on the curve?

Yes, the tangent line can be found at any point on a curve. However, the equation of the tangent line will be different for each point on the curve.

4. What information do I need to find the equation of the tangent line to a curve?

You will need the x-coordinate of the point at which you want to find the tangent line, and the original equation of the curve. From there, you will use the derivative of the curve at the given point to find the slope of the tangent line.

5. Why is finding the equation of the tangent line important?

Finding the equation of the tangent line allows us to understand the behavior of a curve at a specific point. It also helps us find the rate of change of the curve at that point, which is useful in many real-world applications such as physics and economics.

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