- #1

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

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- Thread starter afcwestwarrior
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- #1

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

- #2

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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

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hey i set up the problem help me out mane

- #4

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

- #5

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- #6

tony873004

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

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-x

- #7

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So since you are given points....

- #8

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

- #9

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y = mx + b

y = -1x +3

b = 3, so m = ...

y = -1x +3

b = 3, so m = ...

- #10

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

- #11

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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

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the only part i am having trouble with is the 1+2x-x3, so what do i do with this

- #13

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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

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how did u get 2-3x2

Last edited:

- #15

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[tex]2-3(1)^2=-1[/tex]

- #16

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

- #17

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

- #18

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[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

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- #20

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

- #21

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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

HallsofIvy

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- #23

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i know how u got it

- #24

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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

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