Calling anyone who is good at Calculus, still confused.

[SithSolo1]

Given the equation f(x)=(1-5x)/(1+x^2), find an equation of a line passing through the point (-2,-0.4) that is also tangent to the graph of f ( x ). Note that the point (-2,-0.4) is not on the graph of f ( x ).

Then using f(x)=(1-5x)/(1+x^2), determine an equation for the line normal to the graph of f ( x ) when x = -2.

How in the heck do I do either of these?


Thank You,
-Totaly Failing Calc J.


EDIT: Fixed the equation. I'm stupid, sorry.

 

[Heisenberg]

What is k in your equations? Just an unknown?

 

[SithSolo1]

yeah, just unknown.

ooops, I need to fix it. K should be 5 and 5 should be X.

 

[BigPoppa]

First one:
Derive f(x). Solve for slope (m). Use point-slope formula to make a line through the point (-2, -.4).

Second:
Normal = perpendicular. Plug x=-2 into f(x) to give you a coordinate. (-2, y). Use this point with the derivative from problem 1 to find the slope. Take the negative reciprocal. Use point-slope to creat the line.

 

[SithSolo1]

I don't know point-slope.

I've failed every math since geometry. Alga 2, Alga 3/Trig, Pre-Cal, and now Calc.

 

[Heisenberg]

1) Take the derivative of f(x). Plug in x=-2 to get the slope at that point. Then use point-slope to find an equation for a line through the point (-2,-0.4)

2) Take derivative of f(x). The negative inverse gives you the perpendicular slope. Then do as above.

 

[BigPoppa]

Heisenberg, would we have said anything closer?

Point slope formula: y-y1=m(x-x1). The coordinate you get is your (x1,y1) values. Hope this helps.

 

[TechnoKid]

Originally posted by: SithSolo1
I don't know point-slope.

I've failed every math since geometry. Alga 2, Alga 3/Trig, Pre-Cal, and now Calc.

You don't fail Calc, you fail algebra. If you do not have a strong base of algebra when you go into calc, you won't do well. This is what my engineering and calc teacher tell me.

 

[BigPoppa]

How do you fail a math class yet move on? Use a little grey matter and hold yourself back a year math-wise. Let your skills get upto speed before moving on.

 

[Heisenberg]

Originally posted by: BigPoppa
Heisenberg, would we have said anything closer?

Point slope formula: y-y1=m(x-x1). The coordinate you get is your (x1,y1) values. Hope this helps.

We both said the same thing, and about as straightforward as you can make it. He really should know basic stuff like point-slope by now.

 

[SithSolo1]

You don't fail Calc, you fail algebra. If you do not have a strong base of algebra when you go into calc, you won't do well. This is what my engineering and calc teacher tell me.

That is one of the most intelligent things I've ever heard. It's true too. Going in teh sig(damn, won't fit).

He really should know basic stuff like point-slope by now.

I should but I don't, I can't comprehend higher level math. Teachers, tutors, programs, friends, nothing helps.

 

[TuxDave]

Here's a simple point slope line equation you can use. Given point x1 and y1 and slope m, this is a value equation.

(y-y1)/(x-x1) = m

 

[SithSolo1]

Ok, for the first one I got -(5/(1+x^2))-((2*(1-5*x)*x)/((1+x^2)^2)) for the derivative.
Then I got 0.76 for the slope when I plugged -2 in for x.

Then for the point slope I got:
y-y1=m(x-x1)
y-(-.4)=(0.76)x-(0.76(-2))
y=0.76x-(0.76(-2))+(-0.4)
y=0.76x-1.92

But when I plot it, it is not tanget to the graph of (1-5*x)/(1+x^2). It just cuts straight through the graph. I know there is a 99% chance this is a ID10T error on my part but I don't see where I messed up.

edit: Ok, I know my math is wrong but I still can't figure out where. I used Maple 9 and it spat out 0.76x+3.72 which is tanget to the first equation. How it got that I don't know.

 

[maziwanka]

Originally posted by: Heisenberg
1) Take the derivative of f(x). Plug in x=-2 to get the slope at that point. Then use point-slope to find an equation for a line through the point (-2,-0.4)

2) Take derivative of f(x). The negative inverse gives you the perpendicular slope. Then do as above.

thats cuz this is wrong. the point at f(x=-2) will not have the same slope as the like going through (-2,-0.4). you need to set the derivative of f(x), f'(x) equal to the slope of the line (-0.4-y)/(-2-x) where y=(1-5x)/(1+x^2). then solve for x. that gives u the point where the slopes will be the same.

 

[SithSolo1]

Originally posted by: maziwanka

Originally posted by: Heisenberg
1) Take the derivative of f(x). Plug in x=-2 to get the slope at that point. Then use point-slope to find an equation for a line through the point (-2,-0.4)

2) Take derivative of f(x). The negative inverse gives you the perpendicular slope. Then do as above.

thats cuz this is wrong. the point at f(x=-2) will not have the same slope as the like going through (-2,-0.4). you need to set the derivative of f(x), f'(x) equal to the slope of the line (-0.4-y)/(-2-x) where y=(1-5x)/(1+x^2). then solve for x. that gives u the point where the slopes will be the same.

Whaaaa... You're talking to me like I know math or something.

So -(5/(1+x^2))-((2*(1-5*x)*x)/((1+x^2)^2))=(-0.4-(1-5x)/(1+x^2))/(-2-x) solve for x?

 

[Kyteland]

You are thinking about this wrong. You correctly found the derivative, but you should not plug in x=2 for that equation.

You have two points int he system that you know. One is (-2.-0.4). The other is (x1,y1). You know that (x1,y1) is on the equation f(x)=(1-5x)/(1+x^2). You want to solve for x1 and y1 so plug in the variables to get this: y1=(1-5*x1)/(1+x1^2).

You also know that the derivative goes through the point (x1,y1). The equation for a line is slope = delta y/ delta x. You have two points and the equation for the slope at the point (x1,y1), namely f'(x) = -(5/(1+x^2))-((2*(1-5*x)*x)/((1+x^2)^2)). Plug in x1 to this equation to get -(5/(1+x1^2))-((2*(1-5*x1)*x1)/((1+x1^2)^2)) for the slope. delta y/ delta x = (y1+0.4)/(x1+2) so now you have (y1+0.4)/(x1+2) = -(5/(1+x1^2))-((2*(1-5*x1)*x1)/((1+x1^2)^2)).

You now have 2 equations with two variables and can solve for both x1 and y1 through the substitution method. Once you have x1 and y1 you can solve for the equation of the tangent line and you are done.


The second question is a lot simpler. You have f(x) and are given x so just plug in the value of x to get the point. f(x)=(1-5x)/(1+x^2) and x=-2 so then you can solve for y. You already have the equation for the derivative f'(x) so plug x=-2 in to that equation to get the slope. Then you can use the point-slope formula to get the equation of the tangent line. Once you know this it is easy to find the equation of the normal line (perpendicular) at that point.