Show equation of tangent

[swisscow]

Hi

I can't seem to work out how to show the equation of a tangent line in the graph view. I can see the tangent ok and move it around but not view the equation. I would be grateful for some help,

Many thanks

[nbenm]

Hi,

I don't think it's possible this way, but take a look here. There's another way to do this. It's in french, but some Swiss cows speak french ;-)

[toml_12953]

(11-13-2019 12:23 PM)nbenm Wrote:  Hi,
It's in french, but some Swiss cows speak french ;-)

One of my French teachers said I speak French like a Spanish cow! (comme une vache espagnol) Then again, he was from Québec Canada so what did he know about French?

[pinkman]

I think there is no way to get the formula of the tangent from the graphic view, but you can use the tangent formula in the symbolic view.
Assuming your first function is F1, and you want the tangent for X=3, then the formula is : F2(X)=SLOPE(F1,3)*(X-3)+F1(3)

[swisscow]

Thanks for your help. It's really easy on the Casio fx to do this and it's a big help for my students for the equation to appear on the screen. These workarounds work fine but it's not as straightforward as I'd like. I'm hoping to switch away from TI calls at school but I need to convince others that the prime is the way to go.

Thanks again, and Swiss cows may not speak French, but they muddle by. This one mores in German

[pinkman]

I agree, the equation of the line should be copied in the next empty Fn slot. Anyway, the split view could also help the students.
   

[CyberAngel]

(11-13-2019 03:04 PM)pinkman Wrote:  I think there is no way to get the formula of the tangent from the graphic view, but you can use the tangent formula in the symbolic view.
Assuming your first function is F1, and you want the tangent for X=3, then the formula is : F2(X)=SLOPE(F1,3)*(X-3)+F1(3)

Doesn't work correctly for me.
Could you kindly establish the exact 3rd degree polynomial and
then explain carefully each step on your SLOPE calculation for the plot, please.
Why? What happens here? etc

Thank you!
VPN

[pinkman]

(11-14-2019 10:08 AM)CyberAngel Wrote:  

(11-13-2019 03:04 PM)pinkman Wrote:  I think there is no way to get the formula of the tangent from the graphic view, but you can use the tangent formula in the symbolic view.
Assuming your first function is F1, and you want the tangent for X=3, then the formula is : F2(X)=SLOPE(F1,3)*(X-3)+F1(3)

Doesn't work correctly for me.
Could you kindly establish the exact 3rd degree polynomial and
then explain carefully each step on your SLOPE calculation for the plot, please.
Why? What happens here? etc

Thank you!
VPN

Here are the steps :

Consider (x)->f(x) defined in x1, and assume you're looking for the tangent in x1.
The tangent is a straight line, defined by a slope and a point. Assume its formula is (x)->t(x)
(a) In x1, this slope is f'(x1) => t(x) = f'(x1) * x + b
(b) In x1, this point is (x1, f(x1)) => t(x1) = f(x1)

Combine (a) & (b): f(x1) = f'(x1) * x1 + b
Isolate b: b = f(x1) - f'(x1) * x1
Insert b in t(x): t(x) = f'(x1) * x + f(x1) - f'(x1) * x1
Rewrite t(x): t(x) = f'(x1) * (x - x1) + f(x1)

In my example, f was F1 and x1 was 3.
In the HP Prime, f'(x1) is SLOPE(F1,x1)
=> SLOPE(F1,3)*(X-3)+F1(3)

[CyberAngel]

(11-14-2019 01:26 PM)pinkman Wrote:  

(11-14-2019 10:08 AM)CyberAngel Wrote:  Doesn't work correctly for me.
Could you kindly establish the exact 3rd degree polynomial and
then explain carefully each step on your SLOPE calculation for the plot, please.
Why? What happens here? etc

Thank you!
VPN

Here are the steps :

Consider (x)->f(x) defined in x1, and assume you're looking for the tangent in x1.
The tangent is a straight line, defined by a slope and a point. Assume its formula is (x)->t(x)
(a) In x1, this slope is f'(x1) => t(x) = f'(x1) * x + b
(b) In x1, this point is (x1, f(x1)) => t(x1) = f(x1)

Combine (a) & (b): f(x1) = f'(x1) * x1 + b
Isolate b: b = f(x1) - f'(x1) * x1
Insert b in t(x): t(x) = f'(x1) * x + f(x1) - f'(x1) * x1
Rewrite t(x): t(x) = f'(x1) * (x - x1) + f(x1)

In my example, f was F1 and x1 was 3.
In the HP Prime, f'(x1) is SLOPE(F1,x1)
=> SLOPE(F1,3)*(X-3)+F1(3)

Nice steps! :-)
Aha! - "In the HP Prime, f'(x1) is SLOPE(F1,x1)"

Could you kindly establish the exact 3rd degree polynomial
that you used in the picture or any other
that I can truly test.
Note that a derivative of a 3rd degree polynomial is a 2nd degree polynomial
a straight like is a 1st degree polynomial - one needs f''(x) in terms of derivation

[pinkman]

(11-14-2019 01:55 PM)CyberAngel Wrote:  Could you kindly establish the exact 3rd degree polynomial
that you used in the picture or any other
that I can truly test.
Note that a derivative of a 3rd degree polynomial is a 2nd degree polynomial
a straight like is a 1st degree polynomial - one needs f''(x) in terms of derivation

In the picture I used : F1(X)=X^3+0.5*X^2-2*X-4
I randomly showed the tangent for X=1.07 (right part of the split view).
The SLOPE for x=1.07 is 2.5047 (shown in lower part of display, and also true when calculated from the derivative 3*X^2+X-2 with X=1.07), but to draw the tangent, in the symbolic view you just have to write for F2 : SLOPE(F1,1.07)*(X-1.07)+F1(1.07)

I don't understand your reflexion about the derivative, as a tangent is always a straight line, whatever the function is.

[Wes Loewer]

(11-13-2019 11:28 PM)pinkman Wrote:  I agree, the equation of the line should be copied in the next empty Fn slot. Anyway, the split view could also help the students.

I like your split screen idea. What I have recommended to students in the past is to get the point from the coordinates displayed at the bottom of the screen and get the slope from Fcn/Slope.

[pinkman]

(11-13-2019 05:01 AM)swisscow Wrote:  Hi

I can't seem to work out how to show the equation of a tangent line in the graph view. I can see the tangent ok and move it around but not view the equation. I would be grateful for some help,

Many thanks

I’ve got it!
Use the Geometry App, draw your function and the tangent using symb view, then in num view find the equation.

Or in CAS: equation(tangent(plotfunc(x^3-x,x),1))

[vjc]

It is an old thread, but in case you still need an answer you can see my post at

https://www.hpmuseum.org/forum/thread-17658.html