Calculating Equations of Tangent Lines to a Cubic Function through an External Point

DashingDave shared this question 4 months ago
Answered

I've seen posts in the forums regarding how to have GeoGebra calculate tangent lines to quadratic functions (and designating them as conics) through a point not on the function's curve, but I'm not sure if it's possible to do this with a cubic function.

I've seen an item that can draw tangent lines to a quartic function (https://www.geogebra.org/m/...), so it must be possible, but, when I try to use the tangent function or the shortcut icon/button, I get a tangent line to the curve--but not the one I want.

I've tried typing

A=(1,4)

and then

y=x^3-10x^2+6x-2

and then using the button or the tangent function [tangent(A,p)], p being the label GG gave my cubic function.

The tangent that appears goes through (1,p(1)), not (1,4).

Comments (2)

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You need to do this algebraically, using the steps you usually follow when doing it on paper.

So if P is the (unknown) point in which the tangent touches your cubic f, the tangent through P has equation y - y(P) =f'(x(P)) (x - x(P)).

Then you must tell your tangent to pass through A, and solve this condition to find the point of contact P, and finally create line AP.

See also https://help.geogebra.org/t...

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and tangent(P,f) gives the tangent to f through (x(P),f(x(P)))

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