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

DashingDave shared this question 11 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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