How to graph the tangent to a curve at some point?

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SUMMARY

This discussion focuses on graphing the tangent line to the function x^2 at the point where x equals 3 using graphing software. The derivative of the function, 2x, is evaluated at x = 3 to determine the slope, which is 6. The equation for the tangent line is derived as y = f'(a)(x - a) + f(a), where a is the point of tangency. The user successfully implements this equation in their graphing software to visualize the tangent line.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically derivatives.
  • Familiarity with the function x^2 and its properties.
  • Knowledge of how to use graphing software for plotting functions.
  • Ability to manipulate equations to derive tangent lines.
NEXT STEPS
  • Learn how to implement derivative calculations in graphing software like Desmos or GeoGebra.
  • Explore the concept of higher-order derivatives and their applications.
  • Study the geometric interpretation of derivatives and tangent lines.
  • Investigate other functions and their tangent lines at various points.
USEFUL FOR

Students learning calculus, educators teaching derivatives, and anyone interested in visualizing mathematical concepts using graphing software.

Juwane
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Using a graphing software, I'm trying to graph three things:

1. The function [tex]x^2[/tex].
2. It's derivative [tex]2x[/tex].
3. The tangent to the curve at point 3.

Now I know that that if I want to find the slope of the curve at point 3, I should substitute 3 into the derivative [tex]2x[/tex], which will give 6. What I want is the tangent line, not it's slope, at point 3. What equation should I input into the software to show the tangent to the curve at point 3? Please help.
 
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Asking the program to get the line tangent to a point on the graph probably depends on the program.
Why not use the equation for a tangent line at point (a, f(a))
y = f'(a)(x - a) + f(a)
 
Bohrok said:
Asking the program to get the line tangent to a point on the graph probably depends on the program.
Why not use the equation for a tangent line at point (a, f(a))
y = f'(a)(x - a) + f(a)

It worked! How did you derive this equation?

EDIT: Never mind, I've find out.
 
Last edited:

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