# Calculus – Find the Tangent Line Equation at Point (2,1)

By | October 28, 2016

## Find the Tangent Line Equation at Point (2,1)

In this tutorial students learn how to find the the equation of the line tangent to a function at the given point.  This point is also known as the point of tangency.  The students then verify the results using the TI-84C graphing calculator.  The students will use the GRAPH, DRAW, TANGENT, WINDOW and Y= functionality from the calculator.

## Implicit Function

$x^3+y^3=9$

## Explicit Function

$y=f(x)=(9-x^3)^{\frac {1}{3}}$

## Derivative

$y'=f'(x)=-\frac {x^2}{(9-x^3)^{\frac {2}{3}}}$

## Slope

Substitute in the x value from the point of tangency (2,1)
$y'=f'(2)=-\frac {(2)^2}{(9-(2)^3)^{\frac {2}{3}}}=-4$

## Tangent Line

$y=-4x+9$

## Graphing Tool

Click the link below to view this problem on an interactive online graphing calculator.