1 answer

How do you find the equation of the tangent line to the graph #f(x)=x^3# at the point where x=2?

Question:

How do you find the equation of the tangent line to the graph #f(x)=x^3# at the point where x=2?

Answers

The derivative gives us the general (that is, in terms of #x#) slope of the line
Given #f(x) =x^3#

#(df(x))/(dx) = 3x^2#

At #x=2#
- the slope is #m=3(2^2) = 12#
- and #f(2) = 2^3 = 8 rarr (x,y) = (2,8)#

We can use the slope-point form of the line to write:
#y-8 = 12(x-2)#
or, in standard form:
#12x -y = 16#

.

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