What is the equation of the tangent line of #f(x)=x^2+5 # at #x=5#?

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Answer 1 · 2016-02-24T17:27:59

#y=4x+1/2#

#f(x)=x^2+5#

#f'(x)=2x#

This is equal to #m#, the gradient of the tangent line.

At #x=2:#

#m=2xx2=4#

#f(4)=2xx2+5=9#

So the point is (2,9).

The equation of the tangent line is of the form:

#y=mx+c#

#:.9=2xx9+c#

#:.c=1/2#

So the equation of the tangent line #rArr#

#y=4x+1/2#

The situation is shown here:

graph{(4x+1/2-y)(x^2+5-y)=0 [-56.93, 56.8, -28.45, 28.45]}