How do you find the equation of the line tangent to the graph of #(2x+3)^(1/2)# at the point x=3?

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Answer 1 · 2015-08-22T17:04:03

#y=x/3+2#

First use the chain rule to get the derivative:

Let:

#y=(2x+3)^(1/2)#

#y'=1/2(2x+3)^(-1/2)xx2#

=#(1)/(sqrt(2x+3))#

If #x=3#:

#y'=(1)/(sqrt(9))=1/3#

The equation of the line is of the form:

#y=mx+c#

So #m=1/3#

If #x=3# then #y=sqrt((2xx3)+3)=3#

So #3=1/3xx3+c#

#c=3-1=2#

So the equation#rArr#

#y=x/3+2#