# How do you determine the y-intercept of the tangent line to the curve y=sqrt(x^2+4) at x=3?

to find the tangent line of the curve, take the derivative, to determine the slope of the tangent line.

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y=##sqrt(x^2+4)##
y=##(x^2+4)^(1/2)##
y’=##1/2####(x^2+4)^(-1/2)##(2x)
y’=##x/sqrt(x^2+4)##

because you want the tangent line at x=3, plug it in to y’ to find the slop at that point, since we know that the derivative is the slope at a certain point.

y'(3)=##3/sqrt(3^2+4)##=##3/sqrt(13)##

at x=3, the point on the curve is (3,##sqrt13##)
now, you can create the tangent line
(y-##sqrt13##)= ##3/sqrt13##(x-3)
y-##sqrt13##=##3/sqrt13##x-##9/sqrt13##
y=##3/sqrt13##x-##9/sqrt13##+##sqrt13##

so your y-intercept for the tangent line at x=3 would be (##9/sqrt13##+##sqrt13##)

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