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

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##)

Leave a Reply

Your email address will not be published. Required fields are marked *

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

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##)

Leave a Reply

Your email address will not be published. Required fields are marked *