What is the equation of the tangent line to the function √x (x^2 + 1) at x=1?

Question

Test tomorrow, please help!

Answer 1

y = sqrt(x)*(x^2+1)

y' = sqrt(x) ( 2x) + x^(-1/2)*(x^2+1)

the slope at x=1 is 2 + 2 = 4

at x=1, y = 2, so (1,2) with slope m=4

y = mx+b

2 = 4(1) + b

2 = 4 + b

b = -2

y = 4x - 2