Question

How do you find an equation of the tangent line to the curve `y=x*sqrt(x)` that is parallel to the line `y=1+3*x`?

Answer 1

First find where the derivative of `y=f(x)=x\sqrt{x}=x^{3/2}` has a value of three: `f'(x)=3/2 x^{1/2}`. Setting this equal to 3 gives `x^{1/2}=2` so `x=4`.

The equation of the tangent line to `y=x\sqrt{x}` that is parallel to `y=1+3x` is therefore `y=3(x-4)+f(4)=3(x-4)+8` or `y=3x-4`.