find an equation of the tangent line to the curve y = x √x that is parallel to the line y = 1 + 3x.
y
=f(x)
=x√x
=x3/2
Use the power rule for derivatives to get
f'(x)=(3/2)√x
For L: y=3x+1, the slope is 3.
Solve for x0 in
f'(x0)=3
Find the ordinate
y0=f(x0)
The point of tangency is therefore at (x0,y0).
Now pass a line L1 through (x0,y0) having a slope of 3:
(y-y0)=3(x-x0)
Check that the line has a slope of 3 and pass through the point (x0,y0)