calculus
posted by Sasha .
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
=x^{3/2}
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:
(yy0)=3(xx0)
Check that the line has a slope of 3 and pass through the point (x0,y0)
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