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.

dy/dx = x(1/2)x^(1/2) + x^(1/2)
= x/(2√x) + √x
slope of line = 3
so x/(2√x) + √x = 3
x/(2√x) = 3  √x
square both sides
x^2/(4x) = 9  6√x + x
times 4
x/4 = 9  6√x + x
simplifying ...
8√x = 12+x
square again
64x = 144 + 24x + x^2
x^2  40x + 144 = 0
(x36)(x4) = 0
x = 36 or x = 4
there will be two tangents possible
I will do one, you do the other.
if x=36, y = 216
let y = 3x + b
216 = 3(36) + b
b = 108
y = 3x + 108
etc.
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