Posted by **Sasha** on Wednesday, October 6, 2010 at 9:38pm.

find an equation of the tangent line to the curve y = x √x that is parallel to the line y = 1 + 3x.

- calculus -
**Reiny**, Wednesday, October 6, 2010 at 10:43pm
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

(x-36)(x-4) = 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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