# calculus

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

• calculus -

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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