SOLUTION: Find an equation of the tangent line to the curve y = x ( sq rt x) that is parallel to the line y = 1 + 3x

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Question 350596: Find an equation of the tangent line to the curve y = x ( sq rt x) that is parallel to the line y = 1 + 3x
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Find the derivative of y.
y=x%2Asqrt%28x%29=x%5E%283%2F2%29
dy%2Fdx=%283%2F2%29x%5E%281%2F2%29=%283%2F2%29sqrt%28x%29
The value of the derivative is the slope of the tangent line.
m=%283%2F2%29sqrt%28x%29=3
sqrt%28x%29=2
x=4
Then
y=x%2Asqrt%28x%29=%284%292=8
Use the point-slope form of a line,y-y%5Bp%5D=m%28x-x%5Bp%5D%29 with slope,m=3, and point,(4,8)
y-8=3%28x-4%29
y=3%28x-4%29%2B8
y=3x-12%2B8
highlight%28y=3x-4%29
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Graphical verification showing the function and the two lines.
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