SOLUTION: Consider the curve with equation y^2 = x^3 + 3x^2. a. Find an equation of the tangent line to this curve at the point( 1,2). b. At what points does this curve have horizontal t

Algebra ->  Rational-functions -> SOLUTION: Consider the curve with equation y^2 = x^3 + 3x^2. a. Find an equation of the tangent line to this curve at the point( 1,2). b. At what points does this curve have horizontal t      Log On


   

Question 1023346: Consider the curve with equation y^2 = x^3 + 3x^2.
a. Find an equation of the tangent line to this curve at the point( 1,2).
b. At what points does this curve have horizontal tangents? Justify your answer algebraically using a derivative.
c. Can you find the name of this curve?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Differentiate implicitly,
2ydy=3x%5E2dx%2B6xdx
%282y%29dy=%283x%5E2%2B6x%29dx
dy%2Fdx=%283x%5E2%2B6x%29%2F%282y%29
So at (1,2),
dy%2Fdx=%283%281%29%5E2%2B6%281%29%29%2F%282%282%29%29
dy%2Fdx=%283%2B6%29%2F4
dy%2Fdx=9%2F4
Using the point-slope form,
y-2=%289%2F4%29%28x-1%29
y-2=%289%2F4%29x-9%2F4
y=%289%2F4%29x-9%2F4%2B8%2F4
y=%289%2F4%29x-1%2F4
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b) When the derivative equals zero, the numerator must equal zero so,
3x%5E2%2B6x=0
3x%28x%2B2%29=0
Two solutions,
x=0
However, when x=0, y=0, so then the derivative is undefined since there is a division by zero so this x value is not allowed.
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x%2B2=0
x=-2
So here there are two tangents,
y%5E2=%28-2%29%5E3%2B3%28-2%29%5E2
y%5E2=-8%2B12
y%5E2=4
y=2 and y=-2
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c) Fred or it might be George.