Find Equations for Differentials of 9y^2=x^3+3x^2 at (1,2/3), (-2,2/3), (-3,0)

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The discussion focuses on finding the equations for the differentials of the curve defined by the equation 9y² = x³ + 3x² at specific points: (1, 2/3), (-2, 2/3), and (-3, 0). The derivative is correctly identified as y' = (x² + 2x) / (6y). The slope at (1, 2/3) is confirmed to be 3/4, leading to the equation 9x - 12y - 1 = 0. The slope at (-2, 2/3) is 0, resulting in the equation y = 2/3. However, the slope at (-3, 0) does not exist, which should be represented by the vertical line x = -3.

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If anyone could check this answer for me, it would be greatly appreciated.

Find the equations relating the differentials on the curve [tex]9y^2= x^3 +3x^2[/tex] at the points [tex](1,\frac{2}{3}), (-2,\frac{2}{3}), (-3,0)[/tex]

Here's what I got:

[tex]y'= \frac {x^2+2x}{6y}[/tex]

[tex]m @ (1,\frac{2}{3})= \frac {3}{4}[/tex]
equation: 9x-12y-1=0

[tex]m @ (-2,\frac{2}{3})=0[/tex]
equation: y= 2/3

[tex]m @ (-3,0)[/tex]
does not exist

Thanks in advance.
 
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erik05 said:
If anyone could check this answer for me, it would be greatly appreciated.

Find the equations relating the differentials on the curve [tex]9y^2= x^3 +3x^2[/tex] at the points [tex](1,\frac{2}{3}), (-2,\frac{2}{3}), (-3,0)[/tex]

Here's what I got:

[tex]y'= \frac {x^2+2x}{6y}[/tex] <------ CORRECT

[tex]m @ (1,\frac{2}{3})= \frac {3}{4}[/tex]
equation: 9x-12y-1=0 <----- CORRECT

[tex]m @ (-2,\frac{2}{3})=0[/tex]
equation: y= 2/3 <----- CORRECT

[tex]m @ (-3,0)[/tex]
does not exist <----- Should be the Vertical Line {x = -3}

Thanks in advance.
Everything appears correct except the last item.

~~
 
Thanks for the help.
 

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