Given the equation of the curve:
[tex]y^2=x^3+3x^2...(1)[/tex]
Using implicit differentiation:
[tex]\begin{gathered} 2y\cdot\frac{dy}{dx}=3x^2+6x \\ \\ \frac{dy}{dx}=\frac{3x^2+6x}{2y} \end{gathered}[/tex]
For the points with the slope equal to 0, we have:
[tex]\frac{dy}{dx}=0[/tex]
Using the expression above:
[tex]\begin{gathered} \frac{3x^2+6x}{2y}=0 \\ \\ \frac{3x(x+2)}{2y}=0 \end{gathered}[/tex]
The apparent solutions are x = 0 and x = -2, but if x = 0, then y = 0 (using equation (1)), so we have an indetermination of 0/0. Then, the only solution is x = -2. Using this solution on (1):
[tex]\begin{gathered} y^2=(-2)^3+3(-2)^2=-8+12=4 \\ \\ \Rightarrow y=2\text{ or }y=-2 \end{gathered}[/tex]
And this result is consistent with the graph.
