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That is my exact question. If we graph $x^2$ + $y$ $=$ $0$, do we get a downward opening parabola?

Let me explain what actually got me confused. I know that $y$ = $x^2$ is an upward opening parabola because its leading coefficient is positive, or in other words, its double derivative is positive.

Likewise, $y$ $=$ $-x^2$ is a downward opening parabola because its double derivative is negative.

We can rewrite $y$ = $-x^2$ as $y$ + $x^2$ $=$ $0$.

Now, this equation $y$ + $x^2$ = $0$ should be a downward opening parabola too, because its basically the same equation as $y$ = $-x^2$. But I’m not convinced that it is indeed a downward opening parabola. How do I convince myself?

4d_
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    One way to convince yourself is to compute some specific points on the curve. Note that $(0, 0), (1, -1), (-1, -1)$ all satisfy the equation (e.g. $(-1)^2 + (-1) = 0$). Plot these three points on a Cartesian plane, and note how they indicate a downwards-facing parabola. – Theo Bendit Oct 15 '19 at 07:47
  • But why aren't you convinced? You haven't really explained what got you confused... – YiFan Tey Oct 15 '19 at 08:56

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Perhaps noting that since $x^2\ge0$, $y+x^2=0$ implies that $y$ is not positive.

ajotatxe
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  • Thank you. Perhaps, what got me confused was that the coefficient of $x^2$ is positive. Still I knew that it had to be a downward opening parabola, but wanted to understand it properly. Hence I posted it here – 4d_ Oct 15 '19 at 07:47