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I have the parabola

$$y=-10\cdot x^2+10$$

enter image description here

I want to obtain a parabola which passes in $(-10^{-6},0);(10^{-6},0);(0,10)$. Is it possible to write the equation modifying the above equation?

Thank you for your help.

Gennaro Arguzzi
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  • Sure: consider a general quadratic of the form $y = a x^2 + bx + c$. Individually substitute the three points of interest into the expression; this will yield three equations for the three unknown coefficients $a,b,c$. – okrzysik Jul 09 '16 at 09:34
  • Hi @okrzysik, I would to find the equation without solving a new system. – Gennaro Arguzzi Jul 09 '16 at 09:36
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    Well OK, you must set $c = 10$ and because the curve is symmetric about $x = 0$ we must have $b = 0$. So all that remains is to solve for $a$ in the expression $a(10^{-6})^2 + 10 = 0$. – okrzysik Jul 09 '16 at 09:40
  • So you will need to change the coefficient of the $x^2$ term in your current quadratic. – okrzysik Jul 09 '16 at 09:47

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