Say we had the function $y = -6x^2-3$ and wanted to find the turning point, my first goal is to complete the square, but this is where it all falls apart. Of course I know you can differentiate both sides with respect to $x$ and get $y'=-12x$, which we can solve by setting it equal to $0$ and get $x=0$ for our turning point, which is correct. However I would like to ask is there a way to get this into the normal "vertex" form that we have for quadratics in the form $ax^2+bx+c$ where $b$ isnt equal to $0$?
Asked
Active
Viewed 34 times
0
-
Start with $y=-6(x^2+\frac{1}{2})$. Do you recognize what to do from there inside the bracket? – Paul Dec 05 '22 at 11:01
-
No sorry, usually when I complete the square you have it in the form $p(x^2+qx)$ – Nav Bhatthal Dec 05 '22 at 16:04
-
Why on Earth would you try to complete the square when it is already complete? The whole point of completing the square if you have $b \neq 0$ is to get to the situation where $b=0$, which is what you have here already from the outset... – Hans Lundmark Dec 05 '22 at 16:38
