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Arent quadratic functions supposed to be symmetrical?

Jayden

New member
Jun 6, 2012
21
As we were talking about earlier.

f(x) = x^2 + 3x + 6
f ' (x) = 2x + 3

P1 = (0, f(0))

f ' (0) = 3.

This is not symetrical...

I hope you guys enjoy my question bombarding. As there are alot of maths concepts that make me want to cry. It's funny, half the people in my degree fail their programming units and easily pass maths. However I easily pass my programming units and fail at maths. Sigh..
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
They are symmetrical about the line perpendicular to the directrix, and going through the focus. To find that, I would complete the square. In your case, you get

$$f(x)=x^{2}+3x+6=x^{2}+3x+\frac{9}{4}-\frac{9}{4}+6=\left(x+\frac{3}{2}\right)^{2}+\frac{15}{4}.$$

This is a parabola opening up - its minimum occurs at $x=-3/2$. Plug in that $x$-value to obtain the $f(-3/2)=15/4$. So the minimum occurs at $(-3/2,15/4)$. The parabola is symmetric about the line $x=-3/2$. See here.