Arent quadratic functions supposed to be symmetrical?

[Jayden1]

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 a lot 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]

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.