## gandalfwiz 3 years ago I just learned how to derive the quadratic formula from ax2 bx+c, but I can't figure out how to derive -b/2a (the x coordiante for a vertex). Any ideas?

1. ybarrap

complete the square

2. satellite73

that comes from completing the square and writing $y=ax^2+bx+c$ as $y=a(x+\frac{b}{2a})^2+k$ and since the first term is a perfect square, if "a" is positive the least it can be is 0, and if a is negative the greatest it can be is 0, leaving a max of min of y at k. this occurs when $x=-\frac{b}{2a}$

3. gandalfwiz

I'm confused... where did c go? and where did k come from?

4. cyter

$ax^2+bx+c=0$ $x^2+\frac{b}{a}x+\frac{c}{a}=0$completing th esquare $x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$ $x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}$ $(x+\frac{b}{a})^2=\frac{b^2-4ac}{4a^2}$ $(x+\frac{b}{a})^2=\pm\frac{\sqrt{b^2-4ac}}{2a}$ subtract b/a from both sides $x=-\frac{b}{a}\pm\frac{\sqrt{b^2-4ac}}{2a}$ simplifying to get $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

5. gandalfwiz

Whoah, thanks for that hard work! My brain is currently processing...

6. gandalfwiz

No, wait, I already know how to get the quadratic formula, I want to know how to get -b/2a from it

7. cyter

If you let let the discriminant $b^2-4ac=0$ then $b^2=4ac$ and $x=\frac{-b\pm\sqrt{b^2-b^2}}{2a}\rightarrow\frac{-b}{2a}$

8. JamesJ

Read again what sat73 wrote carefully. He wrote k for all the stuff left over, because it's exact value isn't important. For the record, $$k = (b^2-4ac)/2$$ but as I say that's not critical for the argument. Having showing that the equation equals $a(x +b/2a)^2 + k$ you now ask yourself when does it have a max or min; i.e., where is the vertex. If occurs precisely when the first term is zero, i.e., $a(x +b/2a)^2 = 0$ which means $$x = -b/2a$$

9. cyter

You could also take the first derivative and set it equal to zero to find the critical point $\frac{d}{dx}=2ax+b=0$ $x = -\frac{b}{2a}$

10. precal

careful cyter you are using calculus to explain it and the student might not be in calculus at all (perhaps Algebra I or II)