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gandalfwiz
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?
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 \[x=-\frac{b}{2a}\]
I'm confused... where did c go? and where did k come from?
\[ax^2+bx+c=0\] \[x^2+\frac{b}{a}x+\frac{c}{a}=0\]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 \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Whoah, thanks for that hard work! My brain is currently processing...
No, wait, I already know how to get the quadratic formula, I want to know how to get -b/2a from it
If you let let the discriminant \[b^2-4ac=0\] then \[b^2=4ac\] and \[x=\frac{-b\pm\sqrt{b^2-b^2}}{2a}\rightarrow\frac{-b}{2a}\]
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 \(x = -b/2a\)
You could also take the first derivative and set it equal to zero to find the critical point \[\frac{d}{dx}=2ax+b=0\] \[x = -\frac{b}{2a}\]
careful cyter you are using calculus to explain it and the student might not be in calculus at all (perhaps Algebra I or II)