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gandalfwiz
Group Title
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?
 3 years ago
 3 years ago
gandalfwiz Group Title
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?
 3 years ago
 3 years ago

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ybarrap Group TitleBest ResponseYou've already chosen the best response.0
complete the square
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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 years ago

gandalfwiz Group TitleBest ResponseYou've already chosen the best response.0
I'm confused... where did c go? and where did k come from?
 3 years ago

cyter Group TitleBest ResponseYou've already chosen the best response.1
\[ax^2+bx+c=0\] \[x^2+\frac{b}{a}x+\frac{c}{a}=0\]completing th esquare \[x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{c}{a}+\frac{b^2}{4a^2}\] \[x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^24ac}{4a^2}\] \[(x+\frac{b}{a})^2=\frac{b^24ac}{4a^2}\] \[(x+\frac{b}{a})^2=\pm\frac{\sqrt{b^24ac}}{2a}\] subtract b/a from both sides \[x=\frac{b}{a}\pm\frac{\sqrt{b^24ac}}{2a}\] simplifying to get \[x=\frac{b\pm\sqrt{b^24ac}}{2a}\]
 3 years ago

gandalfwiz Group TitleBest ResponseYou've already chosen the best response.0
Whoah, thanks for that hard work! My brain is currently processing...
 3 years ago

gandalfwiz Group TitleBest ResponseYou've already chosen the best response.0
No, wait, I already know how to get the quadratic formula, I want to know how to get b/2a from it
 3 years ago

cyter Group TitleBest ResponseYou've already chosen the best response.1
If you let let the discriminant \[b^24ac=0\] then \[b^2=4ac\] and \[x=\frac{b\pm\sqrt{b^2b^2}}{2a}\rightarrow\frac{b}{2a}\]
 3 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
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, \(k = (b^24ac)/2\) 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 \(x = b/2a\)
 3 years ago

cyter Group TitleBest ResponseYou've already chosen the best response.1
You could also take the first derivative and set it equal to zero to find the critical point \[\frac{d}{dx}=2ax+b=0\] \[x = \frac{b}{2a}\]
 3 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
careful cyter you are using calculus to explain it and the student might not be in calculus at all (perhaps Algebra I or II)
 3 years ago
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