anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ybarrap
  • ybarrap
complete the square
anonymous
  • anonymous
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 \[x=-\frac{b}{2a}\]
anonymous
  • anonymous
I'm confused... where did c go? and where did k come from?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[ax^2+bx+c=0\] \[x^2+\frac{b}{a}x+\frac{c}{a}=0\]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 \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
Whoah, thanks for that hard work! My brain is currently processing...
anonymous
  • anonymous
No, wait, I already know how to get the quadratic formula, I want to know how to get -b/2a from it
anonymous
  • anonymous
If you let let the discriminant \[b^2-4ac=0\] then \[b^2=4ac\] and \[x=\frac{-b\pm\sqrt{b^2-b^2}}{2a}\rightarrow\frac{-b}{2a}\]
JamesJ
  • 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 \(x = -b/2a\)
anonymous
  • anonymous
You could also take the first derivative and set it equal to zero to find the critical point \[\frac{d}{dx}=2ax+b=0\] \[x = -\frac{b}{2a}\]
precal
  • 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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.