Quadratic Functions - vertex of graph proof












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enter image description here



Can you explain how $x$ becomes $x + b/2a$ and $c$ becomes $4ac - b^2/4a$ all of sudden?



Can you please explain at a Pre-Calculus level, thank you very much.






algebra-precalculus quadratics







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  • If you expand $left(x+frac baright)^2$, and multiply through and add back the constant, you’ll get the original expression.
    – D.R.
    Dec 10 '18 at 3:55
















0














enter image description here



Can you explain how $x$ becomes $x + b/2a$ and $c$ becomes $4ac - b^2/4a$ all of sudden?



Can you please explain at a Pre-Calculus level, thank you very much.






algebra-precalculus quadratics







share|cite|improve this question
























  • If you expand $left(x+frac baright)^2$, and multiply through and add back the constant, you’ll get the original expression.
    – D.R.
    Dec 10 '18 at 3:55














0












0








0







enter image description here



Can you explain how $x$ becomes $x + b/2a$ and $c$ becomes $4ac - b^2/4a$ all of sudden?



Can you please explain at a Pre-Calculus level, thank you very much.






algebra-precalculus quadratics







share|cite|improve this question















enter image description here



Can you explain how $x$ becomes $x + b/2a$ and $c$ becomes $4ac - b^2/4a$ all of sudden?



Can you please explain at a Pre-Calculus level, thank you very much.






algebra-precalculus quadratics




algebra-precalculus quadratics






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share|cite|improve this question













share|cite|improve this question




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edited Dec 10 '18 at 4:06









Andrei

11.2k21026




11.2k21026










asked Dec 10 '18 at 3:48









Ke Ke

1027




1027












  • If you expand $left(x+frac baright)^2$, and multiply through and add back the constant, you’ll get the original expression.
    – D.R.
    Dec 10 '18 at 3:55


















  • If you expand $left(x+frac baright)^2$, and multiply through and add back the constant, you’ll get the original expression.
    – D.R.
    Dec 10 '18 at 3:55
















If you expand $left(x+frac baright)^2$, and multiply through and add back the constant, you’ll get the original expression.
– D.R.
Dec 10 '18 at 3:55




If you expand $left(x+frac baright)^2$, and multiply through and add back the constant, you’ll get the original expression.
– D.R.
Dec 10 '18 at 3:55










1 Answer
1






active

oldest

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0














This technique is called completing the square.






share|cite|improve this answer





















  • Would this not work better as a comment?
    – D.R.
    Dec 10 '18 at 4:04










  • Yeah. That's a respectable position. At one point I had that attitude. but alot of these PSQs get answered in the comments and then never actually answered. My thinking on this issue has changed. I think that if you give an answer (even if it's a hint) you should put it as an answer. And I reserve the right to change my thinking on this.
    – Mason
    Dec 10 '18 at 4:06












  • $$y=ax^2+bx+c$$ $$y/a=x^2+frac{b}ax+frac{c}a$$ $$y/a=x^2+frac{b}ax+frac{b^2}{4a^2}-frac{b^2}{4a^2}+frac{c}a$$ $$y/a=(x+frac{b}{2a})^2+frac{c}a-frac{b^2}{4a^2}$$ $$y/a=(x+frac{b}{2a})^2+frac{4ac}{4a^2}-frac{b^2}{4a^2}$$ $$y/a=left(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a^2}$$ $$y=aleft(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a}$$
    – Christopher Marley
    Dec 10 '18 at 4:48













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1 Answer
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1 Answer
1






active

oldest

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active

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active

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0














This technique is called completing the square.






share|cite|improve this answer





















  • Would this not work better as a comment?
    – D.R.
    Dec 10 '18 at 4:04










  • Yeah. That's a respectable position. At one point I had that attitude. but alot of these PSQs get answered in the comments and then never actually answered. My thinking on this issue has changed. I think that if you give an answer (even if it's a hint) you should put it as an answer. And I reserve the right to change my thinking on this.
    – Mason
    Dec 10 '18 at 4:06












  • $$y=ax^2+bx+c$$ $$y/a=x^2+frac{b}ax+frac{c}a$$ $$y/a=x^2+frac{b}ax+frac{b^2}{4a^2}-frac{b^2}{4a^2}+frac{c}a$$ $$y/a=(x+frac{b}{2a})^2+frac{c}a-frac{b^2}{4a^2}$$ $$y/a=(x+frac{b}{2a})^2+frac{4ac}{4a^2}-frac{b^2}{4a^2}$$ $$y/a=left(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a^2}$$ $$y=aleft(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a}$$
    – Christopher Marley
    Dec 10 '18 at 4:48


















0














This technique is called completing the square.






share|cite|improve this answer





















  • Would this not work better as a comment?
    – D.R.
    Dec 10 '18 at 4:04










  • Yeah. That's a respectable position. At one point I had that attitude. but alot of these PSQs get answered in the comments and then never actually answered. My thinking on this issue has changed. I think that if you give an answer (even if it's a hint) you should put it as an answer. And I reserve the right to change my thinking on this.
    – Mason
    Dec 10 '18 at 4:06












  • $$y=ax^2+bx+c$$ $$y/a=x^2+frac{b}ax+frac{c}a$$ $$y/a=x^2+frac{b}ax+frac{b^2}{4a^2}-frac{b^2}{4a^2}+frac{c}a$$ $$y/a=(x+frac{b}{2a})^2+frac{c}a-frac{b^2}{4a^2}$$ $$y/a=(x+frac{b}{2a})^2+frac{4ac}{4a^2}-frac{b^2}{4a^2}$$ $$y/a=left(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a^2}$$ $$y=aleft(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a}$$
    – Christopher Marley
    Dec 10 '18 at 4:48
















0












0








0






This technique is called completing the square.






share|cite|improve this answer












This technique is called completing the square.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 4:02









Mason

1,9641530




1,9641530












  • Would this not work better as a comment?
    – D.R.
    Dec 10 '18 at 4:04










  • Yeah. That's a respectable position. At one point I had that attitude. but alot of these PSQs get answered in the comments and then never actually answered. My thinking on this issue has changed. I think that if you give an answer (even if it's a hint) you should put it as an answer. And I reserve the right to change my thinking on this.
    – Mason
    Dec 10 '18 at 4:06












  • $$y=ax^2+bx+c$$ $$y/a=x^2+frac{b}ax+frac{c}a$$ $$y/a=x^2+frac{b}ax+frac{b^2}{4a^2}-frac{b^2}{4a^2}+frac{c}a$$ $$y/a=(x+frac{b}{2a})^2+frac{c}a-frac{b^2}{4a^2}$$ $$y/a=(x+frac{b}{2a})^2+frac{4ac}{4a^2}-frac{b^2}{4a^2}$$ $$y/a=left(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a^2}$$ $$y=aleft(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a}$$
    – Christopher Marley
    Dec 10 '18 at 4:48




















  • Would this not work better as a comment?
    – D.R.
    Dec 10 '18 at 4:04










  • Yeah. That's a respectable position. At one point I had that attitude. but alot of these PSQs get answered in the comments and then never actually answered. My thinking on this issue has changed. I think that if you give an answer (even if it's a hint) you should put it as an answer. And I reserve the right to change my thinking on this.
    – Mason
    Dec 10 '18 at 4:06












  • $$y=ax^2+bx+c$$ $$y/a=x^2+frac{b}ax+frac{c}a$$ $$y/a=x^2+frac{b}ax+frac{b^2}{4a^2}-frac{b^2}{4a^2}+frac{c}a$$ $$y/a=(x+frac{b}{2a})^2+frac{c}a-frac{b^2}{4a^2}$$ $$y/a=(x+frac{b}{2a})^2+frac{4ac}{4a^2}-frac{b^2}{4a^2}$$ $$y/a=left(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a^2}$$ $$y=aleft(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a}$$
    – Christopher Marley
    Dec 10 '18 at 4:48


















Would this not work better as a comment?
– D.R.
Dec 10 '18 at 4:04




Would this not work better as a comment?
– D.R.
Dec 10 '18 at 4:04












Yeah. That's a respectable position. At one point I had that attitude. but alot of these PSQs get answered in the comments and then never actually answered. My thinking on this issue has changed. I think that if you give an answer (even if it's a hint) you should put it as an answer. And I reserve the right to change my thinking on this.
– Mason
Dec 10 '18 at 4:06






Yeah. That's a respectable position. At one point I had that attitude. but alot of these PSQs get answered in the comments and then never actually answered. My thinking on this issue has changed. I think that if you give an answer (even if it's a hint) you should put it as an answer. And I reserve the right to change my thinking on this.
– Mason
Dec 10 '18 at 4:06














$$y=ax^2+bx+c$$ $$y/a=x^2+frac{b}ax+frac{c}a$$ $$y/a=x^2+frac{b}ax+frac{b^2}{4a^2}-frac{b^2}{4a^2}+frac{c}a$$ $$y/a=(x+frac{b}{2a})^2+frac{c}a-frac{b^2}{4a^2}$$ $$y/a=(x+frac{b}{2a})^2+frac{4ac}{4a^2}-frac{b^2}{4a^2}$$ $$y/a=left(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a^2}$$ $$y=aleft(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a}$$
– Christopher Marley
Dec 10 '18 at 4:48






$$y=ax^2+bx+c$$ $$y/a=x^2+frac{b}ax+frac{c}a$$ $$y/a=x^2+frac{b}ax+frac{b^2}{4a^2}-frac{b^2}{4a^2}+frac{c}a$$ $$y/a=(x+frac{b}{2a})^2+frac{c}a-frac{b^2}{4a^2}$$ $$y/a=(x+frac{b}{2a})^2+frac{4ac}{4a^2}-frac{b^2}{4a^2}$$ $$y/a=left(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a^2}$$ $$y=aleft(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a}$$
– Christopher Marley
Dec 10 '18 at 4:48




















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