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Question Number 157053 by depressiveshrek last updated on 19/Oct/21
If you want to easily write any quadratic function in vertex form, just use these formulas: f(x)=ax^2 +bx+c if a>0, then: f(x)=(x+(b/2))^2 −((b/2))^2 +c if a<0, then: f(x)=−(x−((b/2)))^2 +((b/2))^2 +c Let′s take some examples: f(x)=x^2 −6x+7 f(x)=(x−(6/2))^2 −((6/2))^2 +7 f(x)=(x−3)^2 −9+7 f(x)=(x−3)^2 −2 Now let′s take the same example, but when a<0: f(x)=−x^2 −6x+7 f(x)=−(x−(−(6/2)))^2 +((6/2))^2 +7 f(x)=−(x+3)^2 +9+7 f(x)=−(x+3)^2 +16 Another example: f(x)=x^2 +4x−5 f(x)=(x+(4/2))^2 −((4/2))^2 −5 f(x)=(x+2)^2 −4−5 f(x)=(x+2)^2 −9 Now let′s also see what happens when a<0: f(x)=−x^2 +4x−5 f(x)=−(x−((4/2)))^2 +((4/2))^2 −5 f(x)=−(x−2)^2 +4−5 f(x)=−(x−2)^2 −1
$$\mathrm{If}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{easily}\:\mathrm{write}\:\mathrm{any}\:\mathrm{quadratic}\:\mathrm{function}\:\mathrm{in}\:\mathrm{vertex}\:\mathrm{form},\:\mathrm{just}\:\mathrm{use}\:\mathrm{these}\:\mathrm{formulas}: \\ $$$$\: \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\: \\ $$$$\mathrm{if}\:{a}>\mathrm{0},\:\mathrm{then}: \\ $$$${f}\left({x}\right)=\left({x}+\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{c} \\ $$$$\: \\ $$$$\mathrm{if}\:{a}<\mathrm{0},\:\mathrm{then}: \\ $$$${f}\left({x}\right)=−\left({x}−\left(\frac{{b}}{\mathrm{2}}\right)\right)^{\mathrm{2}} +\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{c} \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{take}\:\mathrm{some}\:\mathrm{examples}: \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{7} \\ $$$${f}\left({x}\right)=\left({x}−\frac{\mathrm{6}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{6}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{7} \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{9}+\mathrm{7} \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{2} \\ $$$$\: \\ $$$$\mathrm{Now}\:\mathrm{let}'\mathrm{s}\:\mathrm{take}\:\mathrm{the}\:\mathrm{same}\:\mathrm{example},\:\mathrm{but}\:\mathrm{when}\:{a}<\mathrm{0}: \\ $$$$\: \\ $$$${f}\left({x}\right)=−{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{7} \\ $$$${f}\left({x}\right)=−\left({x}−\left(−\frac{\mathrm{6}}{\mathrm{2}}\right)\right)^{\mathrm{2}} +\left(\frac{\mathrm{6}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{7} \\ $$$${f}\left({x}\right)=−\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}+\mathrm{7} \\ $$$${f}\left({x}\right)=−\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{16} \\ $$$$\: \\ $$$$\mathrm{Another}\:\mathrm{example}: \\ $$$$\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{5} \\ $$$${f}\left({x}\right)=\left({x}+\frac{\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{5} \\ $$$${f}\left({x}\right)=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}−\mathrm{5} \\ $$$${f}\left({x}\right)=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{9} \\ $$$$\: \\ $$$$\mathrm{Now}\:\mathrm{let}'\mathrm{s}\:\mathrm{also}\:\mathrm{see}\:\mathrm{what}\:\mathrm{happens}\:\mathrm{when}\:{a}<\mathrm{0}: \\ $$$$\: \\ $$$${f}\left({x}\right)=−{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{5} \\ $$$${f}\left({x}\right)=−\left({x}−\left(\frac{\mathrm{4}}{\mathrm{2}}\right)\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{5} \\ $$$${f}\left({x}\right)=−\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}−\mathrm{5} \\ $$$${f}\left({x}\right)=−\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\: \\ $$
Commented by Tawa11 last updated on 19/Oct/21
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by MJS_new last updated on 19/Oct/21
and if a≠±1? i.e. a=−3 or a=7
$$\mathrm{and}\:\mathrm{if}\:{a}\neq\pm\mathrm{1}? \\ $$$$\mathrm{i}.\mathrm{e}.\:{a}=−\mathrm{3}\:\mathrm{or}\:{a}=\mathrm{7} \\ $$

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