Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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 byTawa11 last updated on 19/Oct/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented byMJS_new last updated on 19/Oct/21

and if a≠±1? i.e. a=−3 or a=7

$$\mathrm{and}\:\mathrm{if}\:{a}\neq\pm\mathrm{1}? \\ $$ $$\mathrm{i}.\mathrm{e}.\:{a}=−\mathrm{3}\:\mathrm{or}\:{a}=\mathrm{7} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com