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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

Ifyouwanttoeasilywriteanyquadraticfunctioninvertexform,justusetheseformulas: f(x)=ax2+bx+c ifa>0,then: f(x)=(x+b2)2−(b2)2+c ifa<0,then: f(x)=−(x−(b2))2+(b2)2+c Let′stakesomeexamples: f(x)=x2−6x+7 f(x)=(x−62)2−(62)2+7 f(x)=(x−3)2−9+7 f(x)=(x−3)2−2 Nowlet′stakethesameexample,butwhena<0: f(x)=−x2−6x+7 f(x)=−(x−(−62))2+(62)2+7 f(x)=−(x+3)2+9+7 f(x)=−(x+3)2+16 Anotherexample: f(x)=x2+4x−5 f(x)=(x+42)2−(42)2−5 f(x)=(x+2)2−4−5 f(x)=(x+2)2−9 Nowlet′salsoseewhathappenswhena<0: f(x)=−x2+4x−5 f(x)=−(x−(42))2+(42)2−5 f(x)=−(x−2)2+4−5 f(x)=−(x−2)2−1

Commented byTawa11 last updated on 19/Oct/21

Great sir

Greatsir

Commented byMJS_new last updated on 19/Oct/21

and if a≠±1? i.e. a=−3 or a=7

andifa≠±1? i.e.a=−3ora=7

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