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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) = {a x}^{2} + b x + c

if a > 0, then :

f (x) = {(x + \frac{b}{2})}^{2} - {(\frac{b}{2})}^{2} + c

if a < 0, then :

f (x) = - {(x - (\frac{b}{2}))}^{2} + {(\frac{b}{2})}^{2} + c

{Let}^{'} s take some examples :

f (x) = x^{2} - 6 x + 7

f (x) = {(x - \frac{6}{2})}^{2} - {(\frac{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} - 6 x + 7

f (x) = - {(x - (- \frac{6}{2}))}^{2} + {(\frac{6}{2})}^{2} + 7

f (x) = - {(x + 3)}^{2} + 9 + 7

f (x) = - {(x + 3)}^{2} + 16

Another example :

f (x) = x^{2} + 4 x - 5

f (x) = {(x + \frac{4}{2})}^{2} - {(\frac{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} + 4 x - 5

f (x) = - {(x - (\frac{4}{2}))}^{2} + {(\frac{4}{2})}^{2} - 5

f (x) = - {(x - 2)}^{2} + 4 - 5

f (x) = - {(x - 2)}^{2} - 1

Commented by Tawa11 last updated on 19/Oct/21

Great sir

Commented by MJS_new last updated on 19/Oct/21

and if a \neq \pm 1 ?

i . e . a = - 3 or a = 7

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