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Rewrite f(x) = 3(x − 2)2 + 1 from vertex form to standard form
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f(x) = 3x2 + 13 f(x) = 3x2−12x + 13 f(x) = 3x2 + 12x − 11 f(x) = 9x2− 36x + 37
Standard form is Ax+Bx=C
that would be a straight line ^^ (vertical line )
\[\huge\rm f(x)=3(x-2)^2+1\] 2 is an exponent
I was thinking it was f(x) = 3x2−12x + 13
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Is that right?
how did you get that ? post your work please
I was guessing because it seems like a guessable answer but I actually have no idea
\[\huge\rm f(x)=3\color{Red}{(x-2)^2}+1\] (x-2)^2 is same as (x-2)(x-2) first we need to foil (x-2)^2
|dw:1446331873636:dw| multiply top terms by x
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