How do you write $y = 3 {(x - 2)}^{2} + 1$ $y = 3(x-2)^2 + 1$ in standard form?

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Answer 1 · 2018-05-01T09:33:40

$y = 3 x^{2} - 12 x + 13$ $y=3x^2-12x+13$

${(x - 2)}^{2} = (x - 2) (x - 2) = x^{2} - 2 x - 2 x + 2 = x^{2} - 4 x + 4$ $(x-2)^2 =(x-2)(x-2) = x^2-2x-2x+2=x^2-4x+4$

$3 {(x - 2)}^{2} = 3 (x^{2} - 4 x + 4) = 3 x^{2} - 12 x + 12$ $3(x-2)^2=3(x^2-4x+4)=3x^2-12x+12$

So $3 {(x - 2)}^{2} + 1 = 3 x^{2} - 12 x + 12 + 1 = 3 x^{2} - 12 x + 13$ $3(x-2)^2+1 =3x^2-12x+12 +1=3x^2-12x+13$

How do you write y=3(x−2)2+1y = 3(x-2)^2 + 1 in standard form?