Y=(x-3)^2+36
Using (a-b)^2=a^2-2ab+b^2, with a=x and b=3
y=(x)^2-2(x)(3)+(3)^2+36
y=x^2-6x+9+36
y=x^2-6x+45
Answer: Option C. y=x^2-6x+45
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There are none. Absolute values are always positive