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x^2 + 6x + 11 can be written as (x + h)^2 + k. Find k.
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Do you know complete square?
no
The technique goes by ax^2+bx+c=0 The coefficient of x^2 must be 1, so factor 'a' out a(x^2+bx/a)+c=0 divide bx/a by 2,then set the x^2 becomes x only,eliminate the x from b/a, then square the whole bracket you ended up with a(x+b/2a)^2+c=0 then, you need to bring the b/2a OUT the brackets and square it, and when you bring it out, it is always negative, be sure that the 'a' will be multiplying the one you bought it out a(x+b/2a)^2-(b/2a)^2(a)+c=0
thanks:)
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