how do i put y=x^2-4x-5 in y=a(x-h)^2+k form?

Answer 1

y = x^2 - 4x - 5 = (x - 4/2)^2 -(4/2)^2 - 5 = (x-2)^2 - 9

Answer 2

y = x^2 - 4x - 5

Well, this is (x-5) (x+1) Let's split the difference between them
and go with (x-2)

(x-2)^2 = x2-4x +4

so y = (x-2)^ -9 = x^2 -4x -5

Answer 3

y=(x-2)^2 -9 the method: Try to consider x^2 -4x as the square and the double product of the square of a binomial and add or substract
what you need to obtain the independent term

Answer 4

I don't need to write down the answer as others have written it down already.

But would like to point out that the key thing you need to know is:

x^2 + ax = (x + a/2)^2 - (a/2)^2