How do you solve using the quadratic formula for #y = x^2 - 4x + 4#?

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Answer 1 · 2015-05-14T15:30:14

The answer is : #y = x^(2) - 4x + 4 = (x-2)^(2)#. Your function has one zero, with #x = 2#.

The usual form of a quadratic function is : #y = ax^(2) + bx + c#

Your function is #y = 1x^(2) - 4x + 4#.

Therefore #a = 1#, #b = -4# and #c = 4#

The quadratic formula gives you the values of #x#, with which your #y = 0#.

#x = (-b +- sqrtDelta)/(2a)#, where #Delta=b^(2) -4ac#.

Since we have a square root, #Delta >=0#. If not, the function don't have any zeros.

Let's calculate the zeros of your function :

#Delta =(-4)^(2)-4*1*4=16-16=0#

#x_1 = (4 +sqrt0)/2 = x_2 = (4-sqrt0)/2=#2

Therefore, #y = x^(2) - 4x + 4 = (x-2)^(2)#. Your function has one zero, with #x = 2#.