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 $x = 2$ .

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