Question

step 1
for ( g(y)=\frac{y - 2}{y^{2}-2y + 4} ), we have
( g(y)=-\frac{y(y - 4)}{(y^{2}2y + 4)^{2}}\times-\frac{(y - 4)y}{(y^{2}-2y + 4)^{2}} ).
step 2
critical numbers occur where ( g(y) ) equals 0 or is undefined. ( g(y) ) is undefined where the quadratic ( y^{2}-2y + 4 ) in the denominator is 0. so, ( g(y) ) is undefined for the following values. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)
( y=)
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Explanation:

Step1: Analyze the quadratic equation

For the quadratic equation \(y^{2}-2y + 4=0\), use the quadratic formula \(y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 1\), \(b=-2\), and \(c = 4\).

Step2: Calculate the discriminant

The discriminant \(\Delta=b^{2}-4ac=(-2)^{2}-4\times1\times4=4 - 16=-12\).

Step3: Determine the nature of the roots

Since \(\Delta=-12<0\), the quadratic equation \(y^{2}-2y + 4 = 0\) has no real solutions.

Answer:

DNE