step 1\nfor ( g(y)=\frac{y - 2}{y^{2}-2y + 4} ), we have\n( g(y)=-\frac{y(y - 4)}{(y^{2}2y +…

step 1\nfor ( g(y)=\frac{y - 2}{y^{2}-2y + 4} ), we have\n( g(y)=-\frac{y(y - 4)}{(y^{2}2y + 4)^{2}}\times-\frac{(y - 4)y}{(y^{2}-2y + 4)^{2}} ).\nstep 2\ncritical 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.)\n( y=)\nsubmit skip (you cannot come back)\nresources\nread it

step 1\nfor ( g(y)=\frac{y - 2}{y^{2}-2y + 4} ), we have\n( g(y)=-\frac{y(y - 4)}{(y^{2}2y + 4)^{2}}\times-\frac{(y - 4)y}{(y^{2}-2y + 4)^{2}} ).\nstep 2\ncritical 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.)\n( y=)\nsubmit skip (you cannot come back)\nresources\nread it

Answer

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