Question

step 1
for $g(y)=\frac{y - 4}{y^{2}-2y + 8}$, we have
$g(y)=\frac{-y^{2}+8y}{(y^{2}-2y + 8)^{2}}$ $-\frac{(y - 8)y}{(y^{2}-2y + 8)^{2}}$
step 2
critical numbers occur where $g(y)$ equals 0 or is undefined. $g(y)$ is undefined where the quadratic $y^{2}-2y + 8$ 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.)
$y=$

Explanation:

Step1: Recall the condition for a rational - function to be undefined

A rational function $\frac{f(y)}{h(y)}$ is undefined when $h(y)=0$. Here, $g^{\prime}(y)=\frac{-(y - 8)y}{(y^{2}-2y + 8)^{2}}$, and we need to find when $y^{2}-2y + 8=0$.

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation $ay^{2}+by + c = 0$ is $y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For the equation $y^{2}-2y + 8=0$, where $a = 1$, $b=-2$, and $c = 8$, we calculate the discriminant $\Delta=b^{2}-4ac=(-2)^{2}-4\times1\times8=4 - 32=-28$. Since $\Delta<0$, the quadratic equation $y^{2}-2y + 8 = 0$ has no real - valued solutions.

Answer:

DNE