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Answer:
(x-8)/(4x-4) x ≠ 1 or 8
Step-by-step explanation:
Common factors can be cancelled, but cannot be ignored when it comes to determining where the function is undefined.
[tex]\dfrac{x^2-16x+64}{4x^2-36x+32}=\dfrac{(x-8)^2}{4(x-8)(x-1)}=\boxed{\dfrac{x-8}{4x-4}}[/tex]
The function is undefined for x=1 and x=8, the values of x that make the original denominator zero.
