1. Answer: C
Step-by-step explanation:
[tex]f(x) = \frac{x-3}{x^{2}-7x+12}[/tex]
[tex]f(x) = \frac{x-3}{(x-3)(x-4)}[/tex]
denominator cannot equal zero so x ≠ 3 and x ≠ 4
Since x - 3 cancels out, then x = 3 is a HOLE
Since x - 4 does not cancel out, then x = 4 is a Vertical Asymptote
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2. Answer: B
Step-by-step explanation:
[tex]\frac{11x^{2}-11}{x}[/tex] ÷ [tex]\frac{11(x^{2}+11)}{32x^{2}-11x}[/tex]
= [tex]\frac{11(x + 1)(x - 1)}{x}[/tex] ÷ [tex]\frac{11(x^{2}+11)}{x(32x-11)}[/tex]
= [tex]\frac{11(x + 1)(x - 1)}{x}[/tex] X [tex]\frac{x(32x-11)}{11(x^{2}+11)}[/tex]
= [tex]\frac{(x + 1)(x - 1)(32x - 11)}{x^{2}+11}[/tex]
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3. Answer: C
Step-by-step explanation:
The parent graph has asymptotes of x = 0, y = 0
The vertical asymptote is shifted to the left 6 units ⇒ x - (-6) ⇒ x + 6
The horizontal asymptote is shifted up 2 units ⇒ +2
So, y = [tex]\frac{4}{x}[/tex]
transformations is: y = y = [tex]\frac{4}{x+6} + 2[/tex]
