Question

Help me pleaseeeeeee

Answer 1

1. Answer: C

Explanation:

`f(x) = (x-3)/(x^(2)-7x+12)`

`f(x) = (x-3)/((x-3)(x-4))`

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

Explanation:

`(11x^(2)-11)/(x)` ÷
`(11(x^(2)+11))/(32x^(2)-11x)`

=
`(11(x + 1)(x - 1))/(x)` ÷
`(11(x^(2)+11))/(x(32x-11))`

=
`(11(x + 1)(x - 1))/(x)` X
`(x(32x-11))/(11(x^(2)+11))`

=
`((x + 1)(x - 1)(32x - 11))/(x^(2)+11)`

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3. Answer: C

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 =
`(4)/(x)`

transformations is: y = y =
`(4)/(x+6) + 2`