What are the vertical asymptote(s) and hole(s) for the graph of y =
X-1/- x² + 6x + 8
asymptote: x = 1 and holes: x = -4,-2
asymptotes: x= -4,-2 and no holes
asymptotes: x = -4,-2 and hole: x = 1
asymptote: x = 1 and no holes

asymptotes: x= -4,-2 and no holes

Step-by-step explanation:

First we need to factor the denominator of the function

y = (x-1) / (x2 + 6x + 8)

x2 + 6x + 8 = 0

delta = b2 - 4ac = 36 - 32 = 4

x1 = (-6 + 2) / 2 = -2

x2 = (-6 - 2) / 2 = -4

So we have that x2 + 6x + 8 = (x+4)(x+2)

So our function is:

y =  (x-1) / (x+4)(x+2)

As there is no common expressions in the numerator and denominator, there are no holes.

The asymptotes are when the denominator is zero, so:

(x+4)(x+2) = 0

x = -4 or x = -2

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.