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Mariulka [41]
3 years ago
6

What are the holes, VA, domain, HA, and Range of n+8/6n? Please help!!

Mathematics
1 answer:
-BARSIC- [3]3 years ago
7 0

Answer:

We show that f(x) n+8/6n = 6 x n = 0

which flips the n+8/1 = 0+8/0-6= x = 3  this is the range.

For the HA we would work left to right.

x goes to positive or negative infinity and is determined by the highest degree terms of the polynomials in the numerator and the denominator. This particular function has polynomials of degree 0 in both the numerator and the denominator

If say n+8 was n+2 then we would use the 2/-2+3 and get 1 and show the hole as the source;

hole : -2+1 as non equal sign. but not in the case of n+8/6n

-2+1 represents 1/3 symmetry.

We see for n+8/6n  with interpreted back into the zero format minus

-0+8/-0-6 we see there is symmetry and can work on the left side of graph and flip over. Where 0 = n+8 and 1=nx6

Step-by-step explanation:

There would be no way of doing the others unless the exponents had been squared ^2

If they were squared then the domain will be (-infinity -3) parenthesis

union of( -3 -2) union of +2 to negative infinity.

There is not a vertical asymptote as the numerator divides into dominator at point 8 as a decimal.

The holes are then closed.

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Answer:

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Step-by-step explanation:

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Once, $\left( \frac{2}{3} \cdot \frac{1}{2} \right)^2=\frac{1}{9} $

$x^2 +\frac{2}{3} x+\frac{1}{9}  = \frac{5}{6}+\frac{1}{9}  $

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$x+\frac{1}{3}  =\pm\sqrt{\frac{17}{18}} }   $

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$\sqrt{\frac{17(18)}{18(18)}}= \sqrt{\frac{306}{324}}=\frac{3\sqrt{34} }{18} =\frac{\sqrt{34} }{6} $

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$x+\frac{1}{3}  =\pm\frac{\sqrt{34} }{6}  $

$x =-\frac{1}{3} \pm\frac{\sqrt{34} }{6}  $

$x_{1}=-\frac{2+\sqrt{34}}{6}$

$x_{2}=-\frac{2-\sqrt{34}}{6}$

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