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zaharov [31]
3 years ago
11

Sketch the asymptotes and graph the function y=6/(x-2)+4

Mathematics
2 answers:
lana [24]3 years ago
6 0
There will be a vertical asymptote when the denominator approaches zero. So the vertical asymptote is about the line x=2.

There will be a horizontal asymptote as x approaches ±oo where the y value will approach 4.  So the horizontal asymptote is about the line y=4
lorasvet [3.4K]3 years ago
6 0

Answer:

Step-by-step explanation:

We have to sketch the asymptotes of the given function y = 6/(x -2) + 4

1) For horizontal asymptotes

In the given function y = 6/(x -2) is in the form of y = \frac{6x^{0} }{(x^{1}-2)}

Degree of x in numerator is lower than denominator therefore horizontal asymptote is y = 0

Since y = 6/(x -2) is shifted 4 upwards so horizontal asymptote for the shifted function will be y = 0+4 = 4

2) For vertical asymptotes

We will put the denominator of the function equal to the zero.

(x - 2) = 0

x = 2 is the vertical asymptote.

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Answer:

<em>Bethany did not do better than Richie because her average score was less than his.</em>

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1800|2\\.\ 900|2\\.\ 450|2\\.\ 225|5\\.\ \ 45|5\\.\ \ \ 9|3\\.\ \ \ 3|3\\.\ \ \ 1|\\\\1880=2\cdot2\cdot2\cdot5\cdot5\cdot3\cdot3=2^3\cdot5^2\cdot3^2\\\\Answer:\ \boxed{2^3\cdot3^2\cdot5^2}

\sqrt[3]{320}\\\\320|2\\160|2\\.\ 80|2\\.\ 40|2\\.\ 20|2\\.\ 10|2\\.\ \ 5|5\\.\ \ 1|\\\\320=2\cdot2\cdot2\cdot2\cdot2\cdo2\cdot5=2^3\cdot2^3\cdot5=2^3\cdot2^3\cdot\\\\\sqrt[3]{320}=\sqrt[3]{2^3\cdot2^3\cdot5}=\sqrt[3]{2^3}\cdot\sqrt[3]{2^3}\cdot\sqrt[3]{5}=2\cdot2\sqrt[3]{5}=4\sqrt[3]5\\\\Answer:\ \boxed{\sqrt[3]{320}=4\sqrt[3]{5}}
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3 years ago
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