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joja [24]
3 years ago
14

What is the solution to 11=c/-5 8?

Mathematics
1 answer:
ss7ja [257]3 years ago
7 0
Is it c/-58, or is it c/-5 blank 8?

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Find the horizontal and vertical asymptotes of​ f(x). ​f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the
miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* <em>Lets the meaning of vertical and horizontal asymptotes</em>

- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes

  of the denominator of a rational function

- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function

 approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

 denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

 denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

 then there is a horizontal asymptote at y = leading coefficient of the

 numerator ÷ leading coefficient of the denominator

* <em>Lets solve the problem</em>

∵ f(x)=\frac{6x}{x+2}

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ <em>The horizontal asymptote can be described by the line y = 6</em>

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ <em>The vertical asymptote can be described by the line x = -2</em>

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