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Thepotemich [5.8K]
3 years ago
12

The following sequence has first five terms as: -20, -17, -14, -11, -8, .......... Find an expression for the nth term of the se

quence
Mathematics
1 answer:
Maslowich3 years ago
4 0

Answer:

a(n) = -20 + 3(n - 1)

Step-by-step explanation:

a(n) = -20 + 3(n - 1)

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133÷7=19

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Show all worked identify the asymptotes and state the end behavior of the function F(x)=6x over x-36
astraxan [27]

Solution

Asymptote:

Vertical Asymptote

- The vertical asymptotes of a rational function are determined by the denominator expression.

- The expression given is:

f(x)=\frac{6x}{x-36}

- The denominator of (x- 36) determines the asymptote line.

- The vertical asymptote defines where the rational function isundefined. Iin order for a rational function to be undefined, its denominator must be zero.

- Thus, we can say:

\begin{gathered} x-36=0 \\ Add\text{ 36 to both sides} \\  \\ \therefore x=36 \end{gathered}

- Thus, the vertical asymptote is

x=36

Horizontal Asymptote:

- The horizontal asymptote exists in two cases:

1. When the highest degree of the numerator is less han the degree of the demnominator. In this case, the horizontal asymptote is y = 0

2. When the highest degee sof the numerator and tdenominator are the same. In this case, the horizontal asymptote is

\begin{gathered} y=\frac{N}{D} \\ where, \\ N=\text{ Coefficient of the highest degree of the numerator} \\ D=\text{ Coefficient of the highest degree of the denominator} \end{gathered}

- For our question, we can see that the highest degrees of the numerator and denominator are the same. Thus, we have the Horizontal Asymptote to be:

y=\frac{6}{1}=6

End behavior:

- The end behavior is examining the y-values of the function as x tendsto negative and positive infinity.

- Thus, we have that:

\begin{gathered} f(x)=\frac{6x}{x-36} \\  \\ \text{ Divide top and bottom by }x \\ f(x)=\frac{6x}{x-36}\times\frac{x}{x} \\  \\ f(x)=\frac{\frac{6x}{x}}{\frac{x-36}{x}}=\frac{6}{1-\frac{36}{x}} \\  \\ As\text{ }x\to-\infty \\ f(-\infty)=\frac{6}{1-\frac{36}{-\infty}}=\frac{6}{1+\frac{36}{\infty}}=\frac{6}{1+0}=6 \\  \\ \text{ Thus, we can say: }x\to-\infty,f(x)\to6 \\  \\ Also, \\ As\text{  }x\to\infty \\ f(\infty)=\frac{6}{1-\frac{36}{\infty}}=\frac{6}{1-0}=6 \\  \\ \text{ Thus, we can also say: }x\to\infty,f(x)\to6 \end{gathered}

Final Answers

Asymptotes:

\begin{gathered} \text{ Vertical:} \\ x=36 \\  \\ \text{ Horizontal:} \\ y=6 \end{gathered}

End behavior:

\begin{gathered} As\text{  }x\to-\infty,f(x)\to6 \\  \\ As\text{  }x\to\infty,f(x)\to6 \end{gathered}

7 0
1 year ago
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