4. `y=(x^3-8)/(x^2+9)`, find Asymptotes of a functionSolution:Asymptotes : Now, find domain of `y=(x^3-8)/(x^2+9)`
`y=(x^3-8)/(x^2+9)`
`x^2+9=0`
`=>x^2+9=0`
`=>x^2=0-9`
`=>x^2=-9`
Domain : all x
Vertical asymptote : none
The highest power in the numerator is 3
The highest power in the denominator is 2
Slant(Oblique) asymptote :
Final Solution
| | `` | `x` | `+` | `0` | | | | | | |
| `color{blue}{x^2+9}` | `` | `x^3` | `+` | `0x^2` | `+` | `0x` | `-` | `8` | | |
| | `` | −`x^3` | | | `+` | −`9x` | | | | `x xx (color{blue}{x^2+9})` |
| | | | `` | `0x^2` | `-` | `9x` | `-` | `8` | | |
| | | | `` | −`0x^2` | | | `+` | −`0` | | `color{green}{0} xx (color{blue}{x^2+9})` |
| | | | | | `-` | `9x` | `-` | `8` | | |
Final answer `= "Quotient" + (color{Magenta}{"Remainder"})/(color{blue}{"Divisor"})`.
`:.` Final answer = `x+0 + (color{Magenta}{-9x-8})/(color{blue}{x^2+9})`
Here, Divisor = `x^2+9`
Dividend = `x^3-8`
Quotient = `x+0`
Remainder = `-9x-8`
Step by step division solution
Step - 1 : 1. Divide the first term of the dividend by the first term of the divisor : `(x^3)/(x^2)=color{green}{x}`
2. Write down the calculated result `color{green}{x}` in the upper part of the table.
3. Multiply it by the divisor `color{green}{x} xx (color{blue}{x^2+9})=color{red}{x^3+9x}`
4. Subtract this result from the dividend
`(x^3+0x^2+0x-8)-(color{red}{x^3+9x})=color{Magenta}{0x^2-9x-8}`
| | `` | `x` | | | | | | | | |
| `color{blue}{x^2+9}` | `` | `x^3` | `+` | `0x^2` | `+` | `0x` | `-` | `8` | | |
| | `` | −`x^3` | | | `+` | −`9x` | | | | `color{green}{x} xx (color{blue}{x^2+9})` |
| | | | `` | `0x^2` | `-` | `9x` | `-` | `8` | | |
Step - 2 : 1. Divide the first term of the dividend by the first term of the divisor : `(0x^2)/(x^2)=color{green}{0}`
2. Write down the calculated result `color{green}{0}` in the upper part of the table.
3. Multiply it by the divisor `color{green}{0} xx (color{blue}{x^2+9})=color{red}{0x^2+0}`
4. Subtract this result from the remainder
`(0x^2-9x-8)-(color{red}{0x^2+0})=color{Magenta}{-9x-8}`
| | `` | `x` | `+` | `0` | | | | | | |
| `color{blue}{x^2+9}` | `` | `x^3` | `+` | `0x^2` | `+` | `0x` | `-` | `8` | | |
| | `` | −`x^3` | | | `+` | −`9x` | | | | `x xx (color{blue}{x^2+9})` |
| | | | `` | `0x^2` | `-` | `9x` | `-` | `8` | | |
| | | | `` | −`0x^2` | | | `+` | −`0` | | `color{green}{0} xx (color{blue}{x^2+9})` |
| | | | | | `-` | `9x` | `-` | `8` | | |
Slant(Oblique) asymptote : `y=x`
Horizontal asymptote : none
This material is intended as a summary. Use your textbook for detail explanation.
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