Sure, let's find the oblique asymptote for the function [tex]\( f(x) = \frac{9x^2 + 36x + 41}{3x + 5} \)[/tex].
To find the oblique asymptote, we need to perform polynomial long division of the numerator [tex]\( 9x^2 + 36x + 41 \)[/tex] by the denominator [tex]\( 3x + 5 \)[/tex].
### Step-by-Step Solution:
1. Setup polynomial long division:
Divide [tex]\( 9x^2 + 36x + 41 \)[/tex] by [tex]\( 3x + 5 \)[/tex].
2. Determine the first term of the quotient:
The leading term in the numerator is [tex]\( 9x^2 \)[/tex] and in the denominator is [tex]\( 3x \)[/tex].
[tex]\[
\frac{9x^2}{3x} = 3x
\][/tex]
So, the first term in the quotient is [tex]\( 3x \)[/tex].
3. Multiply and subtract:
- Multiply [tex]\( 3x \)[/tex] by [tex]\( 3x + 5 \)[/tex]:
[tex]\[
3x \cdot (3x + 5) = 9x^2 + 15x
\][/tex]
- Subtract [tex]\( 9x^2 + 15x \)[/tex] from [tex]\( 9x^2 + 36x + 41 \)[/tex]:
[tex]\[
(9x^2 + 36x + 41) - (9x^2 + 15x) = 21x + 41
\][/tex]
4. Determine the next term of the quotient:
The leading term in the remaining expression [tex]\( 21x + 41 \)[/tex] is [tex]\( 21x \)[/tex]. Divide [tex]\( 21x \)[/tex] by [tex]\( 3x \)[/tex]:
[tex]\[
\frac{21x}{3x} = 7
\][/tex]
So, the next term in the quotient is [tex]\( 7 \)[/tex].
5. Multiply and subtract:
- Multiply [tex]\( 7 \)[/tex] by [tex]\( 3x + 5 \)[/tex]:
[tex]\[
7 \cdot (3x + 5) = 21x + 35
\][/tex]
- Subtract [tex]\( 21x + 35 \)[/tex] from [tex]\( 21x + 41 \)[/tex]:
[tex]\[
(21x + 41) - (21x + 35) = 6
\][/tex]
6. Conclusion:
The quotient of the division is [tex]\( 3x + 7 \)[/tex] and the remainder is [tex]\( 6 \)[/tex]. The quotient [tex]\( 3x + 7 \)[/tex] represents the oblique asymptote of the function.
Therefore, the oblique asymptote of the function [tex]\( f(x) = \frac{9x^2 + 36x + 41}{3x + 5} \)[/tex] is [tex]\( y = 3x + 7 \)[/tex].
The value of [tex]\( k \)[/tex] is [tex]\( 7 \)[/tex].
