Question

if the graph of $f(x) = \frac{9x^2 + 35x + 41}{3x + 5}$ has an oblique asymptote at $y = 3x + k$, what is the value of $k$?

Explanation:

Step1: Perform Polynomial Long Division

Divide \(9x^2 + 35x + 41\) by \(3x + 5\).
First term: \(\frac{9x^2}{3x}=3x\). Multiply \(3x\) by \(3x + 5\): \(9x^2 + 15x\).
Subtract from dividend: \((9x^2 + 35x + 41)-(9x^2 + 15x)=20x + 41\).

Step2: Divide Remaining Term

Next term: \(\frac{20x}{3x}=\frac{20}{3}\)? Wait, no—wait, correct step: Wait, actually, when dividing \(20x + 41\) by \(3x + 5\), the next term is \(\frac{20x}{3x}=\frac{20}{3}\)? No, wait, no—wait, let's do it properly. Wait, actually, the oblique asymptote comes from the quotient when we divide the numerator by the denominator (since degree of numerator is one more than denominator). So let's do the division correctly.

Wait, \(3x + 5\) times \(3x\) is \(9x^2 + 15x\). Subtract from \(9x^2 + 35x + 41\): \(20x + 41\). Now, divide \(20x + 41\) by \(3x + 5\). The next term: \(\frac{20x}{3x}=\frac{20}{3}\), but wait, no—wait, actually, the oblique asymptote is the linear part of the quotient. Wait, no, let's use polynomial long division properly.

Wait, \(3x + 5\) into \(9x^2 + 35x + 41\):

- \(3x\) times \(3x\) is \(9x^2\), \(3x\) times \(5\) is \(15x\). So subtract \(9x^2 + 15x\) from \(9x^2 + 35x + 41\), we get \(20x + 41\).

Now, divide \(20x + 41\) by \(3x + 5\): the coefficient for \(x\) is \(\frac{20}{3}\), but wait, no—wait, actually, the oblique asymptote is the quotient when we divide, ignoring the remainder (as \(x\to\pm\infty\), the remainder term \(\frac{\text{remainder}}{3x + 5}\) approaches 0). Wait, no, let's do the division again. Wait, maybe I made a mistake. Let's use the method of expressing \(f(x)\) as \(3x + k + \frac{r(x)}{3x + 5}\), where \(r(x)\) is the remainder (degree less than 1, so constant).

Let \(f(x)=\frac{9x^2 + 35x + 41}{3x + 5}=3x + k + \frac{c}{3x + 5}\). Multiply both sides by \(3x + 5\):

\(9x^2 + 35x + 41=(3x + k)(3x + 5)+c\)

Expand right side: \(9x^2 + 15x + 3kx + 5k + c=9x^2 + (15 + 3k)x + (5k + c)\)

Equate coefficients:

- For \(x\): \(35 = 15 + 3k\)
- For constant term: \(41 = 5k + c\) (but we don't need \(c\) for the asymptote)

Solve for \(k\) from the \(x\) coefficient:

\(35 - 15 = 3k\)
\(20 = 3k\)? Wait, no—wait, that's not right. Wait, wait, no—wait, in the expansion, \((3x + k)(3x + 5)=9x^2 + (15 + 3k)x + 5k\). So the \(x\) term is \(15 + 3k\), which should equal 35 (from the numerator's \(x\) term). So:

\(15 + 3k = 35\)
\(3k = 35 - 15 = 20\)
\(k=\frac{20}{3}\)? Wait, no, that can't be. Wait, no, wait, I messed up the division. Wait, let's do the long division again.

Divide \(9x^2 + 35x + 41\) by \(3x + 5\):

1. How many times does \(3x\) go into \(9x^2\)? \(3x\) times. Multiply \(3x\) by \(3x + 5\): \(9x^2 + 15x\).
2. Subtract from \(9x^2 + 35x + 41\): \( (9x^2 + 35x + 41) - (9x^2 + 15x) = 20x + 41 \).
3. Now, divide \(20x + 41\) by \(3x + 5\). How many times does \(3x\) go into \(20x\)? \(\frac{20}{3}\) times, but that's a fraction. Wait, no—wait, the oblique asymptote is the linear part of the quotient, so we can also use the method of expressing \(f(x)\) as \(3x + k + \frac{\text{remainder}}{3x + 5}\), where \(k\) is the constant term of the quotient. Wait, no, actually, when we do the division, the quotient is \(3x + k\), and the remainder is a constant (since degree of numerator is 2, denominator is 1, so remainder is degree 0). So let's set \(f(x)=3x + k + \frac{r}{3x + 5}\), where \(r\) is a constant. Then:

\(9x^2 + 35x + 41=(3x + k)(3x + 5)+r\)
Expand right side: \(9x^2 + 15x + 3kx + 5k + r = 9x^2 + (15 + 3k)x + (5k + r)\)

Now, equate coefficients:

- \(x^2\) term: \(9 = 9\) (okay)
- \(x\) term: \(35 = 15 + 3k\)
- Constant term: \(41 = 5k + r\)

From the \(x\) term: \(35 - 15 = 3k \implies 20 = 3k \implies k = \frac{20}{3}\)? Wait, that seems odd. Wait, no, wait, I think I made a mistake in the division. Wait, let's check with another method: synthetic division for linear divisor \(3x + 5\), so root at \(x = -\frac{5}{3}\).

Using synthetic division:

Coefficients of numerator: 9, 35, 41
Root: \(x = -\frac{5}{3}\)

Bring down 9.
Multiply 9 by \(-\frac{5}{3}\): \(-15\). Add to 35: 20.
Multiply 20 by \(-\frac{5}{3}\): \(-\frac{100}{3}\). Add to 41: \(41 - \frac{100}{3}=\frac{123 - 100}{3}=\frac{23}{3}\). Wait, that's the remainder. But the quotient is \(9x + 20\) divided by 3? Wait, no—synthetic division for \(ax + b\) requires dividing by \(a\) at the end. Wait, the divisor is \(3x + 5\), so we can write it as \(3(x + \frac{5}{3})\), so we perform synthetic division for \(x + \frac{5}{3}\), then divide the quotient by 3.

So synthetic division for \(x = -\frac{5}{3}\):

Coefficients: 9, 35, 41
Bring down 9.
9 * (-5/3) = -15. 35 + (-15) = 20.
20 * (-5/3) = -100/3. 41 + (-100/3) = 23/3 (remainder).

Now, the quotient from synthetic division is \(9x + 20\), but since we divided by \(3(x + 5/3)\), we need to divide the quotient by 3: \(\frac{9x + 20}{3}=3x + \frac{20}{3}\). So the quotient is \(3x + \frac{20}{3}\), and the remainder is \(\frac{23}{3}\). Therefore, \(f(x)=3x + \frac{20}{3}+\frac{\frac{23}{3}}{3x + 5}\). As \(x\to\pm\infty\), the remainder term approaches 0, so the oblique asymptote is \(y = 3x + \frac{20}{3}\)? Wait, but that seems like a fraction. Wait, no, wait, let's check the arithmetic again. Wait, 35 - 15 is 20, so 3k = 20, so k = 20/3 ≈ 6.666... But maybe I made a mistake. Wait, let's re-express the function:

\(f(x)=\frac{9x^2 + 35x + 41}{3x + 5}\). Let's write numerator as \(9x^2 + 15x + 20x + 50 - 9\). Wait, 9x² +15x = 3x(3x +5), and 20x +50 = 20(x + 2.5) = 20(x + 5/3 + 5/3 - 5/3)? No, better: 20x + 41 = 20x + 100/3 - 100/3 + 41 = 20(x + 5/3) + (41 - 100/3) = 20(x + 5/3) + 23/3. Wait, no, 3x +5 = 3(x + 5/3), so:

\(f(x)=\frac{9x^2 + 35x + 41}{3(x + 5/3)}=\frac{9x^2 + 15x + 20x + 41}{3(x + 5/3)}=\frac{3x(3x +5) + 20x + 41}{3(x + 5/3)}\). Then, 20x + 41 = 20(x + 5/3) + 41 - 100/3 = 20(x + 5/3) + 23/3. So:

\(f(x)=\frac{3x(3x +5) + 20(x + 5/3) + 23/3}{3(x + 5/3)}=\frac{3x(3x +5)}{3(x + 5/3)}+\frac{20(x + 5/3)}{3(x + 5/3)}+\frac{23/3}{3(x + 5/3)}\)

Simplify:

\(f(x)=3x + \frac{20}{3}+\frac{23}{9(x + 5/3)}\)

Ah, so the oblique asymptote is \(y = 3x + \frac{20}{3}\), so \(k = \frac{20}{3}\)? Wait, but that seems correct. Wait, let's check with actual division:

Divide 9x² +35x +41 by 3x +5:

3x +5 ) 9x² +35x +41 ( 3x + 20/3
9x² +15x
---------
20x +41
20x +100/3
---------
41 - 100/3 = 23/3

Yes, so the quotient is 3x + 20/3, remainder 23/3. Therefore, the oblique asymptote is y = 3x + 20/3, so k = 20/3. Wait, but maybe I made a mistake in the problem statement? Wait, the problem says "oblique asymptote at y = 3x + k". So k is 20/3? Wait, but 20/3 is approximately 6.666... Let me check the arithmetic again. 3x times 3x +5 is 9x² +15x. Subtract from 9x² +35x +41: 20x +41. Then, 3x +5 times 20/3 is 20x + 100/3. Subtract from 20x +41: 41 - 100/3 = (123 - 100)/3 = 23/3. So yes, the quotient is 3x + 20/3, so k = 20/3.

Wait, but maybe the problem has a typo, or I made a mistake. Wait, let's check with another approach: the oblique asymptote of a rational function \(f(x)=\frac{N(x)}{D(x)}\) where deg(N)=deg(D)+1 is found by dividing N(x) by D(x) and taking the quotient (ignoring the remainder). So dividing 9x² +35x +41 by 3x +5, the quotient is 3x + k, so we can write 9x² +35x +41 = (3x +5)(3x + k) + r, where r is the remainder (a constant). Expanding the right side: 9x² + (3k +15)x +5k + r. Equate coefficients:

- x term: 3k +15 = 35 ⇒ 3k = 20 ⇒ k = 20/3.
- Constant term: 5k + r = 41 ⇒ 5*(20/3) + r = 41 ⇒ 100/3 + r = 123/3 ⇒ r = 23/3.

Yes, that's correct. So k = 20/3.

Answer:

\(\frac{20}{3}\)