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3 years ago

If the graph f(x)= 9x^2+37x+41/3x+5 has an oblique asymptote at y=3x+k what is the value of k

Mathematics

2 answers:

ira [324]3 years ago

6 0

Answer:

$k=\frac{22}{3}$

Step-by-step explanation:

The oblique asymptote of

$f(x)=\frac{9x^2+37x+41}{3x+5}$ ,

We perform the long division as shown in the attachment.

The quotient is;

$3x+\frac{22}{3}$

Comparing to 3x+k

Hence the value of k is $\frac{22}{3}$

Archy [21]3 years ago

3 0

Answer:

$\frac{22}{3}$

Step-by-step explanation:

To find out oblique asymptote we divide the polynomials using long division

To find quotient divide the first term. then multiply the answer with 3x+5 and write it down. Subtract it from the top. Repeat the process till we get remainder.

$3x+\frac{22}{3}$

------------------------------

$3x+5$ $9x^2+37x+41$

$9x^2+15x$

-------------------------------------(Subtract)

$22x+41$

$22x+\frac{110}{3}$

------------------------------------(subtract)

$\frac{13}{3}$

Quotient is $3x+\frac{22}{3}$ that is our oblique asympotote

the value of k is 22/3

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