Question

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

Answer 1

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 $\frac{22}{3}$

Answer 2

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