Question

find the horizontal or oblique asymptote of F(x)= -2x^2+3x+6/x+2

Answer 1

f(x)=-2(x^2)+3x+(6/x)+2
fx=2x+3x+(2*3/x)+2
fx={-x*2(x^2)+x(3x)+(2+3)+x*2)/x
result:
fx={-2(x^3)+3(x^2)=2x+6/(x^2)

Answer 2

Answer:

There is no horizontal asymptote for this function but there is an oblique asymptote which is -2x + 7

Step-by-step explanation:

When given a function, it is very easy to check its horizontal asymptote,

When dealing with a function say;  

 $f(x)=\frac{ax^{m} -------}{ax^{n} -------}$

where a and b represents the leading coefficient, m and n represents the degrees, then to find the horizontal asymptote all we need to do is to compare the degree of the polynomials

If m < n then y= 0 That is, if the top degree is less than the bottom degree then the horizontal asymptote is zero(0)

If m = n then y =  

That is, if the top degree is equal to the bottom degree, then the horizontal asymptote  is equal to    where  a and b are coefficients of  

and

respectively.

If m > n then there is no horizontal asympto

te. That is, if the top degree is greater than the bottom degree, then there is no horizontal asymptote.

Which means there is only a slant asymptote or oblique asymptote

To find the slant asymptote, all we need to do is to use long division to divide our polynomials;

 

           -2x + 7______

x + 2    √-2x² + 3x + 6

    -___(-2x - 4x)_____

                   7x  + 6

                  - (7x +14)

___________________

                            -8

Therefore the oblique asymptote is -2x + 7