Question

For the given function, find the vertical and horizontal asymptote(s) (if there are any). f(x) = the quantity x squared plus four x minus three divided by the quantity x minus six

Answer 1

Answer:

Vertical asymptote at x= 6

Oblique asymptote at y=x+10

Step-by-step explanation:

$f(x)= \frac{x^2+4x-3}{x-6}$

To find out vertical asymptote , we take the denominator and set it =0

x-6=0 , so x=6

Vertical asymptote at x= 6

The degree of numerator is 2  and the degree of denominator is 1.

Here,  the degree of numerator is higher than the degree of denominator. So there will be a slant  or oblique asymptote. we can find it by long division.

                                  x + 10      

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

      x-6              x^2  +  4x  -3

                         x^2  - 6x

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

                                  10x    - 3

                                  10x   - 60

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

                                           57

Oblique asymptote at y=x+10

Answer 2

Answer:

vertical asymptote: x=6

Horizontal asymptote: doesn't exist.

Step-by-step explanation:

Given function is $f\left(x\right)=\frac{x^2+4x-3}{x-6}$

To find vertical asymptote, we just set denominator =0 and solve that for x

Given denominator is x-6

then x-6=0

or x=6

Hence vertical asymptote: x=6

-----

We see that degree of numerator = 2

degree of denominator = 1

When degree of numerator is not equal to degree of denominator then horizontal asymptote doesn't exist.

Horizontal asymptote: doesn't exist.