Question

Identify the vertical asymptote and the hole on the graph of the function f(x) = x2+x-6/x2-6x+8.  ANSWER:

Answer 1

Answer: There is an asymptote at x = 4 and 2

and hole at x=2

Explanation:

To find  vertical asymptote  simply equate denominator to zero and find the value of x which is the vertical asymptote.

and hole is the point where numerator and denominator gives equal value

So, to find hole find fractions of numerator and denominator separately the common factor is the hole.

Here after factorisation the given function

$\frac{x^{2}+x-6} {x^{2}-6x+8} =\frac{(x-2)(x+3)}{(x-2)(x-4)}$

Here, (x-2) is the common factor in numerator and denominator is the hole and value of x at denominator after equating to zero is x=2 ,4

Answer 2

Answer:

The vertical asymptote is x=4. The function has whole at x=2.

Step-by-step explanation:

The given function is

$f(x)=\frac{x^2+x-6}{x^2-6x+8}$

Find factor form of numerator and denominator.

$f(x)=\frac{x^2+3x-2x-6}{x^2-4x-2x+8}$

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

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

Cancel out the common factors.

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

Now equate the denominator equal to 0, to find the vertical asymptote.

$x-4=0$

$x=4$

The vertical asymptote is x=4.

Equate the cancel factor equal to zero to find the whole.

$x-2=0$

$x=2$

The function has whole at x=2.

At whole, the value of function is undefined but limit of the function exist.

$f(2)=\frac{2+3}{2-4}=-2.5$

Therefore the limit of the function at x=2 is -2.5.