Question

The graph of f(x)=9(x-6)/x^2-7x+6 has a vertical asymptote at x=??​

Answer 1

`x=1`

Explanation:

For the function
`f(x)=(9(x-6))/(x^2-7x+6)`, you need to factor the denominator.

Find two number that when you add them you get -7 and when you multiply them you get 6. These numbers are -6 and -1.

Then:

`f(x)=(9(x-6))/((x-6)(x-1))`

Simplify:

`f(x)=(9)/((x-1))`

Observe that the denominator is equal to zero in
`x=1`.

Therefore, when x approaches 1, f(x) tends to infinite.

Then the vertical asymptote s:

`x=1`

Answer 2

x=1

Explanation:

First we need to simplify and factorize the denominator as shown below:

`f(x)=(9(x-6))/(x^(2) -7x+6) \\\\ f(x)=(9(x-6))/(x^(2)-6x-x+6 )\\\\ f(x)=(9(x-6))/(x(x-6)-1(x-6))\\\\ f(x)=(9(x-6))/((x-1)(x-6))\\`

Following points must be kept in mind:

• A hole exists at the value at which both numerator and denominator of the rational function becomes zero
• Vertical asymptote exist where only the denominator is zero.

For x=6, both numerator and denominator are zero so we have a hole here. At x=1, only the denominator is zero, so we have a vertical asymptote here.

Thus the vertical asymptote of f(x) is at x=1