Question

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

Answer 1

Answer:

x=1

Step-by-step explanation:

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

$f(x)=\frac{9(x-6)}{x^{2} -7x+6} \\\\ f(x)=\frac{9(x-6)}{x^{2}-6x-x+6 }\\\\ f(x)=\frac{9(x-6)}{x(x-6)-1(x-6)}\\\\ f(x)=\frac{9(x-6)}{(x-1)(x-6)}$

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

Answer 2

Answer:

$x=1$

Step-by-step explanation:

For the function $f(x)=\frac{9(x-6)}{x^2-7x+6}$, 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:

$f(x)=\frac{9(x-6)}{(x-6)(x-1)}$

Simplify:

$f(x)=\frac{9}{(x-1)}$

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:

$x=1$