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

x=1

Step-by-step explanation:

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

Following points must be kept in mind:

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

Step-by-step explanation:

For the function , 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:

Simplify:

Observe that the denominator is equal to zero in .

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

Then the vertical asymptote s:

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.