find the vertical asymptote of the rational function f(x)= 3x-12/4x-2?

Answer 1

Vertical aysmptotes occur when you can't input a certain number in for the x in your equation (because the horizontal x-axis gives you the x=co-ordinate, so that x=a is a horizontal line at a). The only problems you could have with putting a number into your function are if you putting that number in for x causes you to take the square root of a negative or divide by zero. You don't have any square roots, but you are dividing. Depending on your x value, you might be dividing by zero. You're dividing by 4x-2. If this is zero, you're dividing by zero (which you can't do). So if 4x-2 = 0, 4x = 2, x = 0.5, you're doing something that isn't allowed. Instead of having a point on your curve here (and dividng by zero),you have a vertical asymptote at x = 0.5.

Answer 2

the two vertical asymptotes propose that the denominator turns into 0 the two whilst x=2 and x=3. this could ensue if the denominator equals (x-2)(x-3) or x^2-5x+6. The y-intercept being a million potential that as quickly as x is 0, the fraction is a million. We already be attentive to that as quickly as x is 0 the denominator is (-2)(-3) or 6. So, the numerator additionally needs to equivalent 6 whilst x is 0. case in point, x+6. The horizontal asymptote at y=0 potential that the denominator outpaces the numerator as x turns into large (like, infinity). So, purely shop the numerator growing to be extra slowly than x^2. x+6 or 6 could artwork. putting all of it together, how approximately (6) / ((x-2)(x-3)) ?

Answer 3

4x-2 must be zero so x= 1/2 is the equation of the vertical asymptote

y= 3/4 is the equation of the horizontal asymptote

Answer 4

Assume f(x) = (3x - 12) / (4x - 2)

Upon dividing by 4x - 2, obtain :-

y = f(x) = 3/4 - (21/2) / (4x - 2)

y = 3/4 - (21/2) / 4(x - 1/2)

Vertical asymptote when x = 1/2

Answer 5

there is only one vertical asymptote: x= 1/2