What did the asymptote say to the removable discontinuity?
Don't hand that holier than thou line to me.
Asymptotes, removable discontinuities, and intercepts?
2. y = (x^2+4x-5) / (x^2+8x+15) The denominator factors into (x+5)(x+3). So the domain is R except -3 and -5. We can simplify the function to: y= [(x+5)(x-1)] / [(x+5)(x+3)]..............We can cancel (x+5) since this is not zero. y= (x-1)] / (x+3) So we can rewrite the function as: y= (x-1) / (x+3), where x!=-5.........................(*) This function (*) is undefined at x=-3 since it will make the denominator 0, so the function has a vertical asymptote at x=-3. Note that (*) is defined for x=-5 and the corresponding y is 3. But x=-5 is not in the domain of the function, so the graph will have a hole at the point (-5,3). The function has a removable discontinuity at this point. To determine the y-intercept, simply set x to be 0 and solve for y using (*). y= (0-1) / (0+3) = -1/3. y-intercept: (0, -1/3) To determine the x-intercept, simply set y to be 0 and solve for x using (*). 0 = (x-1) / (x+3) 0 = x-1 x = 1 x-intercept: (1, 0)
What are the points of discontinuity, holes, vertical asymptotes, x-intercepts, and horizontal asympotote of?
Points of discontinuity are the the points that would go where the hole is if the hole was filled in. They are found at x coordinates that make both the top and bottom of an equation equal zero, as 2 would be in f(x) = (x^2 - 4)/(x - 2). If x = 2 you have 0/0 but if x is anything else, (x^2 - 4)/(x - 2) is just x + 2 which would be 4 if x = 2. So the point of discontinuity is (2, 4). But yours will never have 0 on top so it has no points of discontinuity or holes. Vertical asymptotes are x = any x value that makes the bottom = 0 but not the top. 3x^2 + 3x - 18 = 3(x^2 + x - 6) = 3(x + 3)(x - 2) which = 0 if x = -3 or x = 2 so it has two VA's. x intercepts are when f(x) = 0 which never happens for yours so there is no x intercept. Horizontal asymptotes only happen when the top degree is less than the bottom, in which case the HA is y = 0 which it is here. Or, when the top degree = the bottom degree then it's y = the top leading coefficient over the bottom leading coefficient. Like if you had f(x) = (3x^2 + 5) / (x^2 - 3x + 1), the HA is y = 3/1
What is the difference between a vertical asymptote and a removable discontinuity?
A vertical asymptotes is a non-removable discontinuity; that is, there is no way to re-define the function such that the function will continuous at that point. On other other hand, holes are removable discontinuities; that is, there IS a way to re-define the function such that it will be continuous at that point. If you know what limits are, think of them like that. A vertical asymptote occurs in a rational function if the denominator is made zero at a certain value of x, but the numerator is non-zero. A hole occurs when a value of x causes both the numerator and denominator to be zero at the same time. With f(x) = [x(x - 2)]/[x(x - 2)(x - 2)], we see that, if x ≠ 0 and x ≠ 2: f(x) = 1/(x - 2). Now, if we tried plugging x = 0 into this new function, we actually get a nice result of -1/2. However, x = 2 still causes division by zero. If we defined f(0) = -1/2, then f(x) = [x(x - 2)]/[x(x - 2)(x - 2)] would be continuous at x = 0, but we cannot do anything about x = 2. I hope this helps!