which of the functions has a vertical asymptote at x=-2?

Question

answers:
x-3/x-2
x/-2
(x+2)(x+5)/x+2
x^2-3x+7/x+2

Answer 1

(x^2-3x+7)/(x+2)

A rational function has a vertical asymptote at a given value of x when that value causes the denominator to approach zero when the numerator does not. A term that approaches zero when x = -2 would be (x + 2), and two of the functions have that term in the denominator. However, (x+2)(x+5)/(x+2) also has (x + 2) in the numerator, so they cancel out and the function is always equal to (x + 5), except for at x = -2 where the function is undefined; it has a point discontinuity at (0, 5), but not a vertical asymptote. The correct answer is (x^2-3x+7)/(x+2). At x = -2, the denominator is equal to 0 but the numerator is equal to 17, causing the function to approach +/- infinity.

Answer 2

Just x^2-3x+7/x+2

(x+3)/(x-2) has one at 2, not -2
x/-2 is continuous for all reals

x^2-3x+7/x+2
This function has an infinite discontinuity at x=-2, which indicates a vertical asymptote. If a function is undefined in such away that the denominator is 0 and the numerator is not zero at that point, then you have a vertical asymptote and infinite discontinuity. This is the only function that this occurs

Now, concerning
(x+2)(x+5)/x+2

This one has a removable discontinuity
it's just y=x+5 for x ≠ 5. meaning it simply the line x+5 with a hole at x =5.

Approaching 5 from the left and right sides y is approaching 3, not infinity.

Answer 3

The first has a vertical asymptote at x=+2.
The second is a straight diagonal line (no vertical asymptote).
The third is equal to x+5 (a diagonal line with no vertical asymptote), except at x=-2, where it is undefined.
*The fourth one has the vertical asymptote at x=-2.

One way to find a vertical asymptote is to find the case where setting x=-2 makes the denominator (the bottom) of the fraction equal to zero, but the numerator (the top) something other than zero.

Another way is to graph these on a graphing calculator or computer. A vertical asymptote is where the value of the function gets very large (or very negative) as it gets closer to (but never quite reaches) a vertical line. It looks like either a spike (for example, if you graph 1/(x^2), there is a vertical asymptote at x=0), or like an upside-down spike (for example, -1/(x^2) at x=0), or like half-and-half (for example, 1/x at x=0).

Answer 4

The 4th function has an asymptote at x=-2, since the divisor "(X+2)" will go to 0 as x approaches -2. The same could be said for the third function, except that it has a factor of "(x+2)" in the numerator as well, which cancels out the denominator, and the funciton evaluates to "3" as x approaches -2. (There will be a discontinuity at x=-2, but not an asymptote.

Answer 5

A vertical asymptote exists when the denominator of a rational function has a restricted domain...in this case, the denominator cannot equal 0.

The third and fourth functions listed have a value of 0 in their denominator if x = -2, so your answers are:

(x+2)(x+5) / (x+2)
(x^2-3x+7) / (x+2)

Note that, even though you can cancel out (x + 2) terms in the third one, the domain of the entire function is still restricted technically.

Answer 6

the last two

this can be found by equating the denominators to zero and finding the value of x, which is the asymptote.

additional: doing this ensures that for all value of x,none of it lies when the denominator is zero as when that happens, the function is undefined

Answer 7

I agree with rkhirsh.