Question

Can someone explain log functions to me?
I don't understand why it has a vertical asymptote.

Answer 1

If a function has a larger exponent in the denominator, then there is always a horizontal asymptote at y = 0, on top of the x axis.

If a function has the same largest exponent in the numerator and denominator, then there is a horizontal asymptote at the number formed by dividing the leading coefficients of the numerator and denominator. So for:
. . . 2x² - 18
y = ------------------ has a horizontal asymptote at y = 2/3 from 2x²/3x²
. . . 3x² - 2x - 1


. . . 6 - 24x²
y = ------------- has a horizontal asymptote at y = -8 from -24x²/3x² = -8
. . . .3x² - 3

If a function has a larger exponent in the numerator than in the denominator, then there is no horizontal asymptote. Instead, there is a slant asymptote whose equation is found by dividing the numerator by the denominator. Discard any remainder from the division.

. . . 2x² - 4x + 5
y = -------------------
. . . . x - 2

. . . _2x________
x - 2) 2x² - 4x + 5
. . . .2x² - 4x
. . . --------------
. . . . . . . . . . . .5

The slant asymptote here is y = 2x.

For vertical asymptotes or holes factor the numerator and denominator. If there are factors in the denominator that do not repeat in the numerator, then when you set them equal to zero and solve, these are places with a vertical asymptote of the form x = number.

If there are factors in the denominator that do repeat in the numerator, then there are holes in the graph at these x values instead of vertical asymptotes. The hole's y value can be calculated by substituting its x value into the reduced, simplified rational expression.

. . . 2x² - 18
y = ------------------ factors into:
. . . 3x² - 2x - 1

. . . 2(x - 3)(x + 3)
y = ---------------------
. . . (3x + 1)(x - 1)

No factors repeat in both the numerator and denominator, so there are no holes. Set the factors in the denominator equal to zero and solve to find the vertical asymptotes.

3x + 1 = 0
3x = -1
x = -1/3 <==ANSWER

x - 1 = 0
x = 1 <==ANSWER


. . .2x² - x - 1
y = --------------- factors into:
. . . .3x² - 3

. . . (2x + 1)(x - 1)
y = ---------------------
. . . .3(x - 1)(x + 1)

This has the factor (x - 1) on top and bottom, so solving x - 1 = 0 gives x = 1 as the location of a hole. The fraction simplifies into:

. . . (2x + 1)
y = ------------- If x = 1, the hole is at (1, ½).
. . . .3(x + 1)

The other factor in the denominator that does not repeat gives a vertical asymptote at x = -1.

Answer 2

does this address log functions? i dont see it

Answer 3

Is it f(x)=log(x)?

Answer 4

yeah

Answer 5

imnot sure how familiar you are with a log function to begin with but:

log (x) = y represents an exponent such that 10^y = x. There is not value defined for the exponent y such that 10^y = 0. You can get closer and closer values as the value of y gets smaller and smaller, but you can never obtain a real value for y such that 10^y = 0

When something is undefined, and we can slide up closer and closer to it, but never reach it .... its called an asymptote.

Since 10^y = 0 is undefined; then so is log(x) = y when x=0