The only way I ever learned to find it was to think about it.
The function f(x) = log(x) only exists of 'x' is positive. As 'x' gets smaller
and smaller, the function asymptotically approaches the y-axis.
y = 4(2x) is an exponential function. Domain: (-∞, ∞) Range: (0, ∞) Horizontal asymptote: x-axis or y = 0 The graph cuts the y-axis at (0, 4)
f(x) = 2*(x-3)*(x+2)/(x-1) for x ≠1
Yes, 2y = 8 is a linear function. 2y = 8 is the same as y=8/2, or y=4, which is a vertical line.
Well, f(x) = 6x + 9 is a function of x {it passes the vertical line test}. but y = 6x + 9 is an equation {linear equation of two variables}
x axis
It will have the same asymptote. One can derive a vertical asymptote from the denominator of a function. There is an asymptote at a value of x where the denominator equals 0. Therefore the 3 would go in the numerator when distributed and would have no effect as to where the vertical asymptote lies. So that would be true.
One point on a logarithmic graph is not sufficient to determine its parameters. It is, therefore, impossible to answer the question.
2x-2/x^2+3x-4
The point you desire, is (1, 0).The explanation follows:b0 = 1, for all b; thus,logb(1) = 0, for all b.On the other hand, logb(0) = -∞,which explains the vertical asymptote at the y-axis.
Yes, the asymptote is x = 0. In order for logarithmic equation to have an asymptote, the value inside log must be 0. Then, 5x = 0 → x = 0.
Yes. Take the functions f(x) = log(x) or g(x) = ln(x) In both cases, there is a vertical asymptote where x = 0. Because a number cannot be taken to any power so that it equals zero, and can only come closer and closer to zero without actually reaching it, there is an asymptote where it would equal zero. Note that transformations (especially shifting the function left and right) can change the properties of this asymptote.
- 2 makes this zero and provides the vertical asymptote. So, from - infinity to - 2 and from - 2 to positive infinity
The domain of the function f(x) = (x + 2)^-1 is whatever you choose it to be, except that the point x = -2 must be excluded. If the domain comes up to, or straddles the point x = -2 then that is the equation of the vertical asymptote. However, if you choose to define the domain as x > 0 (in R), then there is no vertical asymptote.
True
f(x) = bX is not an exponential function so the question makes no sense.
no
asymptote