No they don't. They just stretch for a very long ways horizontally without much increase vertically because the output of the function is the exponent of the input. For example, f(x) = log x when x = 1000, f(x) = 3 because 10^3 = 1000 (10 being the base of common log). Therefore, when you increase x substantially, there is only a small increase in y.
tangent, cosecants, secant, cotangent.
Asymptotes are one way - not the only way, but one of several - to analyze the general behavior of a function.
Yes, to the left (towards minus infinity).Yes, to the left (towards minus infinity).Yes, to the left (towards minus infinity).Yes, to the left (towards minus infinity).
Substitute y = mx + b into the equation and then use the fact that there must a double root (at infinity)
that's simple an equation is settled of asymptotes so if you know the asymptotes... etc etc Need more help? write it
Three types of asymptotes are oblique/slant, horizontal, and vertical
If a hyperbola is vertical, the asymptotes have a slope of m = +- a/b. If a hyperbola is horizontal, the asymptotes have a slope of m = +- b/a.
Many functions actually don't have these asymptotes. For example, every polynomial function of degree at least 1 has no horizontal asymptotes. Instead of leveling off, the y-values simply increase or decrease without bound as x heads further to the left or to the right.
No, it will always have one.
tangent, cosecants, secant, cotangent.
Asymptotes are one way - not the only way, but one of several - to analyze the general behavior of a function.
finding vertical asymptotes is easy. lets use the equation y = (2x-2)/((x^2)-2x-3) since its a rational equation, all we have to do to find the vertical asymptotes is find the values at which the denominator would be equal to 0. since this makes it an undefined equation, that is where the asymptotes are. for this equation, -1 and 3 are the answers for the vertical ayspmtotes. the horizontal asymptotes are a lot more tricky. to solve them, simplify the equation if it is in factored form, then divide all terms both in the numerator and denominator with the term with the highest degree. so the horizontal asymptote of this equation is 0.
Yes. One at y= pi/2 and y=-pi/2
Yes, to the left (towards minus infinity).Yes, to the left (towards minus infinity).Yes, to the left (towards minus infinity).Yes, to the left (towards minus infinity).
there is no horizontal-line test for functions, because people do not do the test that is why !!!
Definition: If lim x->a^(+/-) f(x) = +/- Infinity, then we say x=a is a vertical asymptote. If lim x->+/- Infinity f(x) = a, then we say f(x) have a horizontal asymptote at a If l(x) is a linear function such that lim x->+/- Infinity f(x)-l(x) = 0, then we say l(x) is a slanted asymptote. As you might notice, there is no generic method of finding asymptotes. Rational functions are really nice, and the non-permissible values are likely vertical asymptotes. Horizontal asymptotes should be easiest to approach, simply take limit at +/- Infinity Vertical Asymptote just find non-permissible values, and take limits towards it to check Slanted, most likely is educated guesses. If you get f(x) = some infinite sum, there is no reason why we should be able to to find an asymptote of it with out simplify and comparison etc.
When you plot a function with asymptotes, you know that the graph cannot cross the asymptotes, because the function cannot be valid at the asymptote. (Since that is the point of having an asymptotes - it is a "disconnect" where the function is not valid - e.g when dividing by zero or something equally strange would occur). So if you graph is crossing an asymptote at any point, something's gone wrong.