Veritcal asymptotes are where the denominator of a fraction becomes 0 and the value of f(x) becomes undefined. Set the denominator to 0 and solve.
Horizontal asymptotes are values of f(x) when x→∞ and x→-∞
Two examples:
f(x)=3x² / (x²+1)
Vertical:
Set (x²+1)=0 and solve for x.
x²=-1 has no answers, so there is no vertical asymptote.
Horizontal:
Divide all terms by the highest power of x to eliminate unimportant values
dividing by x², you get 3 / (1 + (1/x²))
As x→∞ then 1/x² vanishes, leaving 3/1=3, so there is an asymptote at y=3
f(x)=(x-3) / (x²+3x)
Vertical:
Set (x²+3x)=0 and solve. x={0,-3} so there are vertical asymptotes at 0 and -3
Horizontal:
Divide out by x²
(x/x² - 3/x²) / (x²/x² + 3x/x²) as x→∞
The terms in the numerator all vanish, making the answer 0, so there is a horizontal asymptote at 0.
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A vertical asymptote also exists for the value of x (assuming a function of x) where the function becomes undefined. For instance:
f(x) = ln (x). The logarithmic function is not defined for x<=0.
Ex: f(x) = ln (x+2)... There would be a vertical asymptote at x= -2.
that's simple an equation is settled of asymptotes so if you know the asymptotes... etc etc Need more help? write it
find the constant difference for a hyperbola with foci f1 (5,0) and f2(5,0) and the point on the hyperbola (1,0).
None.
Asymptotes are one way - not the only way, but one of several - to analyze the general behavior of a function.
Substitute y = mx + b into the equation and then use the fact that there must a double root (at infinity)
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.
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.
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
When you graph a tangent function, the asymptotes represent x values 90 and 270.
find the constant difference for a hyperbola with foci f1 (5,0) and f2(5,0) and the point on the hyperbola (1,0).
None.
music notes
ellipses do have asymptotes, but they are imaginary, so they are generally not considered asymptotes. If the equation of the ellipse is in the form a(x-h)^2 + b(y-k)^2 = 1 then the asymptotes are the lines a(y-k)+bi(x-h)=0 ai(y-k)+b(x-h)=0 the intersection of the asymptotes is the center of the ellipse.
Asymptotes are one way - not the only way, but one of several - to analyze the general behavior of a function.
Substitute y = mx + b into the equation and then use the fact that there must a double root (at infinity)
Not sure what non-verticle means, but a rational function can have up to 2 non-vertical asymptotes,