What's the horizontal & vertical asymptote of this equation??
921915
NOTE: i know this looks long, but i put in a lot of explanation to help on this problem and any others like it. i think (and hope) that youll find very useful! :)\n\n\nyoure right about the vertical asymptote(VA), you just set the denominator equal to zero:\n\nx(x-1) = 0\n\nyou solve it using the zero product theorem.  that says that when the product of two things equals zero, at least one of those individual parts must be equal to zero.  (because the only way to get zero by multiplying is to multiply by zero.  sorry, you probably already knew all that.  anyway, you get\n\nx=0\nand\nx-1=0,   which you solve to get x=1\n\nthat tells you that the function is undefined at x=0 and 1, but those arent necessarily both VAs.  if any part of the denominator can 'cancel' with the a part of the numerator, its considered a 'removable discontinuity' because it cancels out in theory, but you cant really do that because that would totally change the function.\n\nlong story short, an x cancels out, so the discontinuity you got from that piece (x=0) is a hole in the graph, not a VA.  the x-1 cant cancel, so you know that x=1 is your one and only VA.\n\n\nabout the HA . . .\n\nthe degree of the numerator is actually 2.  remember that the degree is the highest power of x.  since you have x^2 in the numerator, its degree is two.\n\nto get the degree of the denominator, you have to look at what the highest power of x would be if you distributed and FOILed it all the way.  if it helps you, go ahead and do that.  all you really have to do is count the x's that will be multiplied together though.  there are 2, which you know will give you x^2, so the degree of the denominator is also 2.\n\nthe rule is that since the degrees of the numerator and denominator are the same, your HA will be y=(leading coeffiecient of the numerator)/(leading coeff. of the denominator)\n\nremember that the leading coeff. is the coeff of the x with the highest power.\n\nfor the numerator thats easy, its 1\n\nfor the denominator, you have you use the distributed version, which is also 1.\n\nfollow the rule, and y=1/1, or y=1\n\n\n\nthere you have it!  final answer . . .\n\nVA: x=1\nHA: y=1\n\n\nof course feel free to send me a message if you have a question :)
788532
Is it x*(e-2) or x^(e-2)?\n\nI don't think it matters anyway, because the bottom fluctuates between 3 and 5 while the top increases indefinitely, i.e. approaches "infinity" if you like.  So there's no limit.
712145
f(x) = ba^x + c\nSince it passes through the point P(1,4)\nf(1) = ba + c = 4\nSince the y-intercept is 2 (ie it passes through (0, 2)\nf(0) = b + a = 2\nThe statement "horizontal asymptote y=2" renders a solution impossible. I think you meant horizontal asymptote y=1\nThen lim x→-∞  f(x) = c → c = 1\nSo b + 1 = 2 → b = 1\nba + c = 4 → a = 3\n\nThus f(x) = 3*x + 1
835719
~this is a horizontal line\nand the equation is x= -5
953056
No asymptote.  It just keeps growing. That is, as x─►±∞, y ─►±∞.
184282
y = 7\n\na horizontal line
934522
y' = [(x-1) - x] / (x-1)^2 = -1/(x-1)^2, so no critical points.  y' < 0 for all x where y is defined, so it's always decreasing.\n\ny'' = 2/(x-1)^3, which is positive when x>1, and negative for x<1.  Therefore the curve is concave up for x>1, and concave down when x<1.\n\nThere is a vertical asymptote at x=1, and lim x->inf x/(x-1) = \nlim x->inf 1/(1-1/x) = 1 (same for x -> -inf) so there is a horizontal asymptote at y=1.
654739
The horizontal position of the projectile is described by x(t) = (v_x)*t + h since the horizontal force is zero. v_x is the initial horizontal velocity and h is the initial horizontal position.\nThe vertical net force is -mg, so -mg=ma, thus -g = a so the projectile is accelerated by -g (the acceleration due to gravity at sea level). Integrating, we get y(t) = -(1/2)gt^2 + (v_y)t + k where v_y is the initial vertical velocity and k is the initial vertical position.\nSolving for t in the first equation, we get t(x) = (x - h)/v_x. Since the times in both equations are the same, we plug this value into y(t) to get y(t(x)) = y(x) = -(1/2)g((x - h)/v_x)^2 + (v_y/v_x)(x - h) + k.\nThus, the path of the projectile is a parabola. Setting y=0 will get you the x-intercepts of the parabola, or where the projectile touches the ground.  Note that the projectile starts above ground and time only runs in one direction, so the projectile only touches ground on one root of the equation.
909100
To have vertical asymptotes at x = 6 and x = -1, the denominator should have factors of (x - 6) and (x + 1), making it x^2 - 5x - 6.  To have a horizontal asymptote of -9, the numerator should be of the same order as the denominator, but with a coefficient for the greatest exponent of x that is -9 times that of the denominator.  The denominator is 2nd order with a coefficient of 1 for x^2, so the numerator can be -9x^2.\n\n-9x^2 / (x^2 - 5x - 6)
960038
Since the foci of this hyperbola both lie on the horizontal line y=7, then the major axis is horizontal and the minor axis is vertical, and so the equation for it is \n\n((x-a)^2)/(M^2) - ((y-b)^2)/(m^2) = 1 \n\nwhere the center of the hyperbola (the midpoint of the segment between the two foci) is (a,b), M is the length of the semi-major axis (half the distance between the vertices),  and m is the length of the semi-minor axis.  If d is half the distance between the foci, then M^2 + m^2 = d^2\nThe center of this hyperbola is (7,2) since it is the midpoint of the segment between the foci.  M = 3 (the distance from the center to each vertex), d = 5 (the distance from the center to each focus), and so m = 4 since 3^2 + 4^2 = 5^2.  \n\nThus this hyperbola has equation \n\n((x-7)^2)/9 - ((y-2)^2)/16 = 1
1144759
I don't know the answer to your question, but here is a good website that might help:\n\nhttp://www.purplemath.com/modules/asymtote3.htm
108513
Asymptotes are vertical lines associated with a graph but not part of a graph.  They are divisions in a graph that I call "territory markers" or "fences".  They are essentially the x values that make any function undefined.  These x values, when graphed, are vertical lines that a graph will  not cross but rather be forced to "walk along".\n\nExample:  f(x)=1/((x)*(x-2))\n\nThe function is obviously undefined at 2 specific x values (set the denominator equal to zero and solve for those x vales to find where the function is undefined), x=0 and x=2.  The graph of the line x=0 and/or x=2 is the graph of a vertical line, or asymptote, that this function cannot cross directly since there is no corresponding y value at that point (the function is undefined at these points, remember?).  The function is forced to skirt the fence line either up to infinity or down to negative infinity.  Graph it and you'll see.
1135263
The height of the trajectory is 1 m (3.05-2.05)\n\nThe horizontal distance is 6.02 m\n\nNow you have ucos(25)t = 6.02\n\nand usin(25)t-0.5*9.8*t^2 = 1\n\nSubstitute t = 6.02/ucos(25) in the 2nd eqn and solv.
86803
0.77(with a little horizontal line right above the last 7)
559279
Correct!\nOnly the second x's in the equation for asymptotes will be y.\nThis is an interesting result. Rectangular hyperbola is the locus of far corner of all rectangles' vertices having a fixed area A. And having the asymptotes as two adjascent sides of the rectangle.\nWith proper substitution, the general eqn can be transformed to ione.
787424
To find the asymptote to the left.  find the limit of f(x) as x approaches -∞.\n\nTo find the asymptote to the right.  find the limit of f(x) as x approaches ∞.\n\nVertical asymptotes usually occur where a value of x gives a division by zero.  Find when the denominator equals zero then take the limit as x approches that value to make sure the value of the function approaches negative or positive infinity.\n\n I hope that helps.
782645
1/x=y+3\nit resembles the equation of the asymptotes of hyperbola