Find the equation of the tangent lines to the graph of the equation y^2-2xy-8=0 at x=1?
968996
y^2-2xy-8=0...........(1)\nor, 2y(dy/dx) - 2 (x dy/dx + y) = 0\nor, y(dy/dx) - x(dy/dx) = y\nor, dy/dx = y/(y-x)\n\nLet us choose an arbitrary pt (1,k) on st line x=1\n\nhence, dy/dx(at (1,k) = k/(k-1).\nequation of the tangent on (1) at (1,k) is,\n\n(y-k) = k/(k-1) (x-1)\nor, (y-k)(k-1) = k (x-1)\nor, yk -k^2 -y +k = kx -k\nor, kx + y(1-k) = 2k - k^2................(2)\n\nfrom (1),\nputting x =1\ny^2-2y-8=0\nor(y-4)(y+2) = 0\nor, y = 4,-2\nhence k can assume the values 4 & -2\nputting the values of k in (2)\n4x + y(1-4) = 2*4 - 4^2\nie, 4x - 3y + 8 =0\n\n&\n-2x + y(1-(-2)) = 2*(-2) - (-2)^2\nie, 2x - 3y = 0\n\nHence the equations are 4x - 3y + 8 =0 & 2x - 3y = 0
774094
- Find the x-intercepts, by setting y=0 and solving the polynomial equation. You may need to factorise\n\n- Find the y-intercepts, by setting x=0 and solving.\n\n- Find stationary points by differentiating the polynomial equation wrt x and setting the differential to 0. These give you the stationary points.\n\n- Differentiate again to find out whether these turning points are maxima, minima or points of inflexion:\nIf d²y/dx²  is positive, then it is a minimum point.  \nIf d²y/dx² is negative, then it is a maximum point  \nIf d²y/dx² is zero, then it could be a max, a min or a point of inflexion.\n\n- Find asymptotes. You may get a horizontal asymptote when x approaches +ve or -ve infinity and a vertical one as y approaches +ve,-ve infinity.\n\nThis is the basic info you need to sketch a graph depending on how complex the polynomial is.
712712
[on] [y=] [(] [x,t,o,n] [+] [2] [)] [/] [x,t,o,n] [enter] [5] [-] [2] [x,t,o,n] [enter] [(-)] [2] [x,t,o,n] [-] [3] [enter] [window] [(-)] [5] [enter] [5] [enter] [1] [enter] [(-)] [5] [enter] [5] [enter] [graph]\n\nYou should have on screen a vertical asymptote at x=0, horizontal asymptote at y=1, a curve going to negative infinity towards 0 on the left, a curve going to positive infinity towards 0 on the right, on the left tangent line y=-2x-3 touches point(-1,-1), and on the right tangent line y=5-2x touches point(1,3).
701505
As others have said, your current equation has x and y being 0, which would be the origin.\n\nIf you meant to type:\n2x - y = 0\n\nThen to solve this, isolate y:\ny = 2x\n\nThis is just a line going through the origin at a slope of 2 through points (1,2), (2,4), (-1, -2), (-2, -4), etc.
597648
y= -4x^2+8x-1\ny = -(4x^2-8x+1)\ny = -[(2x-2)^2 - 4 + 1]\ny = -[(2x-2)^2 - 3]\ny = 3 - (2x-2)^2\n\nSo the vertex is (1,3) and it is a maximum point.
318213
eh? you using simultaneous equations with a straight line equation?\n\ny = mx + c is a straight line equation, where m is the gradient & c is where the line cuts the y axis.\n\nThere for lets say y = -2x + 8:\ny = mx + c\nCuts x-axis Put y = 0\n-2x + 8 = 0\n-2x = -8\nx = -8/-2\nx = 4\n\nHope that was correct :| lol
885928
Find the X intercept by setting Y = 0\n\nx-2 = 0\nx = 2\n\nFind the Y intercept by setting X = 0\n\ny = 0-2\ny = -2\n\nSo the two points to plot are (0,-2) and (2,0) \nThen connect those points with a line
954093
k1ax - k2x^2 = 0\nx(k1a - k2x) = 0\n\nat x = 0 and...\nk1a - k2x = 0\nx = k1a/k2\n\nso at x = 0 and a*k1/k2
771247
for these you need to find the x and y intercepts(the point at which the line crosses the x axis is the x-intercept, and and the point at which the line crosses the y axis is the y-intercept), and then connect the two points to form a line. (you do this for every line.)\n\nfor example: x-y=8. <<is in the Ax+By=C form, where A, B, and C are coefficients\n             to find the x intercept, you put C over A or C/A. in this case it is 8/1=8. That is your x intercept. for the y intercept you put C over B or C/B--in this case  it is 8/-1 or -8. You graph that on a coordinate plane. You connect the two points and get the line x-y=8.\n\n\nYou do the same thing for line 2x+y=7. Your x-intercept is 7/2 and your y-intercept is 7/1 or 7. When you connect those two points and graph that line, there will be a point where those two lines intercect. Find the coordinates of that point and that is your solution.  \n\n\ntry the next one on your own, and you'll get it!!! once you find the x- and y-intercepts it's a piece of cake. good luck!!! need any more help, email me.
271990
x^2-2xy+4xz-8yz=\n(x^2-2xy) + (4xz-8yz)=\nx(x-2y) +4z(x-2y)=\n(x-2y)(x+4z)
199440
find the tangent of the equation x+y=5
820064
x^2+2xy-y^2=7\n\ndy/dx\n2x+2y+2x(dy/dx)-2(dy/dx)=0\n(2x-2)(dy/dx)=-2x-2y\n\ndy/dx= (-2x-2y)/(2x-2)\ndy/dx=-3\n\nd^2y/dx^2\n[(2x-2)(-2-2(dy/dx))-\n(-2x-2y)(2)]/(2x-2)^2\n\nd^2y/dx^2=5\n\ni think thats rite sry bout all the parantheses
389156
actually, you slolve each of the three parenthesis and that tells you when the equation equals zero...\n\nX=-2, x=1 and x=5.\n\nNow look at each interval to determine whether the curve is greater than or less than zero in that interval.  For example, \nless than -2: x=-3 is less than -2.  So substitute x=-3 and solve... (-1)(-4)(-8)=-24 which is less than 0.\nbetween -2 and 1: x=0 is between these numbers.  So substitute x=0 and solve...(2)(-1)(-5)=+10 which is greater than 0.\nbetween 1 and 5: x=2 is between these numbers.  So substitute x=2 and solve....(4)(1)(-3)=-12 which is less than 0. \ngreater than 5: use x=6: (8)(5)(1)=+40\n\nSo the number line should show an open circle at -2 and an arrow pointing left, and open circles at 1 and 5 with a line connecting the two circles at 1 and 5.
795130
y = tan(πx - 1)\nlet u = πx - 1\nnow, dy/dx = (dy/du)(du/dx)\nd(tan(u)/du = sec^2(u)\ndu/dx = π, so \ndy/dx = πsec^2(u)\ndy/dx = πsec^2(πx - 1)
761613
Using implicit differentiation, you get:\n\n-csc^2(y)(dy/dx) = 1 - dy/dx\n-csc^2(y)(dy/dx) + dy/dx = 1\n(dy/dx)[1 - csc^2(y)] = 1\n\nTherefore, dy/dx = 1/[1 - csc^2(y)]\n\nNow plug in your ordered pair to get dy/dx.  Remember csc (y) = 1/sin(y):\n\ndy/dx = 1/[1-1], which is undefined.  Therefore, you cannot write a tangent line for this ordered pair (there is a vertical asymptote at this point)\n\n------------\n\nHope this helps
421517
y-y^1=m(x-x^2)\ny-8=2(x-(-2)\ny-8=2(x+2)\n A is the correct answer.\ny-8=2x+4\ny=2x+4+8\ny=2x+12 >> y form\n(transpose)\n-2x+y=12\n(if the variable x is negative make it positive)\n-(-2x+y=12) \n2x-y=-12 >> standard form
850129
Firstly, determine the slope or gradient of the lines;\n\nLine 1:\n\nm1 = (y2-y1)/(x2-x1)\n     = (3 - -7)/(1 - -4)\n     = 10/5\n     = 2\n\nLine 2:\n\nm2 = (y2-y1)/(x2-x1)\n    = (10 - 6)/(4 - 2)\n    = 4/2\n    = 2\n\nBoth gradients are the same, therefore the lines are parallel. if the product of the gradients (m1m2) was -1 then they would have been perpendicular.\n\nWe can take this further by checking to see if they are the same line.\n\nThe equation y - y1 = m(x - x1) will be the same for both lines if this is the case.\n\nLine 1:\n\ny - 3 = 2(x - 2)\ny = 2x + 1\n\nLine 2:\n\ny - 10 = 2(x - 4)\ny = 2x + 2\n\nThese are not equal, however we now know that line 2 sits 1 unit higher on a graph than line 1, and both lines are parallel to each other.