at what points does the curve have a horizontal tangent?
957082
A curve has an horizontal tangent when its derivative equals 0.\nSo all you have to do is calculate the derivative of your function (which is y'= 6x^2 +6x -36) and solve the equation y' = 0.\nYou'll find two values of x, which are the two points where your curve has an horizontal tangent.
723693
I'm pretty sure you are correct.  I am also working thru Calc III (at the moment, actually).  It helps to picture the gradient as perpendicular to the tangent plane.
724027
sidharth has solved the problem correctly and given you the two values for x that satisfy the conditions of the problem.\n\nHowever, he didn't specify the POINTS on the curve.\nThey are (0,0) and (-2,-2).
258618
This is what I wrote for a very close question to this one. Just ignore the graphing part. I didn't feel like doing the question right now, but this is the way to do it:\n\n"If you need the tangent, take the derivative of the function at the specific point, which will give you the slope. Since you have taken the derivative at some specific point (a,f(a)) you also have a point. Now use the point-slope form to figure out the equation. Then simply get a graph paper and graph it.\n\nExample: f(x)=3tan(2x)\n\nf'(x)=6[sec(2x)]^2 (I assume you can take derivatives.)\n\nYou have a point (a,f(a))\n\nNow use the point-slope form: y-f(a)=f'(x)(x-a)\ny-3tan(2a)=6[sec(2a)]^2*(x-a) is the answer in this case. \n\nOr in function form: y=6[sec(2a)]^2*(x-a)+3tan(2a)\n\nTo graph it, find 2 points on the equation, then draw a straight line between them. That is your tangent."
184282
y = 7\n\na horizontal line
716765
The potential energy at the top is mgR = 2 x10 x R = 20 R.\n\nKinetic energy at the bottom of the curve = 0.5 m U ^2 = 0.5 x 2 x U^2. = U^2.\n\nAs there is no friction  K.E = P.E\n\nU^2 = 20 R.\n\nOn the horizontal path, \n\nInitial speed is U.\n\nFinal speed is V = 0.\n\nDistance traveled = 8 m.\n\nAcceleration = Fricttional Force / mass = -8/ 2 = - 4 m/s^2. ( Negative sign since the force acts opposite to the motion.\n\nUsing V^2 - U^2 = 2as, and also noting that V=0,\n\nU^2 = 2 x 4 x 8 = 64.\n\nU = 8m/s.\n\nThe time of travel is found using V= U+ at.\n\nt = -8/ (-4) = 2 second.\n\nUsing U^2 = 20 R.\n\nR = 64/ 20 = 3.2 m.
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).
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
912852
Let P be (a,b). The slope of the tangent at P is found from the derivative of the curve: -128 / x³, giving the equation of the tangent as\n\n          y - b = ( -128 / x³ ) ( x - a )\n\nNow find where this tangent lines cuts the X and Y axes. Let x=0 and solve for y. Let y=0 and solve for x to get\n\n          ( 0 , b + 128 / a² ) and ( a + a³b / 128 , 0 )\n\n(a, b) is on the curve so b = 64 / a². Use this to simplify the two intercepts to\n\n          ( 0 , 192 / a² ) and ( 3/2 a , 0 )\n\nUse Pythagoras to find the length between these points as the square root of\n\n          f(a) = 192² / a^4 + 9 / 4 a²\n\nNow the problem is to minimize this expression (ignoring the square root, because it is an increasing function). This is a standard minimization calculus problem. Set f '(a) = 0, and check f "(a) > 0 to find that the minimum occurs when a = 4√2, so the required point is \n\n          ( 4√2 , 2 ).
179915
There will be a vertical asymptote at x=0. When x values get increasingly closer to zero from the right, the graph approaches infinity. When the x values approach 0 from the left, the graph approaches negative infinity.\n\nSimilarly, there is also a horizontal asymptote at y=4. As x approaches infinity, the graph approaches 4 (1/infinity + 4 = 4). Also, as x approaches negative infinity, the graph also approaches 4 (1/negative infinitive + 4 = 4).
924809
You can take the deriviative of a function (say f ' (x)) and substitute for x the x-coordinate of a point somewhere on f(x) to get the slope of the tangent line that passes through that point on the curve.\n\nThe tangent line is the best approximation of the curve it is tangent to at the point of tangency.\n\nAlso, you can use point-slope form ( y-yo = m (x-xo) ) to find the equation of the tangent line.
835719
~this is a horizontal line\nand the equation is x= -5
816872
Best to plot this on a piece of graph paper; "+i" is  to the right' , "+j" is upwards, "-i" is to the left, "-j" is downwards, all with respect to the origin.\n\n1. Draw the vectors on the graph paper. "u" is in the second quadrant (i is neg and j is pos). "v" is in the fourth quadrant, with (i is pos, j neg) . There are two answers to this question, since you could go from u to v or from u to v in the opposite direction. I always take the angle from zero to "u" and add it to the angle from 0 to "v".  For "u", the tangent of the angle to the vertical axis is 1/4, then angle = 14%. So the angle from zero degrees to the vert axis = 90, and "u"s angle from zero degrees is 90 + 14 = 104 degrees. For "v", the tangent of the angle to the horizontal axis is 4/3 = 1.333.. and the angle = 53 degrees. Therefore the angle from "u" to "v" is 104 + 53 = 157 degrees. (the other answer is 360 degrees - 157 degrees = 203 degrees if you rotate in the other direction).\n\n2. Again use graph paper. 240 degrees is 30 degrees less that 270 degrees (the vertical axis, down) draw a line 10 graph units long. It will be in the third quadrant where i and j will both be negative. At this point you can use a polar to rectangular converter, or do it by hand. By hand: the i side of the triangle formed with the down axis can be found by saying sin(angle) = i/10, and cos(angle) = j/10. The angle is 30 degrees (from vert downwards). so sin(30) = 0.5, and, cos(30) = 0.866; thus i= 0.5 and j = 0.866 (ignoring pos/neg). Since the vector is in the third quadrant, both i and j are negative.\nAnswer = -0.5i -0.866j\n\nHope this helps!
824813
f(x) = x^m(1-x)^n\nm zeros at  x = 0 and n zeros at x = 1 \nSo x-axis is tangential to the curve at x = 0 and x = 1\n\nf'(x) = mx^(m - 1)(1 - x)^n + x^m * -n(1 - x)^(n - 1)\n= x^(m - 1) * (1-x)^(n - 1) * (m(1 - x) - nx)\n=  x^(m - 1) * (1-x)^(n - 1) * [m - (m + n)x]\n= 0 for stationary points.\n\nSo (m - 1) zeros at x = 0, (n - 1) zeros at x = 1 and a zero at x = m / (m + n)\n\nNow there has to be a turning point between x = 0 and x = 1 as f(x) is a polynomial and therefore continuous and differentiable for all x\n\nSo examine the possibilities\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... - .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... - .......... + ........... + ... Local minimum at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n ............... + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... + .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m .................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... + .......... + ........... + ... Local minimum at (1, 0)
939024
Draw yourself a cartesian graph, vertical line is the y axis, horizontal is the x axis.\n\npick points on the x axis like 1, 2, 3, etc.  plug thos values into the equation and solve for y. then pair the numbers, place points in the graph by taking the x value going straight up (or down if y is a negative number) and place the dot accross from the y value.  do this for each set of points then connect the dots with a line or a curve. for instance in your equationstart with x=1\n\ny= 7(1-5)(1-2)^4(1+.5)(1+2.5)^3\ny= 7(-4)(-1)^4(1.5)(3.5)^3\ny=7(-4)(1)(1.5)(42.875)\ny=-1800.75\n\nso in XY format the first point would go at (1, -1800.75)
271844
If you need the tangent, take the derivative of the function at the specific point, which will give you the slope. Since you have taken the derivative at some specific point (a,f(a)) you also have a point. Now use the point-slope form to figure out the equation. Then simply get a graph paper and graph it.\n\nExample: f(x)=3tan(2x)\n\nf'(x)=6[sec(2x)]^2 (I assume you can take derivatives.)\n\nYou have a point (a,f(a))\n\nNow use the point-slope form: y-f(a)=f'(x)(x-a)\ny-3tan(2a)=6[sec(2a)]^2*(x-a) is the answer in this case. \n\nOr in function form: y=6[sec(2a)]^2*(x-a)+3tan(2a)\n\nTo graph it, find 2 points on the equation, then draw a straight line between them. That is your tangent.
893970
x=t^2 - 1\ny=t^4 - 4t\n\ntangent vector: ( 2t, 4t^3 -4)\nthe tangent vector is horizontal if:4t^3 -4 =0, so t^3=1, so t=1,\nin this case\nx= 0, and y = 1-4=-3\n\nthe tangent vector is vertical if: 2t=0, so t=0\nin this case\nx= -1 and y= 0\n`