How to show that the tangents to the curves y=2x^3 - 2x^2 + 6x + 7 cannot be parallel to the x-axis.?
544525
You need to show that there are no horizontal tangents.\nTake the derivative, set it equal to zero, and show that there are no real values for which the derivative equals zero.
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."
1036703
To find the point of intersection between two curves, put y1=y2 so\n\nLet\ny1 = x^2 + 3x +2\ny2 = 7x -2\n\ny1 = y2\n\nx^2 + 3x + 2 = 7x -2\n\nRearrange:\n\nx^2 + 3x + 2 - 7x + 2 = 0 --->  x^2 - 4x + 4 =0\n\nThis is a second order eq. to solve it\n\nx1,x2 = [-b +- Sqr(b^2 -4 a c )] / (2 a)\n\nwhere a = 1,  b = -4 and c = 4\n\nx1,x2 = [+4 +- Sqr( (-4)^2 - 4 (1) (4) ) ] / (2*1)\n\nx1,x2 = [+4 +- Sqr( 16 - 16) ] / 2\n\nx1,x2 = 4/2 = 2\n\nthen x1=x2= 2  that means there is only one point; to find the corresponding y substitute x in either of the eq.\n\ny = 7 (2) -2 = 12\n\nThen the point (2,12) is the point of tangency because a parabola and a line intersect with two point. Therefore, (2,12) is the tangency point..!
934742
uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
821200
(x²+4x+4) / (2x-3) / (x²-4) / (2x²-7x+6)\n\nYou have 4 terms and three divisions, so it's a little unclear if you mean (a/b) / (c/d) or what.  I will try to help to the extent I can.\n\nLet's assume you mean:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ]\n\nFactor your x² equations to see if you can cancel terms:\n\nx²+4x+4 = (x+2)(x+2)\nx²-4 = (x+2)(x-2)\n2x²-7x+6 = (2x-3)(x-2)\n\nWoohoo - things are looking good!  Now we can rewrite our original problem using these factors:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ] = \n\n[ (x+2)(x+2) / (2x-3) ] / [ (x+2)(x-2) / (2x-3)(x-2) ] =\n \n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\nnotice we can cancel out three terms: (2x-3), (x-2) and (x+2)\n\n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\n(x+2)\n\nHope that helped!\n\n~ ♥ ~
637834
this is paralell to the x axis. It`s 7
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 ).
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`
1068232
Use quotient rule to get the derivative, which, simplified, is:\n\n-30*(2x+3)/((3x^2+9x+7)^2)\n\nSet it equal to zero and solve.  Since the denominator cannot be zero, set the numerator equal to zero\n\n2x+3=0\nx=(-3/2)\n\nI checked it, and it indeed is the solution. \n\nBy the way, you don't have to worry about an undefined points of this function making problems in the graph or in the max/min values in the derivative.  The denominator has imaginary roots.
918737
The equation for the tangent is:\ny = f'(a) * (x-a) + f(a)\n\nYou need to calculate the derivative h'(t) for t=3s and plug that into the equation, adapted to your problem:\n\ny = h'(3) * (t-3) + h(3)\n\nI'll let you do the maths...
865391
I think there is a sign error: the cubic term is supposed to be -2x^3, not 2x^3. Then the other roots are 3 and -1 and the factors are\n\n(x-8i)(x+8i)(x-3)(x+1)
324456
give parameters for x, y, and z.  Otherwise you cannot prove it, as no parameters are given for the equation as you described it.
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
1023653
It is a straight line parallel to X-axis, at a distanceof 3 units frm the X-axis, ie passing through the point (0,3).\n\nThis is because y=3, no matter what the value of x ( _ve, +ve, integral, fractional)
377295
If it has no solutions, then the two lines are parallel but different.  If it has infinitely many solutions, the two lines are actually the same line (think of them as one right on top of another).\n\nWhen you try to solve them algebraically, in both cases the x and y (or whatever two variables are used ) will both cancel out.  If there is no solution, you will end up with two unequal numbers, such as 0 = 3.  If there are infinitely many, you will end up with two equal numbers, such as 6 = 6.
450251
Yes, it is quite possible. First imagine such a curve and take a tangent line at an interior point. Now move that line up slightly so it intersects the original function at two nearby points. Finally let the line curve very slightly between those intersection points and join the resulting curve to (0,0) and (1,1) smoothly. Specific examples should be easy with this intuition.\n\nI'm guessing that any number of intersection points are possible. Just modify this proceedure between intersection points.
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.