consider the algorithm discussed on http://www.cubic.org/docs/hermite.htm\nwhere the curve is modelled using parametric equations.
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If a curve in n dimensions is given parametrically as:\n\n( f1(s), f2(s), ..., fn(s) ) for a <= s <= b\n\nthen you can approximate the arc length by dividing the domain [a,b] into many small pieces a = s0, s1, s2, ..., sk = b and summing the lengths of the chords (secants? straight line segments. whatever) between s_{i} and s_{i+1}.\n\nThe length of the curve is defined as the least upper bound of all such sums (if that set has an upper bound).\n\nUnder suitable hypotheses you can compute this as the definite integral of ``ds'' as s varies from a to b, using the Euclidean rule\n\n(ds)^2 = (df1(s)/ds)^2 + ... + (dfn(s)/ds)^2\n\ni.e., your length is the definite integral from a to b of the square root of the sum of the squares of the derivative of each coordinate function.\n\nFor example for a circle of radius one at the origin parameterized as ( cos(t), sin(t) ) as t varies from a = 0 to b = 2 pi you get ... 2 pi ... which is what you would expect.  :-)  :-)\n\nOr for the arc length of ( (s^3)/3 - s, s^2 ) from s = 5 to s = 8 the length is the integral from 5 to 8 of\n\nsqrt( (deriv (s^3)/3 - s w.r.t. s)^2 + (deriv s^2 w.r.t. s)^2 ) ds\n\n= sqrt( (s^2 - 1)^2 + (2s)^2 ) ds\n= sqrt( s^4 - 2s^2 + 1 + 4s^2 ) ds\n= sqrt( s^4 + 2s^2 + 1 ) ds\n= (s^2 + 1) ds\n\ni.e., integral from s=5 to s=8 of (s^2 + 1) ds.
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The first two answers are completely wrong. The fact that (0, 0) satisfies the equation only means (0, 0) is a point on the parabola, not that it is also the vertex of the parabola. For example, plugging (0, 0) into y = (x - 1)^2 - 1 yields a true equation because the point is on the graph, but the vertex of the parabola is the point (1, -1)!\n\nRemember that if you have a function y = f(x), then y = f(x - k) for some constant k is just the graph of f(x) shifted k units to the right (see what happens to f(k) and f(0) when you subtract k from the argument).\nSimilarly y = f(x) + j will be the graph of f(x) shifted upwards by j units.\nNow, a parabola with vertex at the origin has function f(x) = hx^2 for some scaling constant h. The forms we are looking for are then y = h(x - k)^2 + j where the vertex will have been moved to (k, j). Thus, we just need to adjust our equations to the form above and read (k, j) off of them.\nFor example, your first equation is y = 3x^2 - 4. Adjusted to the form above, we get y = 3(x - 0)^2 + (-4). Thus (k, j) = (0, -4), so this graph's vertex is not at the origin.
470707
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
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a) The curve you need is called a cycloid. Look that up on the web.\n\nb) The parametric form of the cycloid is\n\nx=12(a-sin a)      y=12(1-cos a) where a is then angular displacement. 24pi (circumference) will be the period and it passes through (0,0)
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For a), we are asked to assume this is a parabola, even though on a real bridge the cable does not assume such a shape.\n\nTo solve for k, put in\n\n143 = k * 640^2\nk = 143 / 640^2\n\nDo the computation yourself to at least one decimal place.\n\nFor b), we need to be able to compute the length of a section of a parabola. For that, we need the derivative of the curve,\n\nf'(x) = 2x\n\nThe formula for the length is the integral of \n\nsqrt(1 + 4x^2) \n\nover the desired range. To try to integrate this, we can use trigonometric substitution, defining an angle theta such that \n\n2x = tan(theta)\nx = tan(theta) / 2\n\nNow we take dx/d(theta) = sec^2(theta) / 2\n\nsqrt(1 + 4x^2) = sec^2(theta) / 2
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[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).
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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.
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While Albert has found the correct solution using calculus, the question asked how to find the solution with graphing.  So, if you don't care how you find the answer, use Albert's.  But if you need to graph it to find the solution, you use the function that Albert got V(x)=x(6-2x)(8-2x) and graph the resulting cubic function (explained well by Just Wondering).  You will see a graph that increases from -infinity up to about 24 (on the V), then decreases to about -3.5, and then increases to infinity.  You want to find the x value and V value at that first "hump," or maximum.  If you are using a graphing calculator you can use the "Maximum" function to see that Albert's answer is approximately correct.
505677
I assume this is equal to y?\n\nIf it is, it's definitely a maximum because it's a parabola that opens downward\n\nNow I'll work on the vertex
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f(x,y)= (1-x)^2 + 105 (y- x^2)^2\n\nIt's often used to test out various nonlinear optimization algorithms.
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You can't solve this using traditional techniques since you'll either have a logarithm or an exponential on one of the x variables.  The people claiming Taylor series and techniques like Newton-Raphson approximation are most likely on the right track.  I would use a CAD program, or graph the function y(x)=(1+x)^2-2^x and find the zeroes that way by zooming and such (on at TI-83 or similar).\n\nIn this case, I'll use MathCAD.  The solutions are:\nx={-1.57862, 0, 5.31972} (two of these are approximate, by the way)
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It's curved. Take Newton's second law F=ma, therefore m is proportional to 1/a and therefore the graph will be similar to a y=1/x graph and will be curved.\n\nTake F=ma, so m=F/a, proportionality is about if you double one thing the over doubles, however this is inversely proportional so if you double m you get a half of a, this is assuming that F is a constant.
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D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
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Consider a connection between the polar coordinates of a point and suppose, that connection can be expressed in the form F(r,t)=0 or maybe in the explicit form r = f(t).\nSuch equation is a polar equation of a curve.\n\nWith each solution (ro,to) of the polar equation, corresponds a point with polar coordinates (ro,to).\n\nNow, the cartesian equation of cadiod is (x^2 + y^2 - 2ax)^2 = 4a^2(x^2 + y^2)\n\nSince the cardioid is an epicycloid with one cusp, its parametric equations are\nx(t) = a.cos t (1-cos t)\ny(t) = a.sin t (1-cos t)\n\nand its polar equations are \nr(t) = a(1-cost)\n or \nr(t) = (a/2)(1-cost)
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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 ).
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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).
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the area under the curve f(x)!