How do we get the Turning Point of a curve that has the equation : x^2-4x+1 ??
437876
Take the derivative:\n\nf'(x) = 2x - 4\n\nSet the derivative equal to 0, and solve for x:\n\n0 = 2x - 4\nx = 2\n\nSub this value of x into the original equation to find the turning point:\n\nf(2) = (2)^2 - 4(2) + 1\nf(2) = -3\n\nSo, your turning point is (2, -3).\n\nAnd, since the x-squared term in the equation is positive, you know that this point is a minimum.\n\nDone and done.
856434
First, look at where the curve intersects the line:\n\ncurve y=1/x\nline y=x/4\n\n1/x=x/4\nx=+/- 2\n\nNote that the curve is symmetrical about the origin for x&y positive and x&y negative, so the line intersects twice.\n\nThat's actually somewhat irrelevant, but it helped me visualize the line and the curve.\n\nTake the first derivative of the curve\ndy/dx=-1/x^2\nThis is the instantaneous slope of the curve at any point, which is the tangent by formal definition.\n\nA line perpendicular to the line y=.25x will have a slope of -4.\nSo set the derivative equal to -4 and solve for x\n-4=-1/x^2\nx=+/- 1/2\nsolve for y\nand find the points\n(1/2,2)\n(-1/2,-2)\nThis is where the two lines that are both tangent to the curve and perpendicular to the line intersect the line.\n\none line is y=-4x+4\nthe other is y=-4x -4\n\nj
910766
The only interesting point in this curve will be the inflexion point which you find by doing the derivative twice and equating to zero.\n\nf'(x) = 3x^2 + 1\n\nf"(x) = 6x = 0 <===> x = 0\n\nf(0) = -1\n\nThe curve will have an inflexion point at (0,-1).\n\nf(0) < 0, f(1) > 0 then:\n\nThe curve will also have a single real root in [0,1] (x = 0.6823)
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The graph of y = 1/x^2 can be graphed on a calculator and has even symmetry with the left half looking a lot like the left half of the hyperbola y = 1/x except it cuts upward more sharply.\n\nx = 1/y^2 is just this function's inverse which is found by exchanging the positive x and y axes and the negative x and y axes. \n\nImagine that you have a transparency with the graph on it with your right hand holding the top of the sheet and your left hand holding the bottom of the sheet. You flip the page over in such a way that your right hand is on the right side of the sheet and your left hand is on the left side of the sheet.\n\nSince the original graph has even symmetry, this process is also equivalent to rotating the graph of y = 1/x^2 clockwise by 90 degrees.\n\nSo it's the same kind of curved triangular pattern except about the x-axis rather than the y-axis.
776765
x^2 + (y-5/4)^2 = 1\n\nFirst, the circle must be on the y-axis by symmetry as zanti3 noted.\nSo the equation is in the form x^2 + (y-a)^2 = 1\nTo find the points of intersection with the parabola, y=x^2,\ncombine the equations thusly: \ny + (y-a)^2 = 1\ny + y^2 - 2ay + a^2 = 1\ny^2 + (1-2a)y + (a^2-1) = 0\nor \ny = [2a - 1 ± √(5-4a)] / 2\n\nNow, there is the easy intuitive way to proceed, and the brute force method to proceed.\nThe easy intuitive way is to note that the two values of y represent the intersection points of the circle and parabola.  But by symmetry (and also that the circle is tangent, not twice intersecting for (positive) x), we know that these two y-values must be the same!\nTherefore, \n [2a - 1 + √(5-4a)] / 2 =  [2a - 1 - √(5-4a)] / 2\nor \n√(5-4a) = 0\n5 - 4a = 0\na = 5/4; thus x^2 + (y-5/4)^2 = 1 is the equation!!!\n{(√3/2, 3/4) and (-√3/2, 3/4) are the points of intersection. Note that these points satisfy both equations.}\nFor a graph, see:\nhttp://mudandmuck.com/str2/circle-parabola.JPG\nNote that the circle is indeed "inscribed".\n\nOne brute force method is to take the derivitive of both functions, set them equal to each other and then substitute\ny = [2a - 1 ± √(5-4a)] / 2 for y.  You will get the same answer,\nas follows:\n\n(1) 2x + 2yy' - 2ay' = 0\n2x + y'(2y - 2a) = 0\ny' = 2x/(2a - 2y)\nand\n(2) y' = 2x\nso at the intersection points,\n2x = 2x/(2a - 2y)\n2a - 2y = 1\ny = a - 1/2\n\ncombined with:\ny = [2a - 1 ± √(5-4a)] / 2\na - 1/2 = [2a - 1 ± √(5-4a)] / 2\n2a - 1 = 2a - 1 ± √(5-4a)\n± √(5-4a) = 0\n5 - 4a = 0\na = 5/4
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find the first derivative for the curve, and then substitute the x coordinate for the slope at that point.\n\nfor example, for the graph y = x^2, the first derivative is 2x. to find the slope at where x = 1, sub x = 1 into 2x and the slope at that point is 2.\n\nalternatively, draw a tangent to the point where u want to find the point slope and the slope of the tangent is the point slope. note that this method's accuracy depends on how accurately you draw your tangent
700193
y=1/2x^2 is a "happy" curve with turning point at (0,0)\n\ny=x^2-5x+4 can be written as:\ny=(x-5/2)^2 - 9/4\n\nThis is a "happy" curve with turning point at (-5/2, -9/4)
233874
There are two methods of solving this question.  One method is applying using the calculus, and other method is using Coordinate geometry.  Since u have not mentioned a particular method, easier way is applying the Calculus.\n\nLet (x1, y1) be a point on the curve.\nDifferentiating the given equation to find the slope of the tagent at (x1, y1), we get\ndy/dx = 6x -5\nAt (x1,y1) dy/dx = 6x1-5\n\nGive condition is the slope of the normal at this pt is = 2\nSo the slope of the tangent = -1/2\n\nTherefore, 6x1 - 5 = -1/2 which leads to x1 = 3/4\nSubstituting this value in the given equation we get,\ny1 = 3.(3/4)^2 - 5.(3/4)+1 \nThat is y1 = -17/16\n\nAs asked the abscissa is 3/4
817524
For a quadratic equation, the maximum or minimum will be the vertex. Since the first term of the quadratic is positive, the parabola goes up, so you're looking for the minimum.\n\nThere are 2 methods:\nCalculus (using derivatives):\nf(x) = x2 - 2x - 4\nf'(x) = 2x - 2\nthe maximum/minimum will be where f'(x) = 0.\n2x - 2 = 0\nx = 1\ny = 1^2 - 2*1 - 4 = -5\n\nAlgebra:\nThe vertex can be found using the formula:\n(-b/2a , -(b^2 - 4ac)/4a)\n(-(-2)/2, -(4 - (4*1*-4))/4)\n(1,-5)\n\nCheck: x = 1, y = 1^2 - 2*1 - 4 = -5
737893
Take your quadratic equation.\n\nThe graph cuts the x axis when y = 0\n\nSo simply put your equation = 0, and then solve it either by using factorising or the quadratic formula:\n\n(-b plus or minus sqrt(b^2 +4ac)/2a)\n\nif I remember right, (could be minus 4ac)
41570
A parabola with its base at 0,1.  Go to http://www.coolmath.com/graphit/ Enter x^2+1 and see it for yourself!  :)
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(x^2+4x+1)/(x-2) = x+2+3/(x-2)\n\nx  : x^2+4x+1-x^2-2x =2x+1\n2  : 2x+1 -2x-4 =3
325867
First, I'll assume you mean that this is an ellipse (parenthesis are important):\n((y+2)^2)/25 + ((x-3)^2)/16 = 1\n\nTo graph by hand, start at the origin.  \n1)The center is shifted- to find it's location, set (y+2) = 0 and (x-3) = 0, and solve for x and y. The center is therefore at (-2,3). \n\n2)The half length of the axis (the lines bisecting the ellipse) are given by the denominators of the x and y terms.  To find the length of each axis, take the square root of each denominator.  The "radius" in the y (vertical) direction is sqrt(25) = 5 units long, and in the x direction it is sqrt(16) = 4 units long.  \n\n3)Starting at the center point of the ellipseee, (-2,3), count up half the length of the axis in the y direction, or 5 units, and put a point.  Do the same thing down, and then repeat with the x axis, counting left and right 4 units, and putting points on your graph.  Draw an ellips through these 4 points, and you're done! \n\nThe major axis is therefore the y axis, and is 10 units long, and then minor axis is the x axis, which is 8 units long.\n\nIf you want to graph this on a calculator, you must solve for y:\n((y+2)^2)/25 + ((x-3)^2)/16 -((x-3)^2)/16= 1 -((x-3)^2)/16\n\n(((y+2)^2)/25)*25 = (1 -((x-3)^2)/16)*25\n\n((y+2)^2)^(1/2) = ((1 -((x-3)^2)/16)*25)^(1/2)\n\nNote that raising to the (1/2) power is the same as taking the square root. When you take the square root of a real number, your answer could be negative or positive, so you have 2 solutions, one for the bottom of the ellips, and one for the top:\n\n(y+2) = (((1 -((x-3)^2)/16)*25)^(1/2))\nand\n(y+2) = -(((1 -((x-3)^2)/16)*25)^(1/2))\n\nThe final equations you'd enter into your calculator are:\n\ny = (((1 -((x-3)^2)/16)*25)^(1/2))-2\nand\ny = -(((1 -((x-3)^2)/16)*25)^(1/2))-2
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i don't fucking know.\n\nEDIT: waitaminnit i'm gonna give it a try!\n\nx^3-4x^2-7x+10 = 0\n\nfor x = -1\n-1 -1 +7+10 != 0\nso x != -1\n\nfor x = 1\n1-1-7+10 != 0\nso x != 1 either!! ...goshdarnit.\n\nx^3-4x^2-7x+10 = 0\nx(x^2-4x) - 7x = -10\nx^2-4x - 7 = -10/x (we divided both sides by x)\n..ok...i think i've lost the plot sorry.
900114
hi bob!\nDo you mean ellipse? If so then its equation is: (x/a)^2+(y/b)^2=1, a=4, b=3;\nHammer nails in ellipse focuses with coordinates [-f 0] & [+f 0], now attach the rope ends to these nails, let the rope length be L, keep the rope taught with pencil and draw.\nLet y=0, then x=a (or x=-a which does not matter here), then L=(a+f)+(a-f)=2*a\nLet x=0, then y=b, then f^2+b^2=(L/2)^2, f=sqrt(a*2-b^2)
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Find the coordinates of the points where the straight line meets 'y=1' meets the circle 'x^2+y^2-14x+2y+45=0'.\n\n'x^2+1^2-14x+2(1)+45=0\nx^2 -14x +48 = 0\nthe solutions are:\nx=(14 +- sqrt( 14^2 -4(48) ) /2\nx=14+- sqrt(4) )/2\nx= (14 +2)/2 or x= (14-2)/2\nx=8 or x = 6\n.
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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.