How can I solve for the area under the parabola?
85251
The area under any function can be got by the integration from an initial point to a final point\n\nfor example y(x)=x^2 and we are intrested to solve the area between x=1 to x=3\n\nthe integration = x^3/3 from 1 to 3\nthe area = (3^3/3) -(1^3/3) = 9 - 1/3 = 8.666667
342990
Take the derivative of each equation. Subtract one from the other. That'll get you the area enclosed between the two.
117044
Area equals half the base times the height.  \nA=1/2bh.  Plug in and solve!
133564
The coordinates of the vertex of a parabola \ny = ax^2 + bx + c is\n[-b/2a , (4ac - b^2)/4a]\nJust substitute and you will get\nV(2 , -7)
619069
the parabola will be at the point 2,2 and shaped like a "u"
53126
area = 1/2 x base x height
738071
I assume you want to find the area between y = 0, y = x^3, x = 1, and x = 2.\n\nSince you're trying to find the area under a curve between 2 points on the x axis, just take the integral of the function from x = 1 to x = 2.\nintegral of x^3 = 1/4 * x^4\nf(1) = 1/4 * 1^4 = 1/4\nf(2) = 1/4 * 2^4 = 16/4 = 4\n\nf(2) - f(1) = 3 3/4 (solution!)
592617
the area = (1/2)*(4+x)*(16-(1/4)*(x-4)^2)^(1/2).  you now take the derivative and set it equal to zero and solve for x.  this will give you the length of the forth side that would mazimize the area.  the derivative = (32+4x-x^2)/(2*(48+8x-x^2)^(1/2)) .  when you set this to zero and solve for x, you get that x=8....I got the derivative by applying the product rule of differentiation.  the math is a bit to long to type out in my answer
716583
If you know the vertex then you know\n\na (x-xv)^2=y-yv  , use P to find a, the shape \n\nat P:\na(xp-3)^2 =yp-1\na(-1-3)^2 = 3-1\n16a = 2\na=1/8\n\nSo, the parabola is \n\ny = 1/8*(x-3)^2 + 1\n\nCheck: \nat V  y= 1 ok\nat P y = 1/8(-4)^2 + 1 = 16/8+1 = 3 ok
1076159
Hi:\n\nthere are 6 formulas for a parabola :\n\n1) x^2+y^2= e^2(d+x)^2    ( explain what the variables are below\n as long as e=1 it is a parabola \n\n\n2) r = de/(1-e cos(thea))\n\n3) (y-k)^2=4a(x-h)  make a Y parabola\n\n4)  (x-h)^2=4a(y-k) make a X parabola\n\n5)     Ax^2 + bx+ C= y  ( as long as A does not equal zero)\n\n6 )        A( x-b)^2 + c= y\n\n\n x = x corr.,  y= y corr,  k =distance from y axis,  h= distance from x axis,  d= is the distance from focus to directrix, e = eccentricity  r = radius
801034
Here's the answer: (I'm assuming the top of the box is counted in the area)---------   x = side of square base, h = height, S = surface area, P = perimeter length\n\nS = 2x^2 + 4xh\n\nP = 8x + 4h ---->  h = (P-8x)/4  Substituting,\n\nS = 2x^2 + 4*x*(P-8x)/4 = Px - 6x^2\n\nFor your particular case where P = 120 cm,\n\nS = (120x - 6x^2) cm^2
892318
Take the right triangle to be arranged in the first quadrant.\n\nThe hypotenuse forms the line y = (-4/3)x + 4\n\nTake the dimensions of the rectangle to be x and y, subject to the above constraint.\n\nWe want to maximize A = xy.\n\nWe've solved for y above. This gives:\n\nA(x) = x[(-4/3)x + 4]\nA(x) = -(4/3)x^2 + 4x\n\nIf you're not taking calculus, graph this function and find the x that maximizes the function. It should be at the vertex of a parabola that points down - a clear indication that it is the maximum.\n\nIf you're taking calculus, find A'(x) and set it equal to 0.\n\nA'(x) = (-8/3)x + 4 = 0\n\nx = 4*(3/8) = 1.5\ny = (-4/3)* 1.5 + 4 = 2\n\nThe rectangle measures 1.5 x 2 cm, and has an area of 3 cm^2.
422511
if u kno the base and height.. then area= (base x height)/2\n\nif the triangle is scalar,\nthen :\n\nfirst find s, using hero's formula\n\ns= (a+b+c)/2   .... where a,b,c are the sides of triangle.\n\nthen , area = squareroot(s x (s-a) x (s-b) x (s-c)
600230
Do you mean (x-h)^2? If so, yes, this complies with the general equation of a parabola y = ax^2+c. \nThe value of y is proportional to the square on x, that has to be a parabola.
434174
If the equation is y = 4x^2 + 4x - 5, then:\n\nRewrite the equation and complete the sqaure on the right side of the equation...\n\ny = 4(x^2 + x) - 5\n\ny = 4(x^2 + x + 1/4) - 5 -4(1/4)\n\ny = 4(x + 1/2)^2 - 5 - 1\n\ny = 4(x + 1/2)^2 - 6\n\ny - (-6) = 4(x + 1/2)^2\n\ny - (-6) = 4[x - (-1/2)]^2\n\nVertex for standard equation is in the form:\n\ny - k = +/- a(x - h)^2\nwhere:\nh = x-coordinate of the vertex.\nk = y-coordinate of the vertex.\n+ a = the parabola opens upward.\n- a = the parabola opens downward\na = the parabola will have a wide opening if a > 1.\na = the parabola will have a narrow opening if a < 1.\n\nTherefore vertex is at (-1/2, -6) and a wider opening which is upward.
935174
Let the length be x ft and the width be y ft\n\n2x + 2y = 300\nie x + y = 150\nThus y = 150 - x\n\nArea = xy\n=x(150 - x)\n= 150x - x²\n\nVertex of this parabola occurs when x = -b/2a and at this point Area is maximum as the parabola is concave down. (a < 0)\n\nie x = (-150)/(-2)\n= 75\n\nWhen x = 75 y = 75 so it should be square with sides 75 ft\n\nWhen x = 75, Area = 150*75 - 75² = 5625 sq ft\n\nSo the maximum area occurs when the length (and width) are 75 ft and it equals 5625 sq ft
243669
divde the area by pi, then take the square root.  Multiply that reult by 2 and then by pi and your done.