Free Parabola calculator Calculate parabola foci, vertices, axis and directrix stepbystepGraph y=4(x2)^21 Find the properties of the given parabola Tap for more steps Use the vertex form, , to determine the values of , , and Since the value of is positive, the parabola opens up Opens Up Find the vertex Find , the distance from the vertex to the focus Tap for more stepsProfessionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music WolframAlpha brings expertlevel knowledge and

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Y=4 x^2 y=2-x x=-1 x=1-Jul 22, 18Expression #=x^4y^4# Recall the factorization of the difference of two squares #a^2b^2 = (ab)(ab)# In our example, we will use this factorization twice Note #x^4 =(x^2)^2 and y^4 =(y^2)^2 # Applying the factorization above Expression #= (x^2y^2)(x^2y^2)# Now, the second factor above is also the difference of two squaresY4=2 (x1) Simple and best practice solution for y4=2 (x1) equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it



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About x = 1Find the volume of the solid obtained by rotating theregion bounded by the given curves about the specified line Sketchthe reJul 02, 16Consider the small element width dx as shown, being revolved around the line x = 1 The cross sectional area (csa) of the washer of width dx is the csa of the outer circle minus the csa of the inner ie #dA = pi (1 x dx)^2 pi (1 x)^2# simplify the algebra with #u = 1 x# so we have #dA = pi ((udx)^2 u^2) = pi (2u dx dx^2)#The vertex for this parabola is \displaystyle{\left({1},{1}\right)} Explanation This is an equation for a parabola \displaystyle{y}={3}{\left({x}{1}\right)}^{{2
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Graph x^2y^2=1 Find the standard form of the hyperbola Tap for more steps Flip the sign on each term of the equation so the term on the right side is positive Simplify each term in the equation in order to set the right side equal to The standard form of an ellipse or hyperbola requires the right side of the equation beManipulate c in (x^2 x y y^2) 1 = c;Simple and best practice solution for 2(2xy)=2(xy)4 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework



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Solve by Substitution // Solve equation 2 for the variable y 2 y = 2x 1 // Plug this in for variable y in equation 1 1 (2x1) 4x = 3 1 2x = 4 // Solve equation 1 for the variable x 1 2x = 4 1 x = 2 // By now we know this much y = 2x1 x = 2Step by step solution of a set of 2, 3 or 4 Linear Equations using the Substitution Method y=2x4;y=x1 Tiger Algebra SolverFeb 28, 16See the explanantion This is the equation of a circle with its centre at the origin Think of the axis as the sides of a triangle with the Hypotenuse being the line from the centre to the point on the circle By using Pythagoras you would end up with the equation given where the 4 is in fact r^2 To obtain the plot points manipulate the equation as below Given x^2y^2=r^2 >



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Graph y=2(x1)^24 Find the properties of the given parabola Tap for more steps Use the vertex form, , to determine the values of , , and Since the value of is positive, the parabola opens up Opens Up Find the vertex Find , the distance from the vertex to the focus Tap for more stepsMar , 12/3 cubic units The area we seek lies between the two curves First we need to find the points of intersection between these 4x^2=2x1 x^22x3=0 (x1)(x3)=0=>x=3 and x=1 Interval 3,1 Testing a value in this interval with our two functions, say x=0 4(0)^2=4 2(0)1=1 So, 4x^2 attains greater values and is therefore above 2x1 in this interval We can therefore findResultant((x^2 y^2) 1, ((x 2)^2 (y 1)^2) 4, x) solve (x 2)^2 (y 1)^2 = 4 for x



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