Find the area between the curves y = x{eq}^2 {/eq} x 1 and y = 2x 7 From x = 2 to x = 3 Area Between Functions A planar region can be bounded by two curves(b) 2 x y dx ( y 2 x 2) dy = 0 Here, M = 2 x y, M y = 2x, N = y 2 x 2, and N x = 2 xNow, ( N x M y) / M = ( 2 x 2 x ) / ( 2 x y) = 2 / yThus, μ = exp ( ∫ 2 dy / y ) = y2 is an integrating factor The transformed equation is ( 2 x / y ) dx ( 1 x 2 y2) dy = 0 Let m = 2 x / y, and n = 1 x 2 y2Then, m y = 2 x y2 = n x, and the new differential equation is exactIf x is real, then the minimum value of y = x^2 x 1 / x^2 x 1 is
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Y=x^2+2x+1 in vertex form-Graph y=x^22x1 Find the properties of the given parabola Tap for more steps Rewrite the equation in vertex form Tap for more steps Complete the square for Tap for more steps Use the form , to find the values of , , and Consider the vertex form of a parabolaHere y=2x/(1x^2) Make sense So multiply both side by (1x^2) a quadritic equation will form in which cofficent of x^2 will be y and that of x will be 2 And constant will be 2 For x to be real D greater than or equal to zero So D=2^2–4y^2≥0 now Y^
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The vertex form of a quadratic function is given by y = a(x − h)2 k, where (h,k) is the vertex of the parabola We can use the process of Completing the Square to get this into the Vertex Form y = x2 − 2x 1Y Cos X Pi 3 2; Solve for y y = 1 ± x 1 To decide which sign you have to choose, you have to look what happens with the initial domain x > 1 implies y > − 1 When you swap x and y, you get x > − 1 (which is obvious) and y > 1 The result is, that you need the sign, in other words y = 1 x 1 is the inverse function Share
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Example 3 y = x 2 3 The "plus 3" means we need to add 3 to all the yvalues that we got for the basic curve y = x 2 The resulting curve is 3 units higher than y = x 2 Note that the vertex of the curve is at (0, 3) on the yaxis Next we see how to move a curve left and right Example 4 y = (x − 1) 2Calculadoras gratuitas passo a passo para álgebra, trigonometria e cálculoY = x^2x dy/dx = 2x 1 For maxima or minima put dy/dx = 0 2x1 = 0 => x = 1/2 d2y/dx^2 = 2 , ve There exist minima at x = 1/2 Minimum value = (1/2)^2 (1/2) = 1/4 1/2 = (1–2)/4 = 1/4 , Answer SecondMethod y = x^2 x or, y = (x)^
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Y=2x1 2xy1=0 y=2x1 y=2x1 Both the equations represent the same line so the system has innumerable solutionsThen, f(x)g(x) = 4x 2 4x 1 = 1 Thus deg( f ⋅ g ) = 0 which is not greater than the degrees of f and g (which each had degree 1) Since the norm function is not defined for the zero element of the ring, we consider the degree of the polynomial f ( x ) = 0 to also be undefined so that it follows the rules of a norm in a Euclidean domainX 2 Y 1 2x 3y K;
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Y 4 1 3 X 2 Graph;This equation is in standard form a x 2 b x c = 0 Substitute 1 for a, 1 for b, and 1 − y for c in the quadratic formula, 2 a − b ± b 2 − 4 a c x=\frac {1±\sqrt {1^ {2}4\left (1y\right)}} {2} x = 2 − 1 ± 1 2 − 4 ( 1 − y) Square 1 Square 1 x=\frac {1±\sqrt {14\left (1y\right)}} {2} y=x^22x1 y3x=5 The pair of points representing the solution set of this system of equations is and Answers 2 Get Other questions on the subject Mathematics Mathematics, 1800, warnene17 What expression is equivalent to 42x56y Answers 1 continue
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The function # y = x^2 2x 1 " is in this form "# with a = 1 , b = 2 and c = 1 the xcoordinate of the vertex can be found as follows xcoord of vertex # = b/(2a )= (2)/2 = 1 # substitute x = 1 into equation to obtain ycoord # y = (1)^2 2(1) 1 = 0 # thus coordinates of vertex = (1 , 0) #""# Alternatively factorise as #y = (x 1 )^2#X 2y 5 Y 2x 2 Graphically;X 2 Y 4 5 X Y 2 7 10;
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y = x2 2x 1The right side is a perfect squarey = (x1)^2 The standard form of the parabola is y k = 4a (xh) ^2 Hence the vertex is at (1,0)Found 2 solutions by mananth, ewatrrr Answer by mananth () ( Show Source ) You can put this solution on YOUR website!Y X 2 Z 2 Graph;
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Graph y=x^22x1 Find the properties of the given parabola Tap for more steps Rewrite the equation in vertex form Tap for more steps Complete the square for Tap for more steps Use the form , to find the values of , , and Consider the vertex form of a parabolaY = x^2, x = y ^2;Algebra > Quadratic Equations and Parabolas > SOLUTION y=x^22x1 graph the quadratic function label the vertex and axis of semitry Log On Quadratics solvers Quadratics
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0 votes 1 answer In the following, determine the set of values of k for which the given quadratic equation has real roots (i) 2x^2 kx 2 = 0 (ii) 3x^2 2x k = 0About x = 1Find the volume of the solid obtained by rotating theregion bounded by the given curves about the specified line Sketchthe reNow, y = x± q x2 −4(x2 −7) 2 = x± √ 28−3x2 2 Take the positive root since y(1) = 3 The restriction on x would be that 28−3x2 ≥ 0 Therefore, − s 28 3 < x < s 28 3 5 Problem 15 (xy2 bx2y)dx(xy)x2 dy = 0 First, for this to be exact
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyPLAY Match Gravity Which ordered pairs are in the solution set of the system of linear inequalities?Swap sides so that all variable terms are on the left hand side x^ {2}2x1=y x 2 2 x 1 = y Subtract y from both sides Subtract y from both sides x^ {2}2x1y=0 x 2 2 x 1 − y = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 2 for b, and 1y for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac
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Root plot for y = x 22x1 Axis of Symmetry (dashed) {x}={ 100} Vertex at {x,y} = { 100,0} x Intercepts (Roots) Root 1 at {x,y} = {041, 000} Root 2 at {x,y} = { 241, 000} Solve Quadratic Equation by Completing The Square 22 Solving x 22x1 = 0 by Completing The Square Add 1 to both side of the equation x 22x = 1Let y = x2 x1x2 −x1 ⇒ yx2 yxy = x2 −x1⇒ (x−1)x2 (y1)x(y −1) = 0If x∈ R, thenDiscriminant ≥ 0⇒ (y1)2 −4(y −1)2 ≥0⇒ −3y2 10y−3 ≥0⇒ 3y2 −10y3 ≤0⇒ (3y −1)(y −3) ≤0⇒ 31 ≤ y ≤ 3∴ Range = 31 ,3 Solve the system of equations algebraically Show all of your steps y=x^22x y=3x Math Use a system of equations to find the quadratic function f(x) = ax^2 bx c that satisfies the equations Solve the system using matrices f(−2) = 4, f(1) = −2, f(2) = −12 The answer i got is f(x)=2x^24x8 this is the
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