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 Method 1 Converting to Vertex Form Vertex form can be represented as #y=(xh)^2k# where the point #(h,k)# is the vertex To do that, we should complete the square #y=x^22x2# First, we should try to change the last number in a way so we can factor the entire thing #=># we should aim for #y=x^22x1# to make it look like #y=(x1)^2# Complete the square to rewrite the quadratic function in vertex form y = x ^ 2 2x 8 Answers 3 Get Other questions on the subject Mathematics Mathematics, 1530, southerntouch103 Gretchen is setting up for a banquet she has 300 chairs and needs to distribute them evenly among t tables how many chairs should she put atLearn to complete the square in order to write quadratic equations in vertex form Click Create Assignment to assign this modality to your LMS We have a new and improved read on this topic

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Y=x^2+2x-8 in vertex form-Create your account View this answer The given equation is y =x2−2x−8 y = x 2 − 2 x − 8 To convert this into the vertex form, we have to complete the squares Adding 8 8 on both sidesPlug (1,7) into the equation and define b in terms of a math7=ab8\\b=a1/math (0,8) is a point on the parabola, so by symmetry, (2,8) is also a point Plug (2,8) into the equation and define b in terms of a math8=4a2b8\\b=2a\\/math

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Divide 22\sqrt {y} by 2 The equation is now solved Swap sides so that all variable terms are on the left hand side Factor x^ {2}2x1 In general, when x^ {2}bxc is a perfect square, it can always be factored as \left (x\frac {b} {2}\right)^ {2} Take the square root of1) y=(x−17)(x2) 2)y=(x−17)(x−2) 3)y=(x−6)2 2 4) y=(x−6)2 −2 6 Which equation and ordered pair represent the correct vertex form and vertex for j(x) =x2 −12x7?My initial idea was to take the derivative, and set it to zero y = x^2 8x 10 dy/dx = 2x 8 Setting that to zero gives an x value of 4, and plugging x = 4 into the first equation gives y = 6 Thus, the ve

Subtract y from both sides 2x^ {2}8x1y=0 2 x 2 8 x 1 − y = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 2 for a, 8 for b, and 1y for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a} This equation is in standard form a x 2 b x c = 0X 2 2x 8 = 0 Step 2 Parabola, Finding the Vertex 21 Find the Vertex of y = x 22x8 Parabolas have a highest or a lowest point called the Vertex Our parabola opens up and accordingly has a lowest point (AKA absolute minimum)Y=x^2−16x63 The vertex has an xvalue of b/2a=(16)/2=8 f(8)= 1 Therefore, the vertex is at (8,1) The vertex form is (xh)^2 k, where you change the sign of the x value of the vertex (h) and keep the y component (k)

Y = x2 − 2x − 8 y = x 2 2 x 8 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 x 2 − 2 x − 8 x 2 2 x 8 Tap for more steps Use the form a x 2 b xFree factor calculator Factor quadratic equations stepbystepYou can put this solution on YOUR website!

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Statistics Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability MidRange Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution PhysicsIs it ellipse 2 Find the coordinates of the vertex and the equation of the axis of symmetry for the parabola represented by x^2 4x 6y 10 = 0 vertex (2 , 1) axis of Math For each situation first write and equation in the form of y=mx bSolved by pluggable solver Completing the Square to Get a Quadratic into Vertex Form Start with the given equation Subtract from both sides Factor out the leading coefficient Take half of the x coefficient to get (ie ) Now square to get (ie ) Now add and

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 Click here 👆 to get an answer to your question ️ Express the function y = 2x^2 8x 1 in vertex form Explanation y = x2 − 2x 8 or y = (x2 − 2x 1) −1 8 or y = (x − 1)2 7 Comparing with vertex form of equation f (x) = a(x −h)2 k;(h,k) being vertex we find here h = 1,k = 7,a = 1 ∴ Vertex is at (1,7) and vertex form of equation is y = (x − 1)2 7 graph {x^22x8 3554, 3558, 1778, 1778} Ans Answer link Explanation Changing a quadratic function from standard form to vertex form actually requires that we go through the process of completing the square To do this, we need the x2 and x terms only on the right side of the equation y = x2 2x −8 y 8 = x2 2x −8 8 y 8 = x2 2x −8 8 y 8 = x2 2x

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The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction x^ {2}2xy8=0 x 2 2 x − y − 8 = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 2 for b, and 8y for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a}Find the Vertex Form y=x^22x1 Complete the square for Tap for more steps Use the form , to find the values of , , and Consider the vertex form of a parabola Substitute the values of and into the formula Simplify the right side Tap for more steps Cancel the common factor ofY = (x 2 2x 8) y = (x 2 2 ⋅ x ⋅1 1 218) y = ((x1) 29) y = (x1) 2 9 By comparing this with the vertex form of parabola, we get (h, k) ==> (1, 9) Example 6 y = (x1)(x3) Solution y = x 24x3 y = x 22 ⋅ x ⋅2 2 22 2 3 y = (x2) 243 y = (x2) 21 By comparing this with the vertex form of parabola, we get

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👉 Learn how to graph quadratic equations by completing the square A quadratic equation is an equation of the form y = ax^2 bx c, where a, b and c are c y = x² 4x 12 When y = 0 x² 4x 12 = 0 x² 4x 12 = 0 (x 2)(x 6) = 0 x = 2 or x = 6 The xintercepts are 2 and 6 y = (x² 4x) 12 y = (x² Find the vertex of the function y = –x2 2x 8 It is where the derivative 2x 2 = 0 Since you probably haven't learned about derivatives, do it this way Rewrite is as x^2 2x 8 = (x1)^2 7 Now look at the form of the formula on the right It

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To find the x intercepts find the zeros of the equation y=x^22x1 = (x1)^2 hence the two intercepts are (1,0), (1,0) The vertex is located when the derivative of the equation is zero, y'=0 y'=2x2 y'=0=2x2 => 2x=2 or x =1 The vertex is aFree functions vertex calculator find function's vertex stepbystep This website uses cookies to ensure you get the best experience By using this website, you agree to our Cookie PolicyY=2x^212x14 y=x^2x1 Albebra Find the vertex of the function y = –x2 2x 8 It is where the derivative 2x 2 = 0 Since you probably haven't learned about derivatives, do it this way Rewrite is as x^2 2x 8 = (x1)^2 7 Now look at the form of the formula on

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Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreY = x2 ­ 2x ­ 8 Convert to vertex form by completing the square y = 2x2 ­ 12x 6 Equivalent Equations ­ May be asked which equation has the same solution as the equation in the question Convert the quadratic to standard form y = ­(x 2)2 ­ 7 Convert the quadratic to standard formVertex form of a quadratic equation is y=a(xh) 2 k, where (h,k) is the vertex of the parabola;

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In vertex form as 1) f(x) =(3x6)2 −25 2) f(x) =3(x6)2 −25 3) f(x) =3(x2)2 −1 4) f(x) =3(x2)2 7 5 Which equation is equivalent to y−34 =x(x−12)? the equation of a parabola in vertex form is ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ 2 2y = a(x − h)2 k2 2 ∣∣ ∣ −−−−−−−−−−−−−−−−−−−−− where (h,k) are the coordinates of the vertex and a is a multiplier given the parabola in standard form;y = ax2 bx c then the xcoordinate of the vertex is ∙Algebra Find the Vertex Form y=x^22x8 y = x2 − 2x − 8 y = x 2 2 x 8 Complete the square for x2 −2x−8 x 2 2 x 8 Tap for more steps Use the form a x 2 b x c a x 2 b x c, to find the values of a a, b b, and c c a = 1, b = − 2, c = − 8 a = 1, b = 2, c =

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21 Terms charityjohnson5 Algebra 1 Graphing Quadratics (Vertex & Standard Form) Find the vertex of y = 3 (x 9)^2 1 Find the vertex of y = 1 (x 9)^2 1 Find the vertex of y = 4 (x 7)^2 8 Find the vertex of y = 4 (x 8)^2 7When graphing the parabola in standard form, to find the vertex we use the following method The xvalue of the vertex is calculated as x = b/(2a) The yvalue of the vertex is f(b/(2a)) The xvalue of the vertex also represents the horizontal shift, while the yThe quadratic formula gives two solutions, one when ± is addition and one when it is subtraction 2x^ {2}6xy8=0 2 x 2 6 x − y − 8 = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 2 for a, 6 for b, and 8y for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a}

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Divide 0 0 by − 8 8 Multiply − 1 1 by 0 0 Add 8 8 and 0 0 Substitute the values of a a, d d, and e e into the vertex form a ( x d) 2 e a ( x d) 2 e Set y y equal to the new right side Use the vertex form, y = a(x−h)2 k y = a ( x h) 2 k, to determine the values of a a, h h, and k kHow do i convert quadratic functions from standard to vertex form?Parabola, Finding the Vertex 21 Find the Vertex of y = x 214x58 Parabolas have a highest or a lowest point called the Vertex Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero)

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We want to put it into vertex form y=a (xh) 2 k We can convert to vertex form by completing the square on the right hand side Adding 18 to both sides gives us a perfect square trinomial on the right The vertex of a quadratic equation in vertex form is (h,k), so our vertex is (3,22) The equation is y = x 2 2x 2 To change the expression (x 2 2x) into a perfect square trinomial add (half the x coefficient)² to each side of the expression Here x coefficient = 2 So, (half the x coefficient) 2 = (2/2) 2 = 1 Add and subtract 1 to the expression y = (x 2 2x 2 1 1) y = (x 2 2x 1 2 1) y = (x 2 2x 1 3) y = (x 2 2(1)(x) 1 ²) 3The two answers below are perfect;

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 Vertex Form Of A Quadratic The vertex form of a quadratic is given by y = a(x – h) 2 k, where (h, k) is the vertexThe "a" in the vertex form is the same "a" as in y = ax 2 bx c (that is, both a's have exactly the same value) The sign on "a" tells you whether the quadratic opens up or opens downThink of it this way A positive "a" draws a smiley, and a negativeThe vertex form is a special form of a quadratic function From the vertex form, it is easily visible where the maximum or minimum point (the vertex) of the parabola is The number in brackets gives (trouble spot up to the sign!) the xcoordinate of the vertex, the number at the end of the form gives the ycoordinate This means If the vertex form is , then the vertex is at (hk) How to put a function into vertex form?1 Identify the conic section represented by 9y^24x^2 108y24x= 144?

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But the equation for a parabola can also be written in "vertex form" y = a ( x − h) 2 k In this equation, the vertex of the parabola is the point ( h, k) You can see how this relates to the standard equation by multiplying it out y = a ( x − h) ( x − h) k y = a x 2 − 2 a h x a h 2 k This means that in the standard form, yThe vertex of an equation in vertex form is (h,k), which for our equation is (6,4) Notes Right now our quadratic equation, y=x 2 12x32 is in standard form xintercepts x^2 8x 12 = 0 (x 6)(x 2) = 0 x 6 = 0 or x 2 = 0 x = 6 or x = 2 Vertex The formula for the parabola with vertex (h,k) is y = a(x h)^2 k Complete the square y = x^2 8x 12 = x^2 8x 16 12 16 = (x 4)^2

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 which is in vertex form The graph of this equation is a parabola that opens upward It is translated 2 units to the left and 3 units upward This is the advantage of vertex form The transformations required to draw the graph of the function are easy to spot when the equation is written in vertex formVertex Form Equation & Functions Just as y = mx b is a useful format for graphing linear functions, y = a(x h)^2 k is a useful format for graphing quadratic functions If the equation has the form y = ax 2 bx c, the xcoordinate of the vertex is x = b/(2a) For the problem y = x 2 2x 8, a = 1, b = 2, and c = 8, so, the xcoordinate of the vertex is x = 2/2 or 1

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Select a few x values, and plug them into the equation to find the corresponding y values The x values should be selected around the vertex Tap for more steps Replace the variable x with 0 in the expression f ( 0) = − ( 0) 2 2 ( 0) − 8 Simplify the resultThe vertex of a parabola is the point at the top or bottom of the parabola 'h' is 6, the first coordinate in the vertex 'k' is 4, the

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