[コンプリート！] y=x^2 2x-8 in vertex form 452525-Y=x^2+2x-8 in vertex form

Write the following quadratics in vertex form by completing the square and find the vertex Example 1 y = x 2 2x 4 Solution Using completing the square method, y = (x 2 2xSubtract 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}} {2a} This equation is x = b/2a = 4/4 = 1 y (1) = 4 vertex at (1 , 4) Edit or in vertex form y = 2 (x 1)^2 4 vertical shift down 4 units hope that helped acobdarfq and 8 more users found this answer helpful heart outlined

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Y=x^2+2x-8 in vertex form- While the standard quadratic form is a x 2 b x c = y, the vertex form of a quadratic equation is y = a ( x − h) 2 k In both forms, y is the y coordinate, x is the x coordinate, and a is the constant that tells you whether theRegistered From Posted Saturday 30th of Dec 0933 Hi, I am a senior in high school and need major help in convert to vertex form calculator My math grades are awful and I have decided to do something about it I am looking for some software that will allow me to enter a question and gives detailed step by step solution

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Start with the given quadratic Subtract from both sides Factor out the leading coefficient This step is important since we want the coefficient to be 1 Take half of the x coefficient to get (ie )We can convert to vertex form by completing the square on the right hand side; y 9/8 = 2(x 2 3/2 x 9/16) y 9/8 = 2(x 3/4) 2 y = 2(x 3/4) 2 9/8 The above equation represent the standard form of the equation of parabola The standard form of the parabola with vertex (h, k) and axis of symmetry x = h is y = a(x h) 2 k The vertex form of the equation of parabola is y = 2(x 3/4) 2 9/8 and Vertex (h

 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 Substituting this value back into the equation y = (1) 2 2(1) 8 = 9How to Graph a Parabola of the Form {eq}y=x^2 bx c {/eq} Example 1 Our quadratic equation is {eq}y = x^2 2x 3 {/eq} Step 1 First we need to find the vertex of our parabola The vertex algebra determine the vertex for the function y=2x^25x3 for y = ax^2 bx c, the x value of the vertex is b/ (2a) sub that x into your equation to find the y Another way is to complete the square so for yours, x = 5/4 etc a third way, if you know Calculus, asked by Shay on Math

Our perfect square trinomial factors into two identical binomials, (x6)•(x6) The vertex of an equation in vertex form10) Use your calculator to find the vertex, xintercepts, and yintercept Do they match yours?Regents Exam Questions FIFC8 Vertex Form of a Quadratic Name _____ wwwjmaporg 1 FIFC8 Vertex Form of a Quadratic 1 In the 8 Determine and state the vertex of f(x) =x2 −2x−8 using the method of completing the square 9 Use the method of completing the square to determine the vertex of f(x) =x2 −14x−15 State

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The 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 putSee the answer See the answer See the answer done loadingAnswer to Solved 2) y=x2 – 2x 5 2) y=x2 2x 5 Vertex Form of This problem has been solved!

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C = −4 vertex f orm y = a(x−h)2 k where h = −b 2a and k = −(b2−4ac) 4a h =− −14 2⋅2 = −7 2 k = −(142−4⋅2⋅(−4)) 4⋅2 = −() 8 = −228 8 = −57 2 Answer y =2(x 7 2)2 − 57 2 y = 2 x 2 14 x − 4Free 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 Policy3 must be factored from 3x2 9x x must be factored from 3x2 9x 9 must be factored from 9x 18 3 must be factored from 3x2 18 A Which phrase best describes the translation from the graph y = 6x2 to the graph of y = 6 (x 1)2?

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H)2 k Question Updated PM 1 Answer/Comment yeswey The minimum value of a parabola that opens upward will be its vertex TRUE Added PMComplete the statements below that show y = 8x2 32x 17 being converted to vertex form Factor out the leading coefficient y = 8(x2 4x) 17 1 Form a perfectsquare trinomial y = 8(x2 4x ____) 17 ____ 2 Write in vertex form y = 8(x ____)2 ____Convert to vertex form y=2x^214x4 Convert to vertex form y=2x^214x4 y =2x2 14x −4 a =2;

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Question Write y=2x^212x14 in vertex form Answer by jim_thompson5910 () ( Show Source ) You can put this solution on YOUR website!In comparing the graphs of y = x 2 (red), y = 2x 2 (green), and y = 4x 2 (blue), we see that each parabola opens upward but the larger the value of "a", the steeper (narrower) the graph Thus, when a ³ 1, the parabola opens upward, and as the value of 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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Y = 3(x ) 2 y = 3(x ) 2 This is our equation in vertex form, which tells us that the vertex is at (, ), and also that our parabola opens upwards, since a (3 in this case) is positive To convert from vertex form to standard form, we simply expand vertex form We can confirm that our above equation in vertex form is the same as the original equation in standard form by expanding it yTo find the vertex form of the parabola, we use the concept completing the square method Vertex form of a quadratic function y = a(x h) 2 k In order to find the maximum or minimum value of quadratic function, we have to convert the given quadratic equation in the above formY = a (xh)^2 k is the vertex form equation Now expand the square and simplify You should get y = a (x^2 2hx h^2) k Multiply by the coefficient of a and get y = ax^2 2ahx ah^2 k This is standard form of a quadratic equation, with the normal a, b and c in ax^2 bx c equaling a, 2ah and ah^2 k, respectively

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The vertex form of a quadratic is given by y = a(x – h) 2 k, where (h, k) is the vertex The "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 down 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) 3 Suppose a parabola has a vertex (4,7) and also passes through the point (3,8) Write the equation of the parabola in vertex form f(x)=a(xh)^2k I believe h=4 k=7 Not sure what to do from here Math

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 Methods for solving quadratic equations We find the vertex of a quadratic equation with the following steps Get the equation in the form y = ax2 bx c Calculate b / 2a This is the xcoordinate of the vertex To find the ycoordinate of the vertex, simply plug the value of b / 2a into the equation for x and solve for yY = 2 (x 9) 2 8 is the given equation in vertex form The vertex is (− 9, 8)5 hours ago Substitute (3, 8) for ( x, y) in the vertex form to find a The function of the parabola is 62/87,21 From the figure, the vertex ( h, k) of the parabola is (±3, 2) Substitute ( ±1, 8) for ( x, y) in the vertex form to find a The function of the parabola is Write each function in

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Which equation is y=3(x2)^2(x5)^2 rewritten in vertex form?None y = 3(x 2)² (x 5)² y = 3(x 2)(x 2) (x 5)(x 5)6 unit left 6 unit right 1 unit leftExample 13 Rewrite in y = a (x − h) 2 k form and determine the vertex y = 2 x 2 − 4 x 8 Solution Since a = 2, factor this out of the first two terms in order to complete the square Leave room inside the parentheses to add a constant term

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Section 111 Graphs and Vertex Form In this section, we will explore quadratic functions and, in particular, their vertex and factored forms Example 1111 Maia has a remotecontrolled airplane and she is going to do a stunt dive where the plane divesIn this section, we will explore quadratic functions using graphing technology and learn the vertex and factored forms of a quadratic function's formulaGive the vertex of each function, and graph it How does vertex form compare to the other forms in each problem?

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Question Rewrite the equation y=–2x^24x3 in vertex form Identify the vertex and the axis of symmetry Answer by ewatrrr(243) (Show Source) You can put this solution on YOUR website!We want to put it into vertex form y=a(xh) 2 k;Subtract 3 from both sides x^ {2}2x=y3 Multiply 2 and 1 to get 2 x^ {2}2x1=y31 Divide 2, the coefficient of the x term, by 2 to get 1 Then add the square of 1 to both sides of the equation This step makes the left hand side of the equation a perfect square x^ {2}2x1=y2 Add

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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)To Convert from f (x) = ax 2 bx c Form to Vertex Form Method 1 Completing the Square To convert a quadratic from y = ax 2 bx c form to vertex form, y = a(x h) 2 k, you use the process of completing the square Let's see an example Convert y = 2x 2 4x 5 into vertex form, and state the vertex The result is the same as 1 * 1, which is 1 Next, simplify 2 * 1 The result is 2 Then, add 1 and 2 The answer will be 1 Finally, add 1 and 8 to get the ycoordinate 7 The ordered pair of the vertex is (1,7) Hope this helps!

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Answer Vertex form of quadratic equation is => Y = a(X h)^2 k Let's simplify the given equation Add and subtract coefficient of X^2 on left side Y = 4X^2 8X 4 4 3 => Y = 4(X^2 2X 1) 4 3 => Y = 4(X 1)^2 1 So vertex of this parabola is V = (1,1)Following quadratic functions in the vertex form y = a(x — k Identify the turning point of the quadratic from this form State whether it is a maximum or a minimum l y=x22x 15 2 y = xž 32 3 y 6x 5 4 y = £ 8x 40 qo8\ y = 16K 130 6 75 7 lox 24 Q 8 y 32x 16 21 T I 10 y = —S ( (00 80 ll y = 4x2 56x qVertex form is y = 2 (x 6)^2 – 13 Standard form of given equation is y = 2 x^2 24 x 59 Where, Characteristic Points are Vertex P (6, 13) Yintercept P (0, 59) An online parabola vertex calculator can display a parabola graph with exact values when you substitute the same values for a vertex form equation

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Hi the vertex form of a parabola, where(h,k) is the vertex36 is the value for 'c' that we found to make the right hand side a perfect square trinomial;Substitute the value of the x x coordinate in the given function vy = 2(2)2−8(2)4 v y = 2 ( 2) 2 − 8 ( 2) 4 Solving, we obtain the y y coordinate of the vertex of the parabola vy = 2(4

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 But the equation for a parabola can also be written in "vertex form" y = a(x ?Graph y=x^22x8 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 Use the vertex form, , to determine the values of , , and Since the value of is positive, the parabola opens up Opens UpY = 2x2 –4x –1 STEP 2 Find the vertex y x y 2 1 4 1 1 32 Thus the vertex is (1 ,–3) Since the x –value of the vertex is given by the line of symmetry, we

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Section 132 Quadratic Graphs and Vertex Form Objectives PCC Course Content and Outcome Guide MTH 95 CCOG 5b;Y = x2 − 2x − 8 y = x 2 2 x 8 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 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 = 8 Consider the vertex form of a parabola11) y = (x 3)2 2 12) y = (x 4)2 13) y = x2 3 14) y = x2 3 Convert each function to standard form Give the vertex and yintercept

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The ycoordinate of the vertex is $$ y = f(1) = 1^22\cdot12 = 1 2 2 = 1 $$ The yintercept is $$ y = f(0) = 0^22\cdot02 = 0 0 2 = 2 $$ In this case xintercept doesn't exist since equation $x^22x2=0$ does not has the solutions (use quadratic equation solver to check ) So, in this case we will plot the graph using onlyFrom this chart, we see that the parabola y = x 2 contains the points (3, 9) and (4, 16) On the other hand, he parabola y = 2x 2 contains the points (3, 18) and (4, 32) On the first equation, y = x 2, to move horizontally across the xaxis from x = 3 to x = 4, we move up vertically on the yaxis from y = 9 to y = 16 which is 7 unitsSo, to go from the point (3, 9) to (4, 16), we move over 1

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