# SOLUTION: What are the procedures to write a quadratic equation in vertex form?

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 Question 118862: What are the procedures to write a quadratic equation in vertex form?Found 2 solutions by MathLover1, Edwin McCravy:Answer by MathLover1(6810)   (Show Source): You can put this solution on YOUR website!Follow these steps to convert a parabola from to 1. Factor out an from the first two terms, and leave a space inside. 2. Multiply the coefficient of the term by 3. Square this term and add it to the space we made. 4. Now, we have upset the balance of the equation. We so we have which we need to in order to keep the balance of the equation 5. Now, we can write the part that is in parentheses as This is why we did all of the above steps. 6. Rewrite the terms and simplify a bit. The reason for this will become apparent later on. 7. Now, let’s simply things by calling and Answer by Edwin McCravy(8999)   (Show Source): You can put this solution on YOUR website!What are the procedures to write a quadratic equation in vertex form? ``` First of all, learn what the vertex form is, and how to graph the parabola from that form: The vertex form of y = a(x - h)² + k has vertex (h, k) and goes through the two points, (h-1,k+a) and (h+1,k+a) Example 1: Suppose you started with y = 2x² - 12x + 22 1. Factor the coefficient of x² out of the first two terms, using a bracket, so you can put parentheses inside of it: y = 2[x² - 6x] + 22 2. Out to the side or on scratch paper, complete the square by: (a) Multiplying the coefficient of x by . -6() = -3 (b) Squaring the result of (a). (-3)² = +9 3. Add, then subtract, the result of 2.(b) inside the brackets: y = 2[x² - 6x + 9 - 9] + 22 4. Factor the binomial consisting of the first three terms inside the brackets: y = 2[(x - 3)(x - 3) - 9] + 22 5. If everything has gone right, the two factors of that binomial will be the same, making it a perfect square, so write it as a binomial squared: y = 2[(x - 3)² - 9] + 22 6. Remove the brackets by distributing, remembering to leave the parentheses intact: y = 2(x - 3)² - 18 + 22 7. Combine the last two numerical terms: y = 2(x - 3)² + 4 Compare that to y = a(x - h)² + k Then a = 2, h = 3 and k = 4. So by the above, the graph has vertex (h, k) = (3,4) and it goes through the two points, (h-1,k+a) = (3-1,4+3) = (2,7) and (h+1,k+a) = (3+1,4+3) = (4,7) So plot those three points: and draw a U-shaped graph through them, called a parabola: ---------------------- Example 2: Suppose you started with y = -x² - 8x - 13 y = -(x + 4)² + 3 1. Factor the coefficient of x² out of the first two terms, using a bracket, so you can put parentheses inside of it: y = -1[x² + 8x] - 13 2. Out to the side or on scratch paper, complete the square by: (a) Multiplying the coefficient of x by . 8() = +4 (b) Squaring the result of (a). (+4)² = +16 3. Add, then subtract, the result of 2.(b) inside the brackets: y = -1[x² + 8x + 16 - 16] - 13 4. Factor the binomial consisting of the first three terms inside the brackets: y = -1[(x + 4)(x + 4) - 16] - 13 5. If everything has gone right, the two factors of that binomial will be the same, making it a perfect square, so write it as a binomial squared: y = -1[(x + 4)² - 16] - 13 6. Remove the brackets by distributing, remembering to leave the parentheses intact: y = -1(x + 4)² + 16 - 13 7. Combine the last two numerical terms: y = -1(x + 4)² + 3 Compare that to y = a(x - h)² + k Then a = -1, h = -4 and k = 3. So by the above, the graph has vertex (h, k) = (-4,3) and it goes through the two points, (h-1,k+a) = ( -4-1, 3+(-1) ) = (-5,2) and (h+1,k+a) = ( -4+1, 3+(-1) ) = (-3,2) So plot those three points: and draw a U-shaped graph through them, called a parabola. This is an upside down U graph which will always be the case whenever "a", the coefficient of x² is a negative number: Edwin```