Forms of Quadratic Equations A quadratic function can be expressed in three formats: Standard (or general or expanded) form: Factored form: , where and are the roots of the quadratic equation Vertex form: , where h and k are the x and y coordinates of the vertex, respectively. For example: From: http://www.windberschools.org/153720101122139497/lib/153720101122139497/SMP08ALG_NA_TE2_C12_L03_12.pdf Converting Between Forms of Quadratic Equations To convert the standard (expanded) form to factored form, one could: 1. A "quadratic" is a polynomial that looks like "ax2 + bx + c", where "a", "b", and "c" are just numbers. For the easy case of factoring, you will find two numbers that will not only multiply to equal the constant term " c", but also add up to equal "b", the coefficient on the x-term. For instance: Factor x2 + 5x + 6. I need to find factors of 6 that add up to 5. Since 6 can be written as the product of 2 and 3, and since 2 + 3 = 5, then I'll use 2 and 3. I know from multiplying polynomials that this quadratic is formed from multiplying two factors of the form "(x + m)(x + n)", for some numbers m and n. So I'll draw my parentheses, with an "x" in the front of each: (x )(x ) Then I'll write in the two numbers that I found above: (x + 2)(x + 3) This is the answer: x2 + 5x + 6 = (x + 2)(x + 3) This is how all of the "easy" quadratics will work: you will find factors of the constant term that add up to the middle term, and use these factors to fill in your parentheses. See lots of examples at http://www.purplemath.com/modules/factquad.htm This becomes slightly more complicated when the “ a” coefficient is not 1. To factor a "hard" quadratic, we have to handle all three coefficients, not just the two we handled in the "easy" case, because the leading coefficient (the number on the x2 term) is not 1. The first step in factoring will be to multiply "a" and "c"; then we'll need to find factors of the product "ac" that add up to "b". Factor 2x2 + x – 6. Looking at this quadratic, I have a = 2, b = 1, and c = –6, so ac = (2)(–6) = –12. So I need to find factors of –12 that add up to +1. The pairs of factors for 12 are 1 and 12, 2 and 6, and 3 and 4. Since –12 is negative, I need one factor to be positive and the other to be negative (because positve times negative is negative). This means that I'll want to use the pair "3 and 4", and I'll want the 3 to be negative, because –3 + 4 = +1. Now that I have found my factors, I will use what my students refer to as "box": I will draw a two-by-two grid, putting the first term in the upper left-hand corner and the last term in the lower right-hand corner, like this: Then I will take my factors –3 and 4 and put them, complete with their signs and variables, in the diagonal corners, like this: (It doesn't matter which way you do the diagonal entries; the answer will work out the same either way!) Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved Then I'll factor the rows and columns like this: from the top row from the bottom row from the left column from the right column (Note: The signs for the bottom-row entry and the right-column entry come from the closest term that you are factoring from. Do not forget your signs!) Now that I have factored the box, I can read off my answer from across the top and along the left-hand side: 2x2 + x – 6 = (2x – 3)(x + 2). 2. use the quadratic formula to determine the two roots (more about this later) To convert the factored form (or vertex form) to standard form, one needs to multiply, expand and/or distribute the factors. For example: Vertex to standard: Factored to standard: use an area diagram with the factors as dimensions To convert the standard form to vertex form, one needs a process called completing the square. Completing the Square Say we have a simple expression like x2 + bx. Having x twice in the same expression can make life hard. What can we do? Well, with a little inspiration from Geometry we can convert it, like this: As you can see x2 + bx can be rearranged nearly into a square ... ... and we can complete the square with (b/2)2 In Algebra it looks like this: x2 + bx + (b/2)2 "Complete the Square" = (x+b/2)2 So, by adding (b/2)2 we can complete the square. And (x+b/2)2 has x only once, which is easier to use. Keeping the Balance Now ... you can't just add (b/2)2 without also subtracting it too! Otherwise the whole value would change. So I will show you how to do it properly with an example: Start with: ("b" is 6 in this case) Complete the Square: Also subtract the new term Simplify it and we are done. The result: x2 + 6x + 7 = (x+3)2 - 2 And now x only appears once, and your job is done! From: http://www.mathsisfun.com/alge bra/completing-square.html Examples: Completing the Square when you’re a coefficient (in standard form) is 1. Completing the Square when you’re a coefficient (in standard form) is not 1. Here are the steps for solving 2x2 + 8x +11 2x2 + 8x +11=0 Original equation set to 0 (set this to zero since the y value of 0 is where the xintercepts/solutions are found) 2 2x + 8x= -11 Subtract 11 from both sides 2 2(x + 4x)= -11 Factor out the 2 from the coefficients from the left side of the equation 2(x2 + 4x + 4) = -11+8 Divide the b coefficient by 2 and square it to find the c that makes perfect square trinomial (4/2 = 2; 22=4). Since you added the 4 x2 to left side of the equation to make it a perfect square trinomial, you need to add 4 x2 to the right side of the equation also to keep the equation balanced 2(x+2)2= -3 Simplify the right side of the equation, express the perfect square trinomial on the left as a squared binomial 2 −3 Solve for x. First divide both sides by 2. (x+2) = 2 Take the square root of both sides −3 x+2=√ 2 −3 𝑥 = −2√ 2 Subtract 2 from both sides. These roots do not represent real numbers because the number under the square roots sign is negative. Since every real number is on the x-axis, this is not. Therefore, this equation does not have roots (x-intercepts) and so has no solution. Quadratic Functions What information you have: 1. How to find the 2. solutio ns (also called roots or xintercepts)? 1. Use the Zero Product Property to find the two roots. 1. Find the midpoint of the roots (which will be the xcoordinate of the vertex), plug x back into the original equation to solve for y. Then you have your (x,y) for the vertex. 2. Force the standard form to become a perfect square trinomial (vertex form) by “completing the square”. Solve using square roots. Set f(x) or y equal to zero since y = 0 at the xintercepts. Then solve using square roots. 3. 4. 1. How to find the Graph the standard form equation in the graphing calc and zoom in on the x-intercepts with the table or graph. Use the quadratic equation: Force the standard form to become a perfect square trinomial (vertex form) by “completing the square”. Then look for the vertex in the equation (where h and k are the x and y coordinates of the vertex, respectively). vertex (also called maximum or minimum)? Change the standard form equation into factored form by finding a factorable trinomial = 2. Graph the standard form equation in the graphing calc and zoom in on the vertex with the table or graph. 3. Change to factored form (see above), find the midpoint of the roots (which will be the xcoordinate of the vertex), plug x back into the original equation to solve for y. 1. Look for the vertex in the equation: where h and k are the x and y coordinates of the vertex, respectively. Practice Problems 1. Given the following quadratic Function (given in both standard and factored form), find the coordinates of the vertex and determine if there is a minimum or a maximum. = (x+1) (x+3) 6. Rewrite 7. in general form. Answers