 Page Four If the quadratic equation does not have a left or right shift, it will look like these: The property that all of these functions share is that there is no x term in any of them. It's the x term in the middle that causes the graph to shift left or right. For example: y = x2 + 8x + 16 will be shifted left or right because of the centre term +8x. The problem is that, in this form of the function, we don't know the shift. The function will have to be factored. In addition, the factoring must be a perfect square, so that there are two identical factors: y = x2 + 8x + 16 y = (x + 4)(x + 4) y = (x + 4)2 y = +1(x + 4)2 + 0 which opens up, has steepness factor 1, and is shifted left 4, with no up or down shift Unfortunately, most quadratic functions aren't perfect squares! What are we to do with this example below, which doesn't factor at all, let alone as a perfect square? y = x2 + 6x + 1 What we'll do is force it to become a perfect square! y = x2 + 6x + 1 needs a new third term. We're going to move the 1 out of the way and insert a third term that will make this function a perfect square: y = x2 + 6x + ☐ + 1 - ☐ A number will go in the box to make y = x2 + 6x + ☐ a perfect square. The number must also be subtracted later, to keep the total value of the function the same. The number you add in the box is half the x coefficient, squared y = x2 + 6x + ☐ + 1 - ☐ y = x2 + 6x + 9 + 1 - 9 because half of 6 is 3, and 3 squared gives you 9 Now you can factor the first three terms as a perfect square: y = (x + 3)2 + 1 - 9 y = (x + 3)2 - 8 y = +1(x + 3)2 - 8 which opens up, has steepness factor 1, and after shift left 3 and down 8 has vertex (-3, -8) Let's try another one: y = x2 - 2x - 3 which again is not a perfect square. We'll add in our own third term: y = x2 - 2x + ☐ - 3 - ☐ The box must be half of -2, squared, which is (-1)2 or 1 y = x2 - 2x + 1 - 3 - 1 y = (x - 1)2 - 3 - 1 y = +1(x - 1)2 - 4 which opens up, has shape number 1, and vertex (1, -4) Incidentally, the number you add in is always positive, since it has been squared. Let's introduce a complication: suppose there is a coefficient in front: y = 3x2 - 12x + 7 Remember, we're going to be making the first two terms into a perfect square. Start by factoring out the 3 from just the first two terms: y = 3x2 - 12x + 7 y = 3(x2 - 4x) + 7 Now we're going to add in or own third term inside the brackets: y = 3(x2 - 4x + ☐) + 7 - 3☐ Notice that to keep things balanced, we had to subtract three of the boxes, because the 3 in front of the bracket means that we really added in 3 boxes! Since half of -4 is -2, which when squared gives 4, the next line looks like: y = 3(x2 - 4x + 4) + 7 - 3·4 and now we can factor and simplify: y = 3(x - 2)2 - 5 remember order of operations! y = +3(x - 2)2 - 5 which opens upward much more steeply, with vertex (2, -5) The next example: what happens if the first coefficient won't factor out? Here's how completing the square differs from actual factoring. Not only do we just use the first two terms, but the first coefficient must be factored, even if you get decimals! y = 2x2 - 7x + 3 y = 2(x2 -3.5x) + 3 Now add in the new number, and subtract two of them: y = 2(x2 -3.5x + ☐) + 3 -2☐ You'll need a calculator: half of -3.5 is -1.75; square this to get 3.0625 y = 2(x2 - 3.5x + 3.0625) + 3 -2·3.0625 You may remember that to factor a perfect square, the number in the bracket is the square root of the last term. The square root of 3.0625 is 1.75 y = 2(x - 1.75)2 - 3.125 This opens up, not quite as steeply as the last one, and has vertex (1.75, -3.125) Its axis of symmetry is x = 1.75. The domain is x ε R. The range is y≧-3.125 Another example, with a twist: y = -2x2 + 10x - 1 y = -2(x2 - 5x + ☐) - 1 +2·☐ Notice that because the -2 in front made the box we added really -2☐, in order to keep everyting balanced we had to ADD 2☐ Half of -5 is -2.5, which when squared is 6.25 y = -2(x2 - 5x + 6.25) - 1 +2·6.25 y = -2(x - 2.5)2 + 11.5 This opens downwards, with the same steepness as the previous example, with vertex (2.5, 11.5) |