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1 MA 1165 - Lecture 03 1/21/09 We’ve seen that if we have a quadratic function of the form f(x) = a(x − d)2 + c, then we can tell what the graph is going to look like pretty easily. The a is a stretching and flipping factor, and the vertex will be at (d, c). Now, all we have to do is to get any quadratic function into that form. This turns out to be fairly straightforward. 1 Perfect Square Trinomials Recall that when factoring polynomials, if the two factors are the same, we get a special form. Specifically, (x + a)2 = (x + a)(x + a) = x2 + ax + ax + a2 , and (x + a)2 = x2 + 2ax + a2 . If the x-coefficient is equal to the double the square root of the constant term, then we have a perfect square trinomial. This holds as long as the x2 -coefficient is 1 and the constant term is positive. It’s OK for the middle term to be negative, since (x − a)2 = x2 − 2ax + a2 . Let’s look at a few examples. If we square x + 3, we get (x + 3)2 = x2 + 2 · 3x + 32 = x2 + 6x + 9. If we had started with x2 + 6x + 9, we would look at the 9, its square root is 3, and double 3 is 6. Since that matches the x-coefficient, we have a perfect square trinomial, and it factors as (x + 3)2 . Now, consider x2 − 4x + 4. The constant term is positive 4. Good. Its square root is 2, and double 2 is 4, which matches the x-coefficient. Good. This factors as x2 − 4x + 4 = (x − 2)2 , and it’s “minus 2,” because the x-term is negative. 2 Quiz 03-A Factor the following perfect square trinomials. 1. x2 + 2x + 1 2. x2 − 2x + 1 3. x2 − 6x + 9 4. x2 + 10x + 25 5. x2 − 16x + 64 2 3 COMPLETING THE SQUARE 3 Completing the Square Given a quadratic function of the form (specifically with an x2 -coefficient of 1) f(x) = x2 + 8x + 7, we can complete the square as follows. First, move the 7 to the side. f(x) = x2 + 8x +7 The constant term of a perfect square trinomial would have been half-the-x-term squared. In this case, half of 8 is 4, and that squared is 16. We wanted 16. We’ll put that in, but also subtract it, so we won’t change the function. f(x) = x2 + 8x + 16 − 16 + 7. The first three terms now make a perfect square trinomial, which factors as (x + 4)2 , and −16 + 7 = −9. We have then, f(x) = (x + 4)2 − 9. Now, the (x + 4)2 is a square, so it can’t be negative. The smallest it can be is 0, and this happens at x = −4. Then we’re going to subtract 9. The smallest f(x) can be is −9. The graph of a quadratic function is always a parabola, and here the lowest point is at (−4, −9). It makes sense, therefore, that the graph must look like it does in Figure 1. Figure 1: If the quadratic function has an x2 -coefficient of something other than 1, just factor it out. For example, if we had f(x) = −3x2 + 18x + 9, then factor out the −3 (this may be messy, but the process is exactly the same no matter how bad the numbers look). f(x) = −3 x2 − 6x −3 . Half the middle term squared is 9, so f(x) = −3 x2 − 6x + 9 − 9 − 3 . We get f(x) = −3 (x − 3)2 − 12 . If we want, we can change it to f(x) = −3(x − 3)2 + 48. Since the coefficient of the squared part is negative, the parabola opens down. The biggest −3(x − 3)2 can be is 0, and this happens when x = 3. Then we add 48, so the highest point on the parabola is (3, 48). Part of this graph is shown in Figure 2. 3 4 HOMEWORK 03 Figure 2: 3.1 Quiz 03-B Complete the square on each function. Give coordinates of the vertex (the highest or lowest point on the graph) and whether the parabola opens upward or downward. 1. f(x) = x2 − 4x + 5. 2. f(x) = x2 + 2x − 3. 3. f(x) = 2x2 + 4x − 6 (Hint: Factor the 2 out first). 4 Homework 03 Factor the following perfect square trinomials. 1. x2 − 2x + 1 2. x2 + 8x + 16 3. x2 − 10x + 25 4. x2 − 20x + 100 5. x2 + 4x + 4 Complete the square on each function. Give the coordinates of the vertex (the highest or lowest point on the graph) and whether the parabola opens upward or downward. 6. f(x) = x2 − 6x + 5. 7. f(x) = x2 + 8x − 3. 8. f(x) = x2 − 10x + 2. 9. f(x) = −2x2 + 12x − 6. (Hint: Factor out the −2 first.) 10. f(x) = x2 + 3x + 4. (You’ll get fractions, but that’s OK. Enter your answer like (-3/5,2/3)