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Completing the square Consider the parabola y = x 2 4x +1 . 4 y 3 Complete the square on the quadratic: 2 y = x 2 4x +1 1 = (x 4x) +1 2 2 1 2 () –1 2 1 0 2 3 4 5 x –1 –2 = (x 4x + 4) +1 4 –3 = (x 2) 3 –4 2 2 (Notice how four was added inside the parentheses to complete the square, and subtracted outside to maintain the equality.) Now compare y = (x 2)2 3 with the vertex of the parabola shown in the graph. Can you see a relationship between the two? For a parabola written in the form y = (x - h)2 + k the vertex has coordinates (h, k ). h k vertex (h, k ) The parabola y = (x 2)2 3 has vertex (2, –3). Immediately, you also know that the axis of symmetry of the parabola is x = 2 . Why is this so? Part 1 Parabolas 1 You know that the parabola y = (x h)2 is the same as the parabola y = x 2 shifted h 5 units horizontally. 3 4 2 So y = (x 2)2 is shifted two units to the 1 right, having vertex (2, 0). And y = (x h)2 + k is the parabola y = (x h)2 shifted q units vertically. So y = (x 2)2 3 is y = (x 2)2 shifted y –1 0 1 2 3 4 5 x –1 –2 –3 –4 three units downwards. It has vertex (2, – 3). By writing a parabola in the form y = (x h)2 + k you can quickly determine the coordinates of the vertex. This is just tying together information you learned previously. Follow through the steps in this example. Do your own working in the margin if you wish. Write y = 4 2x x 2 in the form y = (x h)2 + k and hence determine the vertex of this parabola. Draw a sketch of this curve showing the main features. Solution y = 4 2x x 2 y = [x 2 + 2x 4] = [(x 2 + 2x) 4] = [(x 2 + 2x +1) 4 1] (completing the square) = [(x +1)2 5] = (x +1)2 + 5 The vertex is (–1, 5). (The minus in front of the squared term indicates the parabola is facing concave down.) 2 PAS5.3.4 Coordinate geometry From the original y = 4 2x x 2 the y-intercept is 4. You now have enough information to draw a sketch. y 5 4 3 2 1 –4 –3 –2 –1 0 1 2 x –1 –2 In this question you weren’t asked to calculate the x-intercepts. However from the sketch you can see they lie somewhere between –4 and –3, and between 1 and 2. If you needed their values, you could start with y = (x +1)2 + 5 . (x +1)2 + 5 = 0 (x +1)2 = 5 x +1 = ± 5 x = 1± 5 (exact value) x = 3.24or1.24(correct to 2 dec. pl.) Part 1 Parabolas 3 Activity – Completing the square Try these. 1 Write the vertex to y = (x + 7)2 + 5 . _________________________ 2 The parabola y = (x h)2 + k has vertex (–2, 1). What are the values of h and k? _______________________________________ Write y = x 2 6x 3 in the form y = (x h)2 + k and hence 3 determine the vertex of this parabola. _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ _______________________________________________________ Draw a sketch of this curve showing the main features. 3 y 2 1 –1 0 –1 1 2 3 4 5 6 7 8x –2 –3 –4 –5 –6 –7 –8 –9 –10 –11 –12 –13 –14 4 PAS5.3.4 Coordinate geometry