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Parabola. Examples. Give four examples of equations of parabola all having the vertex (2,3), but one is facing upward, one downward, one to the right and one to the left. Identify the axis of symmetry. Solution. Vertex (2,3). Examples (1) – (4).
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Examples Give four examples of equations of parabola all having the vertex (2,3), but one is facing upward, one downward, one to the right and one to the left. Identify the axis of symmetry.
Examples (1) – (4) Identify the parabola having the given equation and its axis of symmetry and graph it.
Example (1)Graph y=x2 - 6x + 7 The graph is facing upward. Why? y= x2 - 6 x + 7 = (x- 3 )2 – 9 + 7 = (x- 3 )2 – 2 The vertex of the parabola is (3 , - 2 ) The intersection with the y-axis is(0,7) y=x2 - 6x + 7 Why? The intersections with the x-axis: (3+√2 , 0 ) and (3-√2 , 0 ) Why?
Example (2)Graph y=2x2 + 2x + 3 The graph is concave upward. Why? y=2[x2 +x] + 3 = 2[(x+½)2–¼] + 3 = 2(x+½)2–½ + 3 = 2(x+½)2 + 5/2 The vertex of the parabola is ( - ½ , 5/2 ) The intersection with the y-axis is (0,3) Why? No intersections with the x-axis: Why?
Example (3)Graph y=-x2 - 4x + 12 The graph is concave downward. Why? y= -(x2 +4x ) + 12=-[(x+ 2 )2 – 4] + 12 = -(x+2)2 + 16 The vertex of the parabola is (-2 , 16 ) The intersection with the y-axis is (0,12) Why? y=-x2 - 4x + 12=-(x2 +4x - 12)=-(x+6)(x-2) The intersections with the x-axis: (2 , 0 ) and (-6 , 0 ) Why?
QuestionDo it now!Do not see the next slide before you do it! Graph y=x2 + 2x - 3
Solutiony=x2 +2x - 3 The graph is concave upward. Why? y=x2 +2x – 3 = = (x+1)2 – 1 – 3 = (x+1)2 - 4 The vertex of the parabola is (-1 , -4 ) The intersection with the y-axis is (0,-3) y = x2 + 2x - 3=(x+3)(x-1) why? The intersections with the x-axis: (-3 , 0 ) and (1 , 0 ) Why?