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x = h. ( h , k ). ( h , k ). x = h. 3.2 Properties of Quadratic Relations. y = ax 2 + bx + c a > 0. If the second difference is positive, the graph opens up and the graph opens down if the second difference is negative.
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x = h (h, k) (h, k) x = h 3.2 Properties of Quadratic Relations y = ax2+bx+c a > 0 If the second difference is positive, the graph opens up and the graph opens down if the second difference is negative. The axis of symmetry is the vertical line which passes through the vertex. y = ax2+bx+ca < 0 If the coordinates of the vertex is (h, k), the equation of the axis of symmetry is x = h.
The y-coordinate of the vertex is called the optimum value of the relation. (h, k) y = k (h, k) The optimum value is called a minimum if the parabola opens up and a maximum if the parabola opens down.
The axis of symmetry is the perpendicular bisector of any horizontal line segment joining two points on the parabola. If the parabola crosses the x-axis, the x-coordinates are called the zeros or x-intercepts.
(2, – 4) Determine the following: a) the coordinates of the vertex (2, – 4) b) the optimum value – 4 c) the equation of the axis of symmetry. x = 2 d) the zeros of the relation. 0 and 4
Example 1: Sketch the graph of y = 3x2 + 12x Start with a table of values.
• Substitute x = – 2 into the original equation to obtain. the vertex • The zeros are at 0 and – 4. • Axis of symmetry (halfway between the zeros). x = – 2
zeros The vertex is at (–2, -12 ) Axis of symmetry x = – 2 Substitute x = – 2 y = 3x2 + 12x y = 3(– 2)2 + 12(– 2) y = – 12
(3, 2) (5, 2) Ex 2: The following points lie on a parabola. Determine the equation of the axis of symmetry. a) (3, 2) and (5, 2) The axis of symmetry lies halfway between 3 and 5.
(–3.25, –2) (2.5, –2) = – 0.375 The equation of the axis of symmetry is x = – 0.375. The following points lie on a parabola. Determine the equation of the axis of symmetry. b) (–3.25, –2) and (2.5, – 2) The axis of symmetry lies halfway between –3.25 and 2.5.
Properties of Quadratic Relations (2) A golf ball is hit in the air. Its height is given by the equation: h = 50t – 5t2, where h is the height in metres and t is the time in seconds. a) What are the zeros of the relation? b) When does the ball hit the ground? c) What are the coordinates of the vertex? d) Graph the relation. e) What is the maximum height of the golf ball? f) After how many seconds does that occur?
Step 1: Set the WINDOW to the following settings. Reminder (–) a) What are the zeros of the relation? Press WINDOW
GRAPH Press Press and enter the equation Y= Press 2nd TRACE Use arrows to cursor to the left and right of the two x-intercepts (or zeros). × Ø
C:\Documents and Settings\Cheryl Ann\My Documents\MPM 2D1\Unit 3\Golf Example.84state a) What are the zeros of the relation? The zeros are 0 and 10. b) When does the ball hit the ground? The ball hits the ground at 10 seconds.
vertex zeros c) What are the coordinates of the vertex? V(5, 125) d) Graph the relation e) What is the maximum height of the golf ball? 125 m f) After how many seconds does that occur? 5 s
Homework: pg 145 #1 – 7, 9 – 15 (for 12 – 15, graph using TOV)