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5.2 Properties of Parabolas. Practice: without graphing, find the minimum or maximum value of the graph. Y=3x 2 + 12x + 8. Hint: -b/2a. Practice: without graphing, find the minimum or maximum value of the graph. Y=-1/4x 2 + 2x - 3. Hint: -b/2a. (4, 1). Practice: Graph y = 2x 2 -4.
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Practice: without graphing, find the minimum or maximum value of the graph • Y=3x2 + 12x + 8 Hint: -b/2a
Practice: without graphing, find the minimum or maximum value of the graph • Y=-1/4x2 + 2x - 3 Hint: -b/2a (4, 1)
Practice: Graph y = 2x2-4 Where is the axis of symmetry? (0,-4) Where does the graph hit the x axis? (Let y=0) + or - Squrt(2) = 1.4, -1.4
Practice: Graph a function of the From y=ax2 + bx +cy = x2 -2x -3 Where is the vertex? (-b/2a) (1,4) Where does the graph cross the y axis? (0,-3) Where does the graph cross the x axis? (let y=0) (-1, 0) and (3, 0)
Real-World Connections • The number of items a company sells frequently is a function of the item’s price. The revenue from sales of the item is the product of the price and the number sold. The number of bicycles that can be sold is modeled by the equation -2.5b + 500. What price will maximize the company’s revenue from bicycle sales? What is the maximum value? Define variables: let’s let R = revenue and b be the price of a bicycle Let (-2.5b + 500) be the number of bicycles sold R = b (-2.5b + 500) R = -2.5b2 + 500b
R = -2.5b2 + 500b • Since a < 0, we have a “sad graph”. Find the value of b at the vertex. This will give you the y value. (hint: -b/2a) B = 100 A price of $100 will gain a maximum revenue of $25,000 Plug into the revenue equation to find the maximum revenue R = $25,000
Introduction to Graphing Vertex Form y = a(x-h)² + k • HAPPY or SAD ? 2 VERTEX = ( h , k ) • T- Chart • Axis of Symmetry
GRAPHING - - Vertex Form (y = a(x-h)² + k ) y = (x+5)² + 2 1) It is happy because a>0 • FIND VERTEX • a =1 h = -5 k = 2 • So V = (h , k) = (-5,2) 3) T-CHART X Y = (x+5)² + 2 -4 • Y = (-4+5)² + 2 = 3 -3 • Y = (-3+5)² + 2 = 6