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Published byJared Johnson Modified over 8 years ago
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Quadratic Functions Lesson 3.3
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Quadratic Function Degree 2 Parabola shaped Can open upward or downward Always has a vertex which is either the max or min Always has exactly one y-intercept Can have 0, 1, or 2 x-intercepts Is symmetric about a line through the vertex called the axis of symmetry
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Forms of Quadratic Function Transformation form: f(x) = a(x – h) 2 + k f(x) = a(x – h) 2 + k Polynomial form: f(x) = ax 2 + bx + c f(x) = ax 2 + bx + c x-Intercept form: f(x) = a(x – s)(x – t) f(x) = a(x – s)(x – t) If a is positive, the graph opens upward
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Transformation {f(x) = a(x – h) 2 + k} Vertex (h, k) x-Intercepts y-intercept ah 2 + k
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Polynomial {f(x) = ax 2 + bx + c}
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x-intercept {f(x) = a(x – s)(x – t)}
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Changing to Polynomial and x-intercept form F(x) = 0.4(x – 3) 2 + 2 .4(x 2 – 6x + 9) + 2 .4(x 2 – 6x + 9) + 2.4x 2 – 2.4x + 5.6 {Polynomial}.4x 2 – 2.4x + 5.6 {Polynomial} b 2 – 4ac = -3.2 so there are no b 2 – 4ac = -3.2 so there are no x-intercepts x-intercepts g(x) = 3x 2 – 3.9x – 43.2 3(x 2 – 1.3x – 14.4) 3(x 2 – 1.3x – 14.4) Quadratic formula step Quadratic formula step 3(x – 4.5)(x + 3.2) 3(x – 4.5)(x + 3.2)
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Changing to Transformation Form f(x) = -3x 2 + 4x – 1 g(x) = 0.3(x -2)(x + 1) Lets look at pg. 168-169 for the work that is done There is also quite a lovely summary of the previous 7 slides on page 169!
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Application Example The owner of a concession stand can sell 150 lunches per day at a price of #3 each. The cost to the owner is $2.25 per lunch. Each $0.25 price increase decreases sales by 30 lunches per day. What price should be charged to maximize profit?
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