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Section 3.6 Quadratic Equations Objectives 1. Find the vertex and standard equation of a parabola 2. Sketch the graph of a parabola 3. Solve applied problem involving maximum or minimum Best way to view this presentation is thru power point XP

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Definition of a quadratic Function: A function f is a quadratic function if f(x) = ax 2 + bx + c Where a, b and c are constants with a ≠ 0 (x,y)yx (-2, -2)-2 (-1, - ½ )- ½ (0, 0 )00 (1, -1/2 )-1/21 (2, - 2 )- 22 Table of variation Solution: Example 1. Sketch the graph of f if (a)f( x ) = x 2 Note: b = c = 0

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Note: Only b = 0 Solution: It is enough to shift the previous graph by 4 units upward as we can see below

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Expressing a Quadratic Equation as f(x) = a ( x – h ) 2 + k Method 1. B y completing the square of the right hand side of the quadratic equation and it will be explained in class. Method 2. ( Zalzali’s Method ) Step 1. Let h = Step 2. Let k = Step 3. Insert the values of h and k in f ( x ) = a( x – h ) 2 + k

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Example 3. Write the following quadratic function in the standard form f(x) = a ( x –h )2 )2 + k Solution: Then f(x) = 3 ( x + 4 ) 2 +2 Class Work 1 Write the following quadratic function f (x) = -x 2 -2x +8 in the standard form f(x) = a ( x –h ) 2 + k Answer: f(x) = - ( x + 1 ) 2 +9

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Standard Equation of a parabola with vertical Axis The graph of the equation f(x) = a ( x –h ) 2 + k For a ≠ 0 is a parabola with vertex V(h, k ) and it has a vertical axis x = h. The parabola opens upward if a > 0 and The parabola opens downward if a < 0. Notes about Vertex V ( h, k ) Note 1. If a < 0, then the vertex V ( h, k ) is a maximum point of the parabola Note 2. If a > 0, then the vertex V ( h, k ) is a minimum point of the parabola

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Graphing Parabolas Strategy: 1. Find the vertex V ( h, k ) 2. Identify if the vertex is maximum or minimum 3. Find x and y intercepts if they exist 4. Plot the vertex and the intercepts 5. Plot the vertical axis ( it is also called axis of symmetry of a parabola ) 6. Connect the points of the parabola and extend the graph Example 4: Graph the parabola Solution: 1. Vertex V ( -1, 9) Check class practice 2. a = -1 < 0 ( Parabola is open downward ) and has a maximum at the vertex 3. x-intercept(s): Set y = 0, then x =-4 and 2 Therefore, points are ( -4, 0 ) and ( 2, 0 ) y-intercept : Set x = 0, then y = 8. Point ( 0, 8 ) x = -1 f(x) = - x 2 – 2x + 9

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Class Work 2 Graph the parabola

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Word Problem Height of a projectile: An object is projected vertically upward from the top of a building with an initial velocity of 144 ft / sec. Its distance s( t ) = - 16t t (a) Find its maximum distance above the ground (b) Find the height of the building. Solution: Height =s(0)

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Homework of section 3.6 Do all homework problems in the syllabus

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