Understanding Quadratic Functions and Their Graphs

1. Lesson 1-7 `Quadratic Functions and their GraphsObject. To define and graph quadratic functions . Axis If a graph has an axis of symmetry when fold the graph along the axis the two halves coincide Vertex Y = ax2+bx + c a> 0

2. The vertex of the parabola is the point where the axis of symmetry intersects the parabola. Y = -3x2 The bigger |a | is the narrower the parabola becomes, the smaller |a| becomes the wider the parabola becomes. The y intercept of y = ax2+ bx +c Is the value of c y = .25x2 Y = ax2+bx+c a < 0 Y = ax2+bx+c a > 0

3. The x – intercepts are the real roots of ax2+bx+c =0. Since the quadratic equation may have two, one,or no real roots. Depending on the value of the discriminant b2 – 4ac we have three possibilities as shown below If b2 – 4ac <0 there are no x –intercepts (0,c) (0,c) If b2 – 4ac >0 There are 2 X- intercepts If b2 – 4ac =0 there is only one x –intercept where the vertex and the x axis meet.

4. The axis of symmetry is a vertical line halfway between the x-intercepts. It passes through the vertex.The equation of the axis is x = - b 2a Example1: Sketch the graph of y=2x2 - 8x + 5. Label the intercepts, axis of symmetry, and the vertex. Step 1: to find the vertex let x =0 y = 2(0)2 -8(0) +5 = y=0+5 So y = 5 Step 2: To find the x- intercepts solve 2x2 -8x+5 = 0 A=2 , B= -8, C = 5

5. Step 2: To find the x- intercepts solve 2x2 -8x+5 = 0 A=2 , B= -8, C = 5 Use X X X

6. The axis of symmetry is the line x = - b = - ( -8 ) = - -8 = 22a 2(2) 4 Axis of Symmetry x = 2 Step 4:Since the vertex is on the axis of symmetry its x value is 2. To find the y value substitute the x with 2 Y = 2(2)2-8(2)+5 Y= 2(4) – 16 +5 = 8+5 – 16 = 13 – 16 y =-3 The vertex (2,-3) (0,5) (4,5) (3,2) (.8,0) If the equation is written in the form of y = a(x –h)2 + k Then the vertex is (h,k) The vertex of the parabola y = 2(x – 3) 2 +7 is (3,7) The vertex of the parabola y = -4(x – 9) 2 +2 is (-9,2) (2,-3)

7. Example .2: Find the vertex of the parabola y = -2x2 +12x + 4 by completing The square. b. Find the x and y intercepts y = -2x2 +12x + 4 = -2(x2 - 6x ) + 4 = -2(x2 - 6x +9) + 4 – (-2)(9) = -2 (x -3)2 + 22 The vertex is (3,22) b.) When x = 0 then y = 4 thus the y intercept is 4 which is the constant term of the equation. To find the x intercept let y = 0 Using the completed square form of the equation -2 (x -3)2 + 22 = 0 -2 (x -3)2 = -22 -----Divide by -2 (x -3)2 = 11 ------take the square root of both sides

9. Example 3: Where does the line y = 2x + 5 intersect the parabola 8 – x2 Set 2x + 5 equal to 8 – x2 and solve for x 2x + 5 = 8 – x2 x2 + 2x - 3 = 0 (x + 3)(x – 1) = 0 x + 3 = 0 or x – 1 = 0 x = -3 x = 1 Substitute x = -3 into the equation y = 2x + 5 to get y = -1 and 1 into the equation to get y = 7. The intersection points are (-3,-1) and (1,7) Use a computer or graphing calculator to graph the equation y=2x + 5 and 8 – x2

10. Example 4: Find an equation of the function whose graph is a parabola with x-intercepts 1 and 4 and y- intercept -8 Any parabola with x-intercepts 1 and 4 has an equation of the form y=a(x-1)(x-4) for some constant a. We use the fact that the parabola contains (0,-8) to find the value of a. -8 = A(0 – 1)(0 – 4) = (-1)(-4) -8 = 4A -2 = A Then the equation is y = -2(x -1)(x-4) or y = -2x2+10x -8

11. Homework 1-26 pg. 41

12. Videos http://www.youtube.com/watch?v=c74m88mbi9k http://www.youtube.com/watch?v=laogqUmKHX0 http://www.youtube.com/watch?v=vIXuqhBI2iM http://www.youtube.com/watch?v=mVIFVXD_z8M&feature=relmfu