Aim: How do transformations affect the equations and graphs of functions? Do Now: Graph y = -.25x2 – 4 and describe some of the important features. Axis of symmetry x = 0 x = 0 Maximum Turning Point (0,-4) Opens down

The absolute value of a determines “fatness”. Dilation of Parabola a > 1 In other words the absolute value of a, the coefficient of x2, is a dilation factor that changes the shape of the parabola a < 1 a = 1 f(x) = ax2 + bx + c The absolute value of a determines “fatness”. As the absolute value of a decreases in value, the shape of the parabola gets fatter or wider. As the absolute value of a increases, the shape of the parabola gets “thinner”. dilation Da

+c -c The Nature of the Parabola A(x, y)  A’(x, y + c) Translation of Parabola: the “c” effect f(x) = ax2 + bx + c y = x2 + 6 (0,6) +c In a quadratic equation “c” tells where the graph crosses the y-axis. y = x2 (0,0) The Nature of the Parabola y = x2 – 3 (0,-3) y = x2 – 5 (0,-5) -c Translation T0,c A(x, y)  A’(x, y + c) f(x)  f(x) + c

Reflection of a parabola On the same set of axes, graph y = .25x2 – 4 and describe your results: Reflection of a parabola Turning point - (0, -4) Axis of symmetry x = 0 y = .25x2 – 4 Observations: (-6,5) (-6,-13) Same turning point & axis of symmetry 9 units (-4,0) (-4,-8) (4,0) (4,-8) other symmetry? 4 units (-2,-3) (-2,-5) y = -4 y = .25x2 – 4 is a reflection of y = -.25x2 – 4 about the horizontal line whose equation is y = -4 y = -.25x2 – 4 Each point on one graph is equidistant from the line of reflection (y = -4) as its image on the other reflected graph.

Graph y = -x2 + 6x – 1 and reflect it about the y-axis Axis of Symmetry x = -6/2(-1) = 3 Table of Values (3,8) (2,7) (4,7) -(0)2 + 6(0) - 1 -1 0,-1 (1,4) (5,4) -(1)2 + 6(1) - 1 4 1,4 -(2)2 + 6(2) - 1 7 2,7 (0,-1) (6,-1) -(3)2 + 6(3) - 1 8 3,8 -(4)2 + 6(4) - 1 7 4,7 -(5)2 + 6(5) - 1 4 5,4 -(6)2 + 6(6) - 1 -1 6,-1 Algebra of Reflection: image of A(x,y) in the y-axis is A’(-x, y) image of B(x,y) in the x-axis is B’(x, -y) To graph: find the coordinates of the image of each of the seven points under a reflection in the y-axis and connect those points with a smooth line.

Graph y = -x2 + 6x – 1 and reflect it about the y-axis Image of A(x, y) in the y-axis is A’(-x, y) f(x)  f(-x) Reflection x = - 3 (5,4) (6,-1) (2,7) (1,4) (0,-1) x = 3 (4,7) (3,8) (-3,8) (x, y)  (-x, y) (-4,7) (-2,7) (0,-1) (0,-1) (-5,4) (-1,4) (1,4) (-1,4) (2,7) (-2,7) (3,8) (-3,8) (-6,-1) (0,-1) (4,7) (-4,7) (5,4) (-5,4) (6,-1) (-6,-1) Note: axis of symmetry is also reflected under the same rules Algebra of Reflection: image of A(x,y) in the y-axis is A’(-x, y) image of B(x,y) in the x-axis is B’(x, -y) To graph: find the coordinates of the image of key points under a reflection in the y-axis and connect those points with a smooth line. What is the equation of the reflected parabola? y = -x2 – 6x – 1

1. 2. 3. 4. Which of the above graphs represent f(x) = x2 + 1? 1 Which represents the image of the parabola f(x) = x2 + 1 under a reflection in the: A. x-axis B. y-axis C. The line y = x D. The line y = -x 3 1 2 4

Graph the translation of y = -x2 + 6x – 1 under a translation that maps (x, y)  (x – 2, y + 1) The image of every point on y = -x2 + 6x – 1 is two units to the left and one unit up. (x, y)  (x – 2, y + 1) (-1,5)  (-2,0) (1,9) (0,-1) (1,4) (2,7) (3,8) (4,7) (5,4) (6,-1) (-2, 0) (-1, 5) (0, 8) (1, 9) (2, 8) (3, 5) (4, 0) (3,8) (4,7) (5,4) (1,4) (0,-1) (6,-1) Algebra of Translation: image of A(x, y)  A’(x + a, y + b); a is the change in horizontal unit distance and b is the change in vertical unit distance To graph: find the coordinates of the image and connect those points with a smooth line.

Vertex Form of Equations of Parabola Graph the following parabola on the same set of axes: y = x2 y = x2 – 5 y = (x + 3)2 - 5 y = (x – 5)2 - 5 y = (x + 7)2 + 1 Describe your findings:

Vertex Form of Equations of Parabola y = (x + 7)2 +1 Vertex Form of Equations of Parabola f(x) = x2 y = (x + 3)2 - 5 y = (x - 5)2 – 5 f(x) = x2 – 5 Trans. Rule? A(x, y)  A’(x , y – 5) f(x) = x2 f(x) = (x + 3)2 – 5 A(x, y)  A’(x – 3, y – 5) f(x) = (x – 5)2 – 5 A(x, y)  A’(x + 5, y – 5) y = x2 – 5 f(x) = (x + 7)2 + 1 A(x, y)  A’(x – 7, y + 1) f(x) = (x – h)2 + k represents a horizontal translation of f(x) = x2, h units to the right if h is positive or h units to left if h is negative and a vertical translation of k units. The coordinate (h, k) is the turning point of the parabola.

Aim: How do transformations affect the equations and graphs of functions? Do Now: Write the equation and sketch the graph of y = x3 after a transformation that translates it 3 units up. …… after a transformation that translates it 3 to the left. …… after both: a transformation that translates it 3 units up and 3 to the left.

Vertical & Horizontal Translations If k and h are positive numbers and f(x) is a function, then f(x) + k shifts f(x) up k units f(x) – k shifts f(x) down k units f(x – h) shifts f(x) right h units f(x + h) shifts f(x) left h units f(x) = (x – h)2 + k - quadratic f(x) = |x – h| + k - absolute value f(x) = (x – h)3 + k - cubic f(x) = (x – h)4 + k - quartic or 4th degree ex. f(x) = (x – 4)2 + 4 is the image of g(x) = x2 after a shift of 4 units to the right and four units up or a translation of T4,4.

Reflections of Functions image of B(x,y) in the x-axis is B’(x, -y) Given f(x): -f(x) is a reflection of f(x) through the x-axis f(x) = (x – 3)2 + 1 f(x) = (x – h)2 + k - parabolic (3, 1) (h, k) is the turning point of the parabola. -f(x) = -[(x – 3)2 + 1]

Reflections of Functions image of A(x,y) in the y-axis is A’(-x, y) Given f(x): f(-x) is a reflection of f(x) through the y-axis f(-x) f(x) = ((-x) – 3)2 + 1 = (x – 3)2 + 1

Reflections of Functions Under reflection in the origin, the Image of P(x, y)  P”(-x, -y) Given f(x): -f(-x) is a reflection of f(x) through the origin = -((-x) – 3)2 + 1) f(x) -f(-x) = (x – 3)2 + 1

Dilations of Functions Given f(x): af(x) is a dilation of f(x) by a factor of a If a > 1, the function is ‘stretched’ vertically 2(f(x)) = 2[(x – 3)2 + 1] .5(f(x)) = .5[(x – 3)2 + 1] If a < 1, the function is ‘stretched’ horizontally f(x) = (x – 3)2 + 1

Transformation of Functions Translation Reflection Dilation f(x) = (x ± h)n ± k n is integer > 1 -f(x) is a reflection of f(x) through the x-axis f(-x) is a reflection of f(x) through the y-axis -f(-x) is a reflection of f(x) through the origin af(x) is a dilation of f(x) by a factor of a If a > 1, the function is ‘stretched’ vertically If a < 1, the function is ‘stretched’ horizontally

Regents Prep

Model Problems Graph: y = (x – 1)2 – 5

Standard to Vertex Form Rewrite the equation of a parabola in vertex form. f(x) = x2 + 4x + 1 f(x) = x2 + 4x + 1 Separate the first two terms from c. f(x) = (x2 + 4x ) + 1 Take 1/2 the coefficient of the b term, square it then add the result to the terms inside the parentheses and subtract it from the constant outside. f(x) = (x2 + 4x ) + 1 + 4 – 4 Rewrite the perfect square trinomial in the parenthesis as a binomial square and add the constants together. f(x) = (x + 2)2 – 3

Translating Absolute Value Functions Graph y = | x | y = | x – 3 | y = | x + 3 | y = | x – 3 | y = | x + 3 | y = | x | y = |x + h| represents a horizontal translation of y = |x|, h units to the left if h is positive or h units to right if h is negative

Translating Absolute Value Functions Graph y = | x | Graph y = | x | + 3 and y = | x | – 2 y = | x | + 3 (0,3) y = | x | (0,0) y = | x | – 2 (0,-2) y = |x| + k represents a vertical translation of y = |x|, k units

Translating Absolute Value Functions Graph y = | x – 3 | + 1 y = | x – 3| + 1 y = | x | (0,0) y = | x – 3 | y = |x – h| + k represents a horizontal shift of h units to the right if h is positive or h units to left if h is negative and a vertical translation of k units

Translating Absolute Value Functions standard form - Graph y = | x – 3 | a = 1, b = 1, c = -3, and d = 0 Find the coordinate of the vertex by evaluating bx + c = 0 bx + c = 0  1(x) – 3 = 0 x = 3 Construct a table of values using the x-value of the vertex and several values to the left and right of it. x |x – 3| y x,y 1 2 3 4 5 3 3 0,3 2 2 1,2 1 1 2,1 Graph the resulting points to form the V-shaped graph 0 1 3,0 1 1 4,1 If a is positive V opens up 2 2 5,2 If a is negative V opens down

The Punted Football The height of a punted football can be modeled by the quadratic function h = - 0.01x2 + 1.18x + 2. The horizontal distance in feet from the point of impact is x, and h is the height of the ball in feet. a. Find the vertex of the graph of the function by completing the square. What is the maximum height of this punt? c. The nearest defensive player from the point of impact is 5 feet away. How high must he get his hand to block the punt?

The Punted Football Really Small Peoples’ Football League x = 5 h = - 0.01x2 + 1.18x + 2 x = 5 4’ 2’

The Punted Football (x - 59)2 = 3681 x - 59 = ±60.67 x – 59 = 60.67 h = -x2 + 118x + 200 Rewrite the quadratic for h = 0: x2 – 118x = 200 Take 1/2 the coefficient of the linear term & square it. This is now the c term. = 3481 Add the c term to both sides of equation x2 – 118x = 200 + 3481 + 3481 (x - 59)2 = 3681 Binomial Squared Find square root of both sides Solve for x x - 59 = ±60.67 x – 59 = 60.67 x – 59 = -60.67 x = 119.67 x = -1.67