Translations & Transformations QUADRATIC FUNCTIONS Translations & Transformations

Identifying Parts of a Quadratic Function A function that can be represented in the form of f(x) = ax2 + bx + c is called a quadratic function where a 0. The greatest exponent of the variable x is 2. The most basic quadratic function is f(x) = x2, which is the parent quadratic function. The graph of a quadratic function is a curve called a parabola. The point at which the parabola turns directions is called the vertex. The vertical line that passed through the vertex and divides the parabola in half is called the axis of symmetry.

Graphing the Parent Quadratic Function f(x) = x2 Create a table of values The vertex of the parent function f(x) = x2 is (0,0) and the axis of symmetry is x = 0. x f(x) = x2 Point -3 f(-3) = (-3)2 = -2 f(-2) = (-2)2= -1 f(-1) = (-1)2 = f(0) = (0)2 = 1 f(1) = (1)2 = 2 f(2) = (2)2 = 3 f(3) = (3)2 =

Graphing g(x) = ax2 when a > 0 The graph of g(x) = ax2 is a vertical stretch or vertical compression of its parent function f(x) = x2. When a > 1 the graph of g(x) is narrower than the parent function f(x) and there is a vertical stretch. When 0 < a < 1 (which means a is a fraction) the graph of g(x) is wider than the parent function f(x) and there is a vertical compression. When a > 0, the graph g(x) opens upward, and the function has a minimum value that occurs at the vertex of the parabola. Domain – The domain of a quadratic function is all real numbers. Range – The range of g(x) = ax2, where a > 0, is the set of all real numbers greater than or equal to the minimum value. The range is y > 0.

Graphing Vertical Stretch & Compression Parent Function Vertical Stretch Vertical Compression x f(x) = x2 Point -3 -2 -1 1 2 3 x f(x) =2x2 Point -2 -1 1 2 x f(x)=½x2 Point -4 -2 2 4

Graphing g(x) = ax2 when a < 0 ( a is negative) When a < 0 the graph g(x) opens downward, and the function has a maximum value that occurs at the vertex of the parabola. The graph g(x) = -x2 is a reflection of the graph f(x) = x2 in the x axis. Domain – The domain of the quadratic function is still all real numbers. Range – The range of g(x) = -ax2 is the set of real numbers less than or equal to the maximum value. The range is y < 0. Graph the following quadratic functions. g(x) = - 2x2 g(x) = - ½ x2

Transforming Quadratic Functions Vertical Translations A vertical translation of a quadratic function is a shift of the parabola up or down with no change in the shape of the parabola. The graph of the function g(x) = x2 + k is the graph of f(x) = x2 translated vertically. If k > 0, the graph of f(x) = x2 is translated k units up. If K < 0, the graph of f(x) = x2 is translated k units down. Graph each function and identify the minimum value, the range and axis of symmetry. g(x) = x2 + 2 g(x) = x2 - 5

Transforming Quadratic Functions Horizontal Translations A horizontal translation of a quadratic function is a shift of the parabola left or right with no change in the shape of the parabola. The graph of the function f(x – h)2 is the graph of f(x) = x2 translated horizontally. If h > 0, the graph of f(x) = x2 is translated h units to the right. If h < 0, the graph of f(x) = x2 is translated h units to the left. Graph each function and identify the vertex and axis of symmetry. g(x) = (x – 1)2 g(x) = (x + 3)2

Vertex Form of a Quadratic Function g(x) = a(x – h)2 + k The vertex form gives us a lot of information about the graph a quadratic function. (h,k) is the vertex of the parabola x = h is the axis of symmetry k is the vertical shift of the parabola h is the horizonal shift of the parabola a is the vertical stretch or compression of the parabola. If a>0 the parabola is a “smile”(opens upward) and if a < 0, the parabola is a “frown” (opens downward and is a reflection in the x-axis). If a > 0, the minimum value is k, and the range is y > k. If a < 0, the maximum value is k, and the range is y < k.

Identify the vertex, axis of symmetry, the minimum or maximum value of the parabola and the range. Describe the transformation and graph the function. Example1: g(x) = -3(x + 1)2 – 2 Vertex: (-1, -2) Axis of Symmetry: x = -1 Maximum value: y = -2 Range: y < -2 This graph is a reflection of the parent function f(x) = x2 in the x-axis with a vertical stretch multiplied by a factor of 3 shifted two units downward and 1 unit to the left.

Creating the graph of the parabola using the vertex Create a chart. Use the vertex as the middle point on the chart. Find two points on each side of the vertex, and substitute the values in the function to get corresponding y values. Graph the points. x g(x) = -3(x + 1)2 - 2 Point -3 -3(-3 + 1)2 – 2 = -2 -3(-2 + 1)2 – 2 = -1 (-1,-2) -3(0 + 1)2 – 2 = 1 -3 (1 + 1)2 – 2 =

g(x) = -(x – 2)2 + 4 g(x) = 2(x + 3)2 - 1 x g(x) = 2(x – 1)2 - 7 Point Identify the vertex, axis of symmetry, the minimum or maximum value of the parabola and the range. Describe the transformation and graph the function. Example 2 - g(x) = 2(x – 1)2 – 7 Vertex: Description: Axis of Symmetry: Minimum Value: Range: Practice Problems g(x) = -(x – 2)2 + 4 g(x) = 2(x + 3)2 - 1 x g(x) = 2(x – 1)2 - 7 Point

Standard Form of a Quadratic Function f(x) = ax2 + bx + c The standard form of a quadratic equation is y = ax2 + bx + c. We can use the standard form to find the x-value of the vertex (the axis of symmetry) by using the equation x = -b/2a. The y value of the vertex can be found by substituting the x-value back into the original equation. Find the axis of symmetry and the vertex of the following quadratic equation. y = 3x2 + 6x + 11 y = 3(-1)2 + 6(-1) + 11 x = -6/2(3) y = 3 - 6 + 11 x = -1 y = 8 The axis of symmetry is x = -1 and the vertex is (-1,8)

Let’s Practice ! Find the axis of symmetry and the vertex for the equation below. Example 1: y = -2x2 + 12x - 16 Find the axis of symmetry and the vertex for the equation below. Example 2: y = 2x2 + 2x - 4

Changing Vertex Form to Standard Form y = a(x – h)2 + k to y = ax2 + bx + c Example 1: y = 4(x – 6)2 + 3 y = 4(x – 6)(x – 6) + 3 y = 4(x2 – 12x + 36) + 3 y = 4x2 – 48x + 144 + 3 y = 4x2 – 48x + 147 Standard form is y = 4x2 – 48x + 147 Practice Problem y = 2(x + 5) + 3 Complete pg. 725 #12 - 17 Example 2: y = -3(x + 2)2 – 1 y = -3(x + 2)(x + 2) – 1 y = -3(x2 + 4x + 4) – 1 y = -3x2 – 12x – 12 – 1 y = -3x2 -12x – 13 Standard form is y = - 3x2 – 12x – 13 Practice Problem y= -3(x – 7)2 + 2

Writing a Quadratic Equation in Standard form from a Table Step 1: Use the vertex in the table to write the quadratic equation in vertex form. Vertex Vertex Form: y = a(x –(-3))2 Vertex Form is y = a(x + 3)2 x y -6 9 -4 1 -3 -2

Step 2: Substitute a point from the table for x and y in the vertex form to find a. y = a(x + 3)2 1 = a( -2 +3 )2 1 = a Vertex Form: y = (x + 3)2 y = (x + 3)(x + 3) Standard Form: y = x2 + 6x + 9 x y -6 9 -4 1 -3 -2

Use the tables below to write the quadratic equations in standard form. 1. 2. x y 13 -1 1 -2 -3 -4 x y -27 -1 -12 -2 -7 -3 -4