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Functions and Their Graphs Chapter 12

12 Functions and Their Graphs 12.1 Relations and Functions 12.2 Graphs of Functions and Transformations 12.3 Quadratic Functions and Their Graphs 12.4 Applications of Quadratic Functions and Graphing Other Parabolas. 12.5 The Algebra of Functions 12.6 Variation

12.3 Quadratic Functions and Their Graphs In this section, we will study quadratic functions in more detail and see how the rules we learned in Section 12.2 apply specifically to these functions. Quadratic functions can be written in other forms as well. One common form is f (x)= a(x ‒ h)2 + k. An example is f (x)= 2(x ‒ 3)2 + 1. We will study the form f (x)= a(x ‒ h)2 + k first since graphing parabolas from this form comes directly from the transformation techniques we learned earlier.

Graph a Quadratic Function by Shifting the Graph of f (x)= x2 Example 1 Solution y If we compare g(x) to f (x)= x2 what do the constants that have been added to g(x) tell us about transforming the graph of f (x)? right 2 right 2 down 1 g(x) = (x ‒ 2) 2 ‒ 1 x Vertex (0,0) Vertex (2,-1) Shift f(x) Down 1 Shift f (x) right 2. Every parabola has symmetry. If we were to fold the paper along the y-axis, one half of the graph would fall exactly on the other half. Axis of symmetry of g(x)

Graph f (x)= a(x ‒ h)2 + k Using the Vertex, Axis of Symmetry, and Other Characteristics. When a quadratic function is in the form f (x)= a(x ‒ h)2 + k, we can read the vertex directly from the equation. Furthermore, the value of a tells us whether the parabola opens upward or downward and whether the graph is narrower, wider, or the same width as y = x2

Example 2 Solution Here is the information we can get from the equation. 1). h = ‒ 3 and k = ‒ 2. The vertex is (‒3,‒2) 2). The axis of symmetry is x = ‒ 3 3). . Since a is positive, the parabola opens upward. 4). Since and , the graph of is wider than the graph of

Example 2-continued Solution To graph the function, start by putting the vertex on the axes. Then, choose a couple of values of x to the left or right of the vertex to plot more points. We can read the x-intercepts from the graph: (-5,0)and (-1,0). To find the y-intercept, let x=0 and solve for y. The y-intercept is

Example 3 Solution Here is the information we can get from the equation. 1). h = 1 and k = 5. The vertex is (1,5) 2). The axis of symmetry is x = 1 3). . Since a is negative, the parabola opens downward. 4). Since , the graph of is the same width as

Example 3-continued Solution To graph the function, start by putting the vertex on the axes. Then, choose a couple of values of x to the left or right of the vertex to plot more points. We can read the y-intercepts from the graph: (0,4) To find the x-intercepts, let y=0 and solve for x. Substitute 0 for f(x) Subtract 5 Divide by -1. Square Root Property The x-intercepts are Add 1

Graph f (x)= ax2 + bx + c by Completing the Square

Example 4 Graph . Begin by completing the square to rewrite it in the form f (x)= a(x ‒ h)2 + k. Include the intercepts. Solution Step 1: The coefficient of x2 is .Multiply both sides of the equation [including the g(x)] by -2 so that the coefficient of x2 will be 1. Step 2: Separate the constant from the variable terms using parentheses. Multiply by ‒ 2. Distribute.

‒ 2g(x)=(x2 ‒ 8x +16)+12 ‒16 ‒ 2g(x)=(x2 ‒ 8x +16) ‒ 4 Example 4-continued Step 3: Complete the square for the quantity in parentheses. (‒ 4)2 = 16 Add 16 inside the parentheses and subtract 16 from the 12. ‒ 2g(x)=(x2 ‒ 8x +16)+12 ‒16 ‒ 2g(x)=(x2 ‒ 8x +16) ‒ 4 Step 4: Factor the expression inside the parentheses. ‒ 2g(x)= ( x ‒ 4)2 ‒ 4 Step 5: Solve the equation for g(x) by dividing by ‒ 2. ‒ 2g(x) = ( x ‒ 4)2 ‒ 4 2 2 2

ii) The axis of symmetry is x= 4 Example 4-continued From we can see that The vertex is (4, 2). ii) The axis of symmetry is x= 4 iii) , so the parabola opens downward. iv) Since the graph of g(x) will be wider than Find some other points on the parabola. Using the axis of symmetry, we can see that the x-intercepts are (6, 0) and (2, 0) and that the y-intercept is (0, 6).

h = 3. Then the y-coordinate, k, of the vertex is k = f (3) Graph f (x)= ax2 + bx + c Using Example 5 Graph f (x)= x2 ‒ 6x + 3 using the vertex formula. Include the intercepts. Solution a = 1, b = ‒ 6, c = 3. Since a = +1, the graph opens upward. The x-coordinate, h, of the vertex is h = 3. Then the y-coordinate, k, of the vertex is k = f (3) f (x)= x2 ‒ 6x + 3 f (3)= 32 ‒ 6(3) + 3 = 9 ‒ 18 + 3 = 6 The vertex is (3,‒6). The axis of symmetry is x = 3.

0 = x2 ‒ 6x + 3 Example 5-continued Find more points on the graph of f (x)= x2 ‒ 6x + 3 ,then use the axis of symmetry to find other points on the parabola. To find the x-intercepts, let f (x)=0 and solve for x. 0 = x2 ‒ 6x + 3 Solve using the quadratic formula. x-intercepts: Simplify. y – intercept: (0,3)