1. Copyright © Cengage Learning. All rights reserved. 4 Quadratic Functions

2. Copyright © Cengage Learning. All rights reserved. 4.7 Graphing Quadratics from Standard Form

3. 3 Objectives  Graph a quadratic from standard form.  Graph quadratic inequalities in two variables.

4. 4 Graphing from Standard Form

5. 5 Because quadratic functions are often written in standard form rather than in vertex form, it is important that we learn to graph from the standard form. Since all quadratics have a vertex and a vertical intercept, these two will be the starting points for graphing. A quadratic function in standard form f (x) = ax 2 + bx + c gives us several key pieces of information about the graph. The value of a in the standard form affects the graph of the quadratic the same way that it affects the vertex form. Also a will affect whether the graph is wide or narrow.

6. 6 Graphing from Standard Form If a is positive, the parabola will open upward. If a is negative, the parabola will open downward. Also a will affect whether the graph is wide or narrow. The value c will also give us information about the graph. When we substitute zero into the input variable for the function, the output will be c.

7. 7 Graphing from Standard Form Let f (x) = ax 2 + bx + c and x = 0 f (0) = a(0) 2 + b(0) + c f (0) = c which means that (0, c) will be the vertical intercept for the graph.

8. 8 Graphing from Standard Form Another important relationship for parabolas is that between the vertex and the standard form of a quadratic function. There are several ways to see this relationship, one being to consider the quadratic formula and symmetry.

9. 9 Graphing from Standard Form From the graph, the two x-intercepts are located at the x-values given by the quadratic formula. The symmetry of the graph indicates that the x-value of the vertex must be midway between these two points. If we take the two x-values and add them together, the square root portions will cancel out because one is positive and the other is negative. Then we divide by 2 to get the average.

10. 10 Graphing from Standard Form Horizontal intercepts are the x-values. Add them and divide by 2. The radical parts cancel.

11. 11 Graphing from Standard Form This result indicates that the x-value for the vertex will be. Another way to discover this relationship is to consider that the only point on a parabola without a symmetric point is the vertex. With this information, the only way to come to a single solution from the quadratic formula is to have the square root after the plus/minus symbol be zero. Reduce.

12. 12 Graphing from Standard Form This way, we are not adding or subtracting anything to or from the remaining parts of the formula. If we consider the square root to be zero, we are left with This x-value will then represent the one input value that does not have a symmetric point. Hence, the vertex has the input value.

13. 13 Graphing from Standard Form The vertex is a point on the graph, so find the y-value of the point. To find the y-value of the vertex, simply evaluate the equation at. The vertical intercept, vertex, and symmetry will all help us to graph quadratic functions. The process will be basically the same as graphing from vertex form. We will find the vertex and vertical intercept, and we will use the axis of symmetry to find symmetric points.

14. 14 Graphing from Standard Form Now that we know how to solve quadratic equations, we will also find the horizontal intercepts, if there are any, and use them as an additional pair of symmetric points. In general, at least five points should be plotted for a graph of a parabola. If we want more details, we can always find more symmetric points by plugging in values for x and finding y.

15. 15 Graphing from Standard Form

16. 16 Example 2 – Graphing quadratic functions Sketch a graph of the following: a. f (x) = 2x 2 – 12x + 5 b. f (x) = –0.5x 2 + 4x – 10 Solution: a. f (x) = 2x 2 – 12x + 5 Step 1: Determine whether the graph opens up or down. Because a is positive in this quadratic, we know that the parabola will open upward.

17. 17 Example 2 – Solution Step 2: Find the vertex and the equation for the axis of symmetry. This function is in standard form for quadratic, so we can use the formula to find the x-value of the vertex. x = 3 Now that we have the input value for the vertex, we can substitute it into the function and find the output value for the vertex. cont’d

18. 18 Example 2 – Solution f (x) = 2x 2 – 12x + 5 f (3) = 2(3) 2 – 12(3) + 5 f (3) = 18 – 36 + 5 f (3) = –13 The vertex is (3, –13). The axis of symmetry is the vertical line through the vertex. The axis of symmetry is x = 3. Step 3: Find the vertical intercept. Because this quadratic is in standard form, we know that the vertical intercept has an output value equal to the constant c. In this case, the vertical intercept is (0, 5). cont’d

19. 19 Example 2 – Solution Step 4: Find the horizontal intercepts (if any). Horizontal intercepts will happen when the output variable is equal to zero. Substitute zero for the output variable and solve. f (x) = 2x 2 – 12x + 5 0 = 2x 2 – 12x + 5 a = 2 b = –12 c = 5 cont’d Use the quadratic formula.

20. 20 Example 2 – Solution x = 5.55 x = 0.45 The horizontal intercepts are (5.55, 0) and (0.45, 0). cont’d

21. 21 Example 2 – Solution Step 5: Plot the points you found in steps 2 through 4. Plot their symmetric points and sketch the graph. (Find an additional pair of symmetric points if needed.) cont’d

22. 22 Example 2 – Solution We now have the vertex (3, –13), the vertical intercept (0, 5), and the horizontal intercepts (5.55, 0) and (0.45, 0). Plotting these points and using the axis of symmetry to plot their symmetric points, we get the following graph. cont’d

23. 23 Example 2 – Solution Sketching a smooth curve through the points gives us the following graph. cont’d

24. 24 Example 2 – Solution b. f (x) = –0.5x 2 + 4x – 10 Step 1: Determine whether the graph opens up or down. Because a is negative in this quadratic, we know that the parabola will open downward. Step 2: Find the vertex and the equation for the axis of symmetry. This function is in standard form for a quadratic, so we can use the formula to find the x-value of the vertex. x = 4 cont’d

25. 25 Example 2 – Solution Now that we have the input value for the vertex, we can substitute it into the function and find the output value for the vertex. f (x) = –0.5x 2 + 4x – 10 f (4) = –0.5(4) 2 + 4(4) – 10 f (4) = –8 + 16 – 10 f (4) = –2 The vertex is (4, –2). The axis of symmetry is the vertical line through the vertex. The axis of symmetry is x = 4. cont’d

26. 26 Example 2 – Solution Step 3: Find the vertical intercept. Because this quadratic is in standard form, we know that the vertical intercept has an output value equal to the constant c. In this case, the vertical intercept is (0, –10). Step 4: Find the horizontal intercepts (if any). Horizontal intercepts will happen when the output variable is equal to zero, so substitute zero for the output variable and solve. cont’d

27. 27 Example 2 – Solution f (x) = –0.5x 2 + 4x –10 0 = –0.5x 2 + 4x –10 a = – 0.5 b = 4 c = –10 There are no real solutions when this function is set equal to zero, so there are no horizontal intercepts. cont’d Set the function equal to zero. Use the quadratic formula. A negative discriminant indicates no real solutions.

28. 28 Example 2 – Solution Step 5: Plot the points you found in steps 2 through 4. Plot their symmetric points and sketch the graph. (Find an additional pair of symmetric points if needed.) We now have the vertex (4, –2) and the vertical intercept. There are no horizontal intercepts, so we will want an additional pair of symmetric points to give us more points to sketch our curve through. cont’d

29. 29 Example 2 – Solution To find another point, we can pick an additional input value on one side of the axis of symmetry and find the output value. x = 2 f (2) = –0.5(2) 2 + 4(2) – 10 f (2) = –4 (2, –4) cont’d Pick any input value we don't already have and substitute to find the output value.

30. 30 Example 2 – Solution Plotting all the points that we have found and using the axis of symmetry to plot their symmetric points, we get the following graph. cont’d

31. 31 Example 2 – Solution We now have the vertex (4, –2) and two sets of symmetric points: (2, –4), (6, –4) and (0, –10), (8, –10). Sketching a smooth curve through these points gives us the final graph. cont’d

32. 32 Graphing Quadratic Inequalities in Two Variables

33. 33 Graphing Quadratic Inequalities in Two Variables Graphing quadratic inequalities can be done by using the same techniques as graphing linear inequalities. First, graph the inequality as if it were an equality using a solid Curve if there is an “equal to” part and a dashed curve if it is only “less than” or “greater than” but not “equal to.” This separates the graph into two sections: points that are above the parabola and points that are below the parabola.

34. 34 Graphing Quadratic Inequalities in Two Variables Then decide which section satisfies the inequality and, therefore, should be shaded. So pick a point that is not on the curve and check to see whether it satisfies the inequality. Finally, shade the side of the curve that does satisfy the inequality.

35. 35 Example 4 – Graphing quadratic inequalities Graph the following inequalities. a. y < x 2 + 3x – 10 b. y –2x 2 + 11x – 12

36. 36 Example 4(a) – Solution a. First, we will sketch the graph of the quadratic using a dashed curve because there is no “equal to” part.

37. 37 Example 4(a) – Solution Doing so separates the grid into two sections: the points above the parabola and the points below the parabola. cont’d

38. 38 Example 4(a) – Solution Now if we pick (0, 0) as a test point, we find that it does not satisfy the inequality, so we will shade the points below the parabola. cont’d

39. 39 Example 4(b) – Solution First, we will sketch the graph using a solid line because it is a “greater than or equal to” inequality. cont’d

40. 40 Example 4(b) – Solution If we pick (0, 0) for our test point, we find that it does satisfy the inequality, so we should shade above the curve. cont’d