Name Period Date: Topic: 7-5 Graphing ( ) Essential Question: What is the vertex of a parabola, and what is its axis of symmetry? Standard: F-IF.7a Objective: Graph linear and quadratic functions and show intercepts, maxima, and minima. To graph parabolas whose equations have the form ( ) and to find the vertices and axes of symmetry. This is called the vertex form of the quadratic equation. To graph the equation first make a table of pairs of values of x and v that satisfy the equation. Then plot the graph of each ordered pair of coordinates, as shown in the figure below. x y 3 9 2 4 1 1 0 0 1 1 2 4 3 9 Summary
If you plotted more points, you would see that they all lie on the smooth curve shown in the figure below. This curve, called a parabola, is the graph of. In the figure notice that if the point (x, y) is on the parabola, then ( x, y), its "mirror image" across the y-axis, is also on the parabola. This property can also be seen in the table where the coordinate pairs are mirror images. Because of this property, the y-axis is called the axis of symmetry, or simply the axis of the parabola. The vertex of a parabola is the point where the parabola crosses its axis. In the case of, the vertex is the origin. The graph of, shown in the figure below, is a mirror image or congruent copy of the graph of, If the graph of is reflected across the x-axis, then the result is the graph of. 2
The two figures below show the effect of the value of a on the graph of an equation of the form. The graph of opens upward if and downward if. The larger is, the "narrower" the graph is. By graphing pairs of quadratic equations on the same axes, such as those in the six figures that follow, you may investigate the methods for graphing parabolas described in this lesson. A computer or a graphing calculator may be helpful. The two figures below illustrate the following method for graphing a parabola whose equation has the form ( ) To graph ( ), slide the graph of horizontally h units. If, slide it to the right; if slide it to the left. The graph has vertex (h, 0) and its axis is the line. y 1 2 (x 3)2 y 1 2 x ( 3) 2 3
Exercise 1: Fill in the table. Then graph, and ( ). Use the grid below. ( ) 3 1 2 0 1 1 0 2 1 3 2 4 3 5 4
The two figures below illustrate the following method for graphing a parabola whose equation has the form To graph ( ), slide the graph of vertically k units. If, slide it upward; if, slide it downward. The graph has vertex (0, k) and its axis is the line (the y- axis). 1 1 y 3 2 x2 y ( 3) 2 x2 The two figures below illustrate the following method for graphing a parabola whose equation has the form ( ) To graph ( ), slide the graph of horizontally h units and vertically k units. The graph has vertex (h, k) and its axis is the line. y 2 1 2 (x 4)2 y 3 1 (x + 2)2 2 5
Exercise 2: Fill in the table. Then graph, and. Use the grid below. + 3 2 1 0 1 2 3 6
Exercise 3: Fill in the table. Then graph, and ( + ). Use the grid below. ( + ) + 3 2 1 5 4 3 0 2 1 1 2 0 3 1 7
Exercise 4: What are the coordinates of the vertex of the graph in Exercise 1? What is the equation of the axis of symmetry of the graph in Exercise 1? Label the vertex and draw the axis of symmetry on the graph in Exercise 1. Exercise 5: What are the coordinates of the vertex of the graph in Exercise 2? What is the equation of the axis of symmetry of the graph in Exercise 2? Label the vertex and draw the axis of symmetry on the graph in Exercise 2. Exercise 6: What are the coordinates of the vertex of the graph in Exercise 3? What is the equation of the axis of symmetry of the graph in Exercise 3? Label the vertex and draw the axis of symmetry on the graph in Exercise 3. 8
Example 1: Graph ( + ). Label the vertex and axis. Solution Since, the parabola opens downward. Since and the vertex is ( ). The axis of symmetry is the line. Calculate a few convenient ordered pairs and plot the corresponding points. Also plot their images by reflection across the axis. Now draw the parabola by connecting the points with a smooth curve. x y 0 1 1 4 2 5 9
Intercepts: When graphing an equation in the coordinate plane, it is usually helpful to know the intercepts of the graph. The y-coordinate of a point where a graph crosses the y-axis is called the y-intercept. The x-coordinate of a point where a graph crosses the x-axis is called an x-intercept. A parabola may have no x-intercepts, one x-intercept, or two x-intercepts, as illustrated below. No x intercepts y intercept = 2 One x intercept y intercept = 1 Two x intercepts y intercept = 3 10
Determining the x-intercepts: 1. Set. ( ) 2. Transform the quadratic equation into standard form. + + 3. Find the discriminant where, + Discriminant < 0 If the discriminant is negative, the quadratic equation has no real roots, and the quadratic function has no x-intercepts. Discriminant = 0 If the discriminant is zero, the quadratic equation has one real root, and the quadratic function has one x-intercept. Discriminant > 0 If the discriminant is positive, the quadratic equation has two real roots, and the quadratic function has two x-intercepts. 11
Example 2: Graph + ( + ). Label the vertex and axis. Find all intercepts. Solution 1. Since, the parabola opens upward. Since and the vertex is ( ) The axis of symmetry is the line. 2. To find the y-intercept, set and solve for y. + ( + ) + y-intercept Therefore, the graph crosses the y-axis at ( ). Since ( ) is on the graph, so is its mirror image across the axis of symmetry, ( ). 3. To find any x-intercepts, set. Since + ( + ) + ( + ), the x-intercepts are, and 4. Plot the vertex ( ) and the intercepts. Then complete the curve using symmetry. The graph is shown on the next page. 12
Exercise 7: Find the vertex, axis of symmetry, y-intercept and all x-intercepts (if any) for the following quadratic functions: 1. + ( + ) 13
Exercise 7, continued: 2. ( ) 3. + ( ) 14
Example 3: Find an equation ( ) of the parabola having vertex (1, 2) and containing the point (3, 6). Solution Substitute (1, 2) for (h, k) in the equation ( ). ( ) ( ) + ( ) Since the parabola contains the point (3, 6), the coordinates of this point must satisfy the equation, + ( ) an equation of the parabola is + ( ) Exercise 8: 1. Find an equation ( ) of the parabola having vertex (5, 4) and containing the point (3, 8). 15
Exercise 8, continued: 2. Find an equation ( ) of the parabola having vertex ( 5, 4) and containing the point ( 4, 2). 3. Find an equation ( ) of the parabola having vertex ( 5, 4) and containing the point ( 1, 6). Class work: p 330 Oral Exercises: 1-17 Homework: p 331 Written Exercises: 1-30, P 332 Mixed Review: 1-6 16