Presentation is loading. Please wait.

Presentation is loading. Please wait.

Vertex Form of Quadratics

Similar presentations


Presentation on theme: "Vertex Form of Quadratics"— Presentation transcript:

1 Vertex Form of Quadratics

2 Vertex Form (Vertical Parabola):
We already know: We are learning today: Vertex Form (Vertical Parabola): Or sometimes written as: *The only difference between the two is which side k is on. *The MOST IMPORTANT thing to pay attention to is the signs for h and k

3 Why is it called vertex form?
Opens up: positive a Opens down: negative a Vertex (-h,k) Vertex (-h,-k) *h is the x value of the vertex and is opposite of the sign shown in the equation *h is the x value and k is the y value of the vertex and is opposite of the sign shown in the equation

4 Let’s make connections:
Graph: Down Direction: (2, 4) Vertex:

5 Let’s make connections:
Graph: Down Direction: (-5,-3) Vertex:

6 Let’s make connections:
Graph: Up Direction: (-2, -1) Vertex:

7 Vertex Form (Horizontal Parabola):
sometimes written as: *The difference between vertical and horizontal parabola is: x and y are switched h and k are switched h still corresponds with x, and k still corresponds with y

8 Opens Right: positive a
Vertex Form (Horizontal Parabola): Opens Right: positive a Opens Left: negative a Vertex (h,-k) Vertex (-h,-k) *k is the y value of the vertex and is opposite of the sign shown in the equation *h is the x value and k is the y value of the vertex and is opposite of the sign shown in the equation

9 Let’s make connections:
Graph: right Direction: (2, 7) Vertex:

10 Let’s make connections:
Graph: left Direction: (-1,-3) Vertex:

11 Let’s make connections:
Graph: right Direction: (5,4) Vertex:

12 Determine which way the parabola opens
and where the vertex is. Opens: Up Vertex: Opens: Right Vertex: Opens: Left Vertex: Opens: Up Vertex: Opens: Right Vertex: Opens: Down Vertex:

13 Warm Up Determine which way the parabola opens and find the vertex.
Direction: Vertex: Direction: Vertex: Direction: Vertex:

14 Objectives Write the standard equation of a parabola and its axis of symmetry. Graph a parabola and identify its focus, directrix, and axis of symmetry.

15 Vocabulary focus of a parabola directrix

16 In Chapter 5, you learned that the graph of a quadratic function is a parabola. Because a parabola is a conic section, it can also be defined in terms of distance.

17 A parabola is the set of all points P(x, y) in a plane that are an equal distance from both a fixed point, the focus, and a fixed line, the directrix. A parabola has a axis of symmetry perpendicular to its directrix and that passes through its vertex. The vertex of a parabola is the midpoint of the perpendicular segment connecting the focus and the directrix.

18 The distance from a point to a line is defined as the length of the line segment from the point perpendicular to the line. Remember!

19

20

21 Notice: The distance between the focus/parabola and the directrix/parabola is the same.

22 Given either the focus or directrix, find the value of the other from the graph.
Focus: (0,2) Directrix: ______ Focus: (4,0) Directrix: ______

23 Given either the focus or directrix, find the value of the other from the graph.
Focus: ______ Directrix: y=0 Focus: ______ Directrix: x=-2

24 Example 1: Using the Distance Formula to Write the Equation of a Parabola
Use the Distance Formula to find the equation of a parabola with focus F(2, 4) and directrix y = –4. PF = PD Definition of a parabola. Distance Formula. Substitute (2, 4) for (x1, y1) and (x, –4) for (x2, y2).

25 Example 1 Continued Simplify. Square both sides. (x – 2)2 + (y – 4)2 = (y + 4)2 (x – 2)2 + y2 – 8y + 16 = y2 + 8y + 16 Expand. Subtract y2 and 16 from both sides. (x – 2)2 – 8y = 8y Add 8y to both sides. (x – 2)2 = 16y Solve for y.

26 Example 2 Use the Distance Formula to find the equation of a parabola with focus F(0, 4) and directrix y = –4. PF = PD Definition of a parabola. Distance Formula Substitute (0, 4) for (x1, y1) and (x, –4) for (x2, y2).

27 Check It Out! Example 1 Continued
Simplify. x2 + (y – 4)2 = (y + 4)2 Square both sides. x2 + y2 – 8y + 16 = y2 + 8y +16 Expand. Subtract y2 and 16 from both sides. x2 – 8y = 8y x2 = 16y Add 8y to both sides. Solve for y.

28 Previously, you have graphed parabolas with vertical axes of symmetry that open upward or downward. Parabolas may also have horizontal axes of symmetry and may open to the left or right. The equations of parabolas use the parameter p. The |p| gives the distance from the vertex to both the focus and the directrix.

29

30 Example 3: Writing Equations of Parabolas
Write the equation in standard form for the parabola. Step 1 Because the axis of symmetry is vertical and the parabola opens downward, the equation is in the form y = x2 with p < 0. 1 4p

31 Step 3 The equation of the parabola is . y = – x2
Example 3 Continued Step 2 The distance from the focus (0, –5) to the vertex (0, 0), is 5, so p = –5 and 4p = –20. Step 3 The equation of the parabola is y = – x2 1 20 Check Use your graphing calculator. The graph of the equation appears to match.

32 Example 4: Writing Equations of Parabolas
Write the equation in standard form for the parabola. vertex (0, 0), directrix x = –6 Step 1 Because the directrix is a vertical line, the equation is in the form The vertex is to the right of the directrix, so the graph will open to the right.

33 Step 2 Because the directrix is x = –6, p = 6 and 4p = 24.
Example 4 Continued Step 2 Because the directrix is x = –6, p = 6 and 4p = 24. Step 3 The equation of the parabola is x = y2 1 24 Check Use your graphing calculator.

34 Example 5 Write the equation in standard form for the parabola. vertex (0, 0), directrix x = 1.25 Step 1 Because the directrix is a vertical line, the equation is in the form of The vertex is to the left of the directrix, so the graph will open to the left.

35 Example 5 Continued Step 2 Because the directrix is x = 1.25, p = –1.25 and 4p = –5. Step 3 The equation of the parabola is Check Use your graphing calculator.

36 The vertex of a parabola may not always be the origin
The vertex of a parabola may not always be the origin. Adding or subtracting a value from x or y translates the graph of a parabola. Also notice that the values of p stretch or compress the graph.

37

38 Example 6: Graphing Parabolas
Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola Then graph. y + 3 = (x – 2)2. 1 8 Step 1 The vertex is (2, –3). Step , so 4p = 8 and p = 2. 1 4p 8 =

39 Example 6 Continued Step 3 The graph has a vertical axis of symmetry, with equation x = 2, and opens upward. Step 4 The focus is (2, –3 + 2), or (2, –1). Step 5 The directrix is a horizontal line y = –3 – 2, or y = –5.

40 Example 7 Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola. Then graph. Step 1 The vertex is (1, 3). Step , so 4p = 12 and p = 3. 1 4p 12 =

41 Example 7 Continued Step 3 The graph has a horizontal axis of symmetry with equation y = 3, and opens right. Step 4 The focus is (1 + 3, 3), or (4, 3). Step 5 The directrix is a vertical line x = 1 – 3, or x = –2.

42 Example 8 Find the vertex, value of p axis of symmetry, focus, and directrix of the parabola. Then graph. Step 1 The vertex is (8, 4). Step , so 4p = –2 and p = – 1 4p 2 = –

43 Example 8 Continued Step 3 The graph has a vertical axis of symmetry, with equation x = 8, and opens downward. Step 4 The focus is or (8, 3.5). Step 5 The directrix is a horizontal line or y = 4.5.

44 Light or sound waves collected by a parabola will be reflected by the curve through the focus of the parabola, as shown in the figure. Waves emitted from the focus will be reflected out parallel to the axis of symmetry of a parabola. This property is used in communications technology.

45 Example 4: Using the Equation of a Parabola
The cross section of a larger parabolic microphone can be modeled by the equation What is the length of the feedhorn? x = y2. 1 132 The equation for the cross section is in the form x = y2, 1 4p so 4p = 132 and p = 33. The focus should be 33 inches from the vertex of the cross section. Therefore, the feedhorn should be 33 inches long.

46 The equation for the cross section is in the form
Check It Out! Example 4 Find the length of the feedhorn for a microphone with a cross section equation x = y2. 1 44 The equation for the cross section is in the form x = y2, 1 4p so 4p = 44 and p = 11. The focus should be 11 inches from the vertex of the cross section. Therefore, the feedhorn should be 11 inches long.

47 Lesson Quiz 1. Write an equation for the parabola with focus F(0, 0) and directrix y = 1. 2. Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola y – 2 = (x – 4)2, 1 12 then graph. vertex: (4, 2); focus: (4,5); directrix: y = –1; p = 3; axis of symmetry: x = 4


Download ppt "Vertex Form of Quadratics"

Similar presentations


Ads by Google