Download presentation

1
**6.6 Analyzing Graphs of Quadratic Functions**

2
**Graphing the parabola y = f (x) = ax2**

Consider the equation y = x2 Axis of symmetry: x = ( y=x2 is symmetric with respect to the y-axis ) x – 2 – 1 1 2 y 4 4 1 1 (–2, 4) (–1, 1) (0, 0) (1, 1) (2, 4) When a > 0, the parabola opens upwards and is called concave up. The vertex is called a minimum point. y Vertex(0, 0) x

3
**For the function equation y = x2 , what is a ?**

a = 1 . What if a does not equal 1? Consider the equation y = – 4x2 . What is a ? a = – 4 x – 2 – 1 1 2 y y – 4 – 4 – 16 – 16 (–2, –16) (–1, –4) (0, 0) (1, –4) (–4, –16) x When a < 0, the parabola opens downward and is called concave down. The vertex is a maximum point.

4
**Properties of the Parabola f (x) = ax2**

The graph of f (x) = ax2 is a parabola with the vertex at the origin and the y axis as the line of symmetry. If a is positive, the parabola opens upward, if a is negative, the parabola opens downward. If a is greater than 1 (a > 1), the parabola is narrower then the parabola f (x) = x2. If a is between 0 and 1 (0 < a < 1), the parabola is wider than the parabola f (x) = x2.

5
**Graphing the parabola y = f (x) = ax2 + k**

Consider the equation y = – 4x2 – 3 . What is a ? a = – 4 Graphical Approach: Numerical Approach: x – 2 – 1 1 2 -4x2 -16 -4 -4x2-3 -19 -7 -3 x Vertex(0, -3) Algebraic Approach: y = – 4x2 – 3

6
**The graph y = – 4x2 is shifted down 3 units.**

Vertex(0, -3) y = – 4x2 – 3 . In general the graph of y = ax2 + k is the graph of y = ax2 shifted vertically k units. If k > 0, the graph is shifted up. If k < 0, the graph is shifted down. (P. 267)

7
**Consider the equation y = – 4(x – 3)2 . What is a ? **

a = – 4. What effect does the 3 have on the function? The axis of symmetry is x = 3. Numerical Approach: y x – 2 – 1 1 2 –4 x2 -16 -4 3 -36 x -4 (x-3) 2 -100 -64 -36 -16 -4 Axis of symmetry is shifted 3 units to the right and becomes x = 3 y= – 4x2 y= – 4(x – 3)2

8
**Vertex Form of a Quadratic Function**

The quadratic function f(x) = a(x – h)2 + k, a The graph of f is a parabola . Axis is the vertical line x = h. Vertex is the point (h, k). If a > 0, the parabola opens upward. If a < 0, the parabola opens downward.

9
**6.6 Analyzing Graphs of Quadratic Functions**

A family of graphs – a group of graphs that displays one or more similar characteristics. Parent graph – y = x2 Notice that adding a constant to x2 moves the graph up. Notice that subtracting a constant from x before squaring it moves the graph to the right. y = x2 + 2 y = (x – 3)2 y = x2

10
**Vertex Form (x – h)2 + k – vertex form**

Each function we just looked at can be written in the form (x – h)2 + k, where (h , k) is the vertex of the parabola, and x = h is its axis of symmetry. (x – h)2 + k – vertex form Equation Vertex Axis of Symmetry y = x2 or y = (x – 0)2 + 0 (0 , 0) x = 0 y = x2 + 2 or y = (x – 0)2 + 2 (0 , 2) y = (x – 3)2 or y = (x – 3)2 + 0 (3 , 0) x = 3

11
**Graph Transformations**

As the values of h and k change, the graph of y = a(x – h)2 + k is the graph of y = x2 translated. | h | units left if h is negative, or |h| units right if h is positive. | k | units up if k is positive, or | k | units down if k is negative.

12
**Graph a Quadratic Function in Vertex Form**

Analyze y = (x + 2) Then draw its graph. This function can be rewritten as y = [x – (-2)]2 + 1. Then, h = -2 and k = 1 The vertex is at (h , k) = (-2 , 1), the axis of symmetry is x = -2. The graph is the same shape as the graph of y = x2, but is translate 2 units left and 1 unit up. Now use the information to draw the graph. Step 1 Plot the vertex (-2 , 1) Step 2 Draw the axis of symmetry, x = -2. Step 3 Find and plot two points on one side of the axis symmetry, such as (-1, 2) and (0 , 5). Step 4 Use symmetry to complete the graph.

13
**Graph Transformations**

How does the value of a in the general form y = a(x – h)2 + k affect a parabola? Compare the graphs of the following functions to the parent function, y = x2. a. b. c. d. a b y = x2 d c

14
**Since you know the vertex, use vertex form! y=a(x-h)2+k **

Example: Write a quadratic function for a parabola with a vertex of (-2,1) that passes through the point (1,-1). Since you know the vertex, use vertex form! y=a(x-h)2+k Plug the vertex in for (h,k) and the other point in for (x,y). Then, solve for a. -1=a(1-(-2))2+1 -1=a(3)2+1 -2=9a Now plug in a, h, & k!

15
Standard 9 Write a quadratic function in vertex form Vertex form- Is a way of writing a quadratic equation that facilitates finding the vertex. y – k = a(x – h)2 The h and the k represent the coordinates of the vertex in the form V(h, k). The “a” if it is positive it will mean that our parabola opens upward and if negative it will open downward. A small value for a will mean that our parabola is wider and vice versa. 15

16
**( ) Standard 9 Write a quadratic function in vertex form**

Write y = x2 – 10x + 22 in vertex form. Then identify the vertex. y = x2 – 10x + 22 Write original function. y + ? = (x2 –10x + ? ) + 22 Prepare to complete the square. Add –10 2 ( ) = (–5) 25 to each side. y + 25 = (x2 – 10x + 25) + 22 y + 25 = (x – 5)2 + 22 Write x2 – 10x + 25 as a binomial squared. y + 3 = (x – 5)2 Write in vertex form. The vertex form of the function is y + 3 = (x – 5)2. The vertex is (5, –3). ANSWER 16

17
**Write a quadratic function in vertex form**

EXAMPLE 1 Write a quadratic function in vertex form Write a quadratic function for the parabola shown. SOLUTION Use vertex form because the vertex is given. y – k = a(x – h)2 Vertex form y = a(x – 1)2 – 2 Substitute 1 for h and –2 for k. Use the other given point, (3, 2), to find a. 2 = a(3 – 1)2 – 2 Substitute 3 for x and 2 for y. 2 = 4a – 2 Simplify coefficient of a. 1 = a Solve for a.

18
EXAMPLE 1 Write a quadratic function in vertex form ANSWER A quadratic function for the parabola is y = (x – 1)2 – 2.

19
EXAMPLE 1 Graph a quadratic function in vertex form 14 Graph y – 5 = – (x + 2)2. SOLUTION STEP 1 Identify the constants a = – , h = – 2, and k = 5. Because a < 0, the parabola opens down. 14 STEP 2 Plot the vertex (h, k) = (– 2, 5) and draw the axis of symmetry x = – 2.

20
EXAMPLE 1 Graph a quadratic function in vertex form STEP 3 Evaluate the function for two values of x. x = 0: y = (0 + 2)2 + 5 = 4 14 – x = 2: y = (2 + 2)2 + 5 = 1 14 – Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry. STEP 4 Draw a parabola through the plotted points.

21
GUIDED PRACTICE for Examples 1 and 2 Graph the function. Label the vertex and axis of symmetry. 1. y = (x + 2)2 – 3 2. y = –(x + 1)2 + 5

22
GUIDED PRACTICE for Examples 1 and 2 12 3. f(x) = (x – 3)2 – 4

Similar presentations

OK

1 Copyright © Cengage Learning. All rights reserved. 2 Polynomial and Rational Functions.

1 Copyright © Cengage Learning. All rights reserved. 2 Polynomial and Rational Functions.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on different types of dance forms productions Ppt on number system for class 9 Ppt on non biodegradable waste recycling Ppt on new year resolutions Ppt on road accidents in kenya Ppt on life of guru gobind singh ji Ppt on english grammar in hindi Ppt on conservation of momentum equations Ppt on save water free download Ppt on nutrition in plants for class 7