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![Graph a Quadratic Function in Vertex Form Analyze y = (x + 2) 2 + 1. 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 = x 2 , 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.](https://image.slidesharecdn.com/6-6analyzinggraphsofquadraticfunctions-100629103202-phpapp02/75/6-6-analyzing-graphs-of-quadratic-functions-12-2048.jpg)
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6.6 analyzing graphs of quadratic functions
This document discusses analyzing and graphing quadratic functions. It defines key terms like vertex, axis of symmetry, and vertex form. It explains that the graph of y=ax^2 is a parabola, and how the value of a affects whether the parabola opens up or down. It also describes how to graph quadratic functions in vertex form by plotting the vertex and axis of symmetry, and using symmetry.
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6.6 analyzing graphs of quadratic functions
- 1. 6.6 Analyzing Graphs of Quadratic Functions
- 2. Graphing the parabola y = f ( x ) = ax 2 Consider the equation y = x 2 0 1 4 1 (–1, 1) (0, 0) (1, 1) (2, 4) y x 4 (–2, 4) Axis of symmetry: x = 0 ( y=x 2 is symmetric with respect to the y-axis ) Vertex (0, 0) When a > 0 , the parabola opens upwards and is called concave up. The vertex is called a minimum point. y 2 1 0 – 1 – 2 x
- 3. For the functionequation y = x 2 , what is a ? a = 1 . What if a does not equal 1? Consider the equation y = – 4 x 2 . 0 – 4 – 4 (0, 0) – 16 y – 16 (–2, –16) (–1, –4) (1, –4) (–4, –16) x When a < 0 , the parabola opens downward and is called concave down. The vertex is a maximum point. What is a ? a = – 4 y 2 1 0 – 1 – 2 x
- 4. Properties of theParabola f ( x ) = ax 2 The graph of f ( x ) = ax 2 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 ) = x 2 . If a is between 0 and 1 (0 < a < 1), the parabola is wider than the parabola f ( x ) = x 2 .
- 5. Graphing the parabola y = f ( x ) = ax 2 + k Algebraic Approach: y = – 4 x 2 – 3 Numerical Approach: Graphical Approach: Consider the equation y = – 4 x 2 – 3 . What is a ? a = – 4 x Vertex (0, -3) -16 -4 0 -4 -16 -4 x 2 -19 -7 -3 -7 -19 - 4 x 2 - 3 2 1 0 – 1 – 2 x
- 6. y = – 4 x 2 – 3 . x Vertex (0, -3) y = – 4 x 2 In general the graph of y = ax 2 + k is the graph of y = ax 2 shifted vertically k units. If k > 0, the graph is shifted up. If k < 0, the graph is shifted down. (P. 267) The graph y = – 4 x 2 is shifted down 3 units.
- 7. a =– 4. What effect does the 3 have on the function? y x y = – 4 x 2 y = – 4( x – 3) 2 Consider the equation y = – 4( x – 3) 2 . What is a ? The axis of symmetry is x = 3. Numerical Approach: Axis of symmetry is shifted 3 units to the right and becomes x = 3 -16 -4 0 -4 -16 – 4 x 2 2 1 0 – 1 – 2 x -4 -16 -36 -64 -100 - 4 ( x- 3) 2 -36 3 0
- 8. Vertex Form ofa Quadratic Function The quadratic function f ( x ) = a ( x – h ) 2 + k, a 0 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 Graphsof Quadratic Functions A family of graphs – a group of graphs that displays one or more similar characteristics. Parent graph – y = x 2 Notice that adding a constant to x 2 moves the graph up . Notice that subtracting a constant from x before squaring it moves the graph to the right . y = x 2 y = x 2 + 2 y = (x – 3) 2
- 10. Vertex Form Eachfunction 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 x = 3 ( 3 , 0 ) y = (x – 3 ) 2 or y = (x – 3 ) 2 + 0 x = 0 ( 0 , 2 ) y = x 2 + 2 or y = (x – 0 ) 2 + 2 x = 0 ( 0 , 0 ) y = x 2 or y = (x – 0 ) 2 + 0 Axis of Symmetry Vertex Equation
- 11. Graph Transformations Asthe values of h and k change, the graph of y = a(x – h) 2 + k is the graph of y = x 2 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 QuadraticFunction in Vertex Form Analyze y = (x + 2) 2 + 1. 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 = x 2 , 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 Howdoes 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 = x 2 . a. b. c. d. y = x 2 a b d c
- 14. Example: Write aquadratic 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 Writea 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.
- 16. Standard 9 Writea quadratic function in vertex form Write y = x 2 – 10 x + 22 in vertex form. Then identify the vertex. y = x 2 – 10 x + 22 Write original function. y + ? = ( x 2 – 10 x + ? ) + 22 Prepare to complete the square. y + 25 = ( x 2 – 10 x + 25 ) + 22 y + 25 = ( x – 5) 2 + 22 Write x 2 – 10 x + 25 as a binomial squared. y + 3 = ( x – 5) 2 Write in vertex form. Add – 10 2 2 ( ) = (–5) 2 = 25 to each side. The vertex form of the function is y + 3 = ( x – 5) 2 . The vertex is (5, – 3). ANSWER
- 17. EXAMPLE 1 Writea 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 = 4 a – 2 Simplify coefficient of a . 1 = a Solve for a .
- 18. EXAMPLE 1 Writea quadratic function in vertex form A quadratic function for the parabola is y = ( x – 1) 2 – 2 . ANSWER
- 19. EXAMPLE 1Graph a quadratic function in vertex form Graph y – 5 = – ( x + 2) 2 . SOLUTION STEP 1 STEP 2 Plot the vertex ( h , k ) = ( – 2, 5) and draw the axis of symmetry x = – 2. 14 Identify the constants a = – , h = – 2, and k = 5. Because a < 0, the parabola opens down. 14
- 20. EXAMPLE 1Graph a quadratic function in vertex form STEP 3 Evaluate the function for two values of x . 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. x = 0 : y = ( 0 + 2) 2 + 5 = 4 14 – x = 2 : y = ( 2 + 2) 2 + 5 = 1 14 –
- 21. GUIDED PRACTICE forExamples 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 forExamples 1 and 2 3. f ( x ) = ( x – 3) 2 – 4 12