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1 6.6 Analyzing Graphs of Quadratic Functions. 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)

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Presentation on theme: "1 6.6 Analyzing Graphs of Quadratic Functions. 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)"— Presentation transcript:

1 1 6.6 Analyzing Graphs of Quadratic Functions

2 Graphing the parabola y = f (x) = ax 2 Consider the equation y = x (–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. x – 2 – 1012 y

3 For the function equation y = x 2, what is a ? a = 1. What if a does not equal 1? Consider the equation y = – 4x 2. 0– 4 (0, 0) – 16 y (–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. x – 2 – 1012 y What is a ? a = – 4

4 Properties of the Parabola 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 Algebraic Approach: y = – 4x 2 – 3 x – 2 – x x2-3-4x Numerical Approach: Graphical Approach: Graphing the parabola y = f (x) = ax 2 + k Consider the equation y = – 4x 2 – 3. What is a ? a = – 4 x Vertex (0, -3)

6 y = – 4x 2 – 3. x Vertex (0, -3) y = – 4x 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 = – 4x 2 is shifted down 3 units.

7 a = – 4. What effect does the 3 have on the function? y x y= – 4x 2 y= – 4(x – 3) 2 Consider the equation y = – 4(x – 3) 2. What is a ? The axis of symmetry is x = 3. x – 2 – 1012 –4 x (x-3) Numerical Approach: Axis of symmetry is shifted 3 units to the right and becomes x = 3

8 Vertex Form of a 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 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 = 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 y = (x – 3) 2

10 10 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 EquationVertexAxis of Symmetry y = x 2 or y = (x – 0) (0, 0)x = 0 y = x or y = (x – 0) (0, 2)x = 0 y = (x – 3) 2 or y = (x – 3) (3, 0)x = 3

11 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 = 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 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)] –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 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 = x 2. a. b. c. d. y = x 2 ab d c

14 14 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)) =a(3) =9a Now plug in a, h, & k!

15 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 T he h and the k represent the coordinates of the vertex in the form V(h, k). T he 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 16 Standard 9 Write a quadratic function in vertex form Write y = x 2 – 10x + 22 in vertex form. Then identify the vertex. y = x 2 – 10x + 22 Write original function. y + ? = (x 2 –10x + ? ) + 22 Prepare to complete the square. y + 25 = (x 2 – 10x + 25) + 22 Add – ( ) = (–5) 2 = 25 to each side. y + 25 = (x – 5) Write x 2 – 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

17 17 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 18 EXAMPLE 1 Write a quadratic function in vertex form A quadratic function for the parabola is y = (x – 1) 2 – 2. ANSWER

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

20 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) = – x = 2: y = (2 + 2) = – 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 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 22 GUIDED PRACTICE for Examples 1 and 2 3. f(x) = (x – 3) 2 –


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