Presentation is loading. Please wait.

Presentation is loading. Please wait.

Section 2.2 Quadratic Functions

Similar presentations


Presentation on theme: "Section 2.2 Quadratic Functions"— Presentation transcript:

1 Section 2.2 Quadratic Functions

2 Graphs of Quadratic Functions

3 Quadratic Functions Quadratic functions of the standard form
f(x) = ax2 + bx + c The graph of a quadratic function is called a parabola. Parabola’s are “U” shaped. Parabolas have reflection symmetry with respect to the line called the axis of symmetry. If folded over this line the two halves match exactly.

4 Graphs of Quadratic Functions Parabolas
Vertex Minimum -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 -10 x y Maximum Axis of symmetry -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 -10 x y Domain will always be: (-∞, ∞)

5 Graphing Quadratic Functions in Vertex Form

6 Vertex Form of a Quadratic
f (x) = a(x – h)2 + k, a ≠ 0, where (h, k) is the vertex of the parabola. The line of symmetry is x = h. If a > 0, then the parabola opens up. If a < 0, then the parabola opens down. Find x-intercepts by solving f (x) = 0. Find the y-intercepts by solving f (0). Plot the intercepts, the vertex, and any additional points necessary. Connect these points with a smooth “U” shaped curve.

7 Seeing the Transformations
f (x) = (x - 5) 2 + 3 f(x) = x2 a > 0 so the parabola opens upward and the vertex is a minimum Vertex: (5, 3) Vertex: (0, 0) Up 3 Right 5 Axis of symmetry x = 5

8 Seeing the Transformations
Axis of symmetry x = -8 a < 0 so the parabola opens downward and the vertex is a maximum Vertex: (-8, -4) Vertex: (0, 0) f(x) = - (x + 8)2 - 4 Left 8 f(x) = -x2 Down 4 Page 313 Problems 1-4

9 a < 0 so parabola opens down, and vertex is a maximum
Using Vertex Form -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 -10 x y a < 0 so parabola opens down, and vertex is a maximum

10 Now change into standard form.
Using Vertex Form -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 -10 x y Now change into standard form. Find the DOMAIN. Find the RANGE.

11 Then convert to standard form. Example 1
Graph the quadratic function f (x) = -(x+2)2 + 4 and state the vertex, axis of symmetry, x- and y-intercepts. Then convert to standard form. Example 1 (-2, 4) (-4, 0) (0, 0) x = -2 Find the DOMAIN. Find the RANGE.

12 Graph the quadratic function f (x) = (x – 3)2 – 4
Example 2 Graph the quadratic function f (x) = (x – 3)2 – 4 (0, 5) (6, 5) x = 3 (1, 0) (5, 0) (3, - 4) Find the DOMAIN. Find the RANGE. Page 313 Problems 17-26

13 Graphing Quadratics in the Standard form f (x) = ax2 + bx + c

14 Standard Form We identify the vertex of a parabola whose equation is in the standard form by completing the square. See next slide for finished equation

15 Ta Da!! Vertex Form From Standard Form

16 Finding the Vertex from Standard Form
When the equation is f (x) = ax2 + bx + c, the vertex is… You would then proceed to graph the equation the same way as before. Determine if it opens up or down. Plot the vertex. Find the x-intercepts and y-intercepts and plot them. Connect the points with your smooth curve.

17 Using the form f(x) = ax2 + bx + c
-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 -10 x y Vertex form: f (x) = a(x – h)2 + k f (x) = 1 (x - - 1) 2 + 0 f (x) = (x + 1) 2 a > 0 so parabola opens up, vertex is a minimum Find the DOMAIN. Find the RANGE.

18 a < 0 so parabola opens down, vertex is a maximum
Example 3 Find the vertex, axis of symmetry, x-intercepts and y-intercept, open up/down of the function f (x) = - x2 - 3x + 7 a < 0 so parabola opens down, vertex is a maximum f (x) = -1 (x + 1.5) Find the DOMAIN. Find the RANGE.

19 Quadratics are Polynomials Domain: (-∞, ∞) Parabola opens Down Range:
Graph the function f (x) = - x2 - 3x Use the graph to identify the domain and range. (-1.5, 9.25) Quadratics are Polynomials Domain: (-∞, ∞) (-3, 7) (0, 7) Parabola opens Down Range: (-∞, 9.25] Page 313 Problems 27-38

20 Minimum and Maximum Values of Quadratic Functions

21 Minimum and Maximum Values of Quadratic Functions
Consider the quadratic function f (x) = ax2 + bx + c or f (x) = a(x-h)2 + k If a > 0, then f has a minimum that occurs at the vertex. If a < 0, then f has a maximum that occurs at the vertex. In each case, the value gives the location of the minimum or maximum value. To find the domain, remember a quadratic is a polynomial so it will ALWAYS be (-∞, ∞). The Range depends if the vertex is a minimum or maximum. If it is a minimum the Range will be… If it is a maximum, the Range will be…

22 Domain is (-∞, ∞). Range is (-∞, -14/3].
Example 5 For the function f (x) = - 3x2 + 2x – 5 Without graphing determine if it has a min or max, then find it. Identify the function’s domain and range. a = -3, so parabola opens downward and vertex is a maximum. Domain is (-∞, ∞). Range is (-∞, -14/3]. Page 313 Problems 39-44

23 For each quadratic function…
Sketch the graph precisely and accurately. State the vertex. State the axis of symmetry. State the Domain. State the Range. State the x-intercepts. State the y-intercept. Domain: (-∞, ∞) Range: [-4, ∞)

24 Graphing Calc – Min/Max
Type the equation into f (x) = −3x2 + 2x – 5 Hit b and select option 6: Analyze Graph, then choose option 3: Maximum because our graph opens down. Move the finger to the left (lower bound) of the vertex and click (or press enter). Repeat for the upper bound, but on the right side of the vertex. What happens?

25 Applications of Quadratic Functions

26 Thinking is needed! 45. Vertex (-1, -2) Domain (-∞, ∞) Range [-2, ∞)
Complete problems on page 314. 45. Vertex (-1, -2) Domain (-∞, ∞) Range [-2, ∞) 46. Vertex (-3, -4) Domain (-∞, ∞) Range (-∞, -4] 47. Vertex (10, -6) Domain (-∞, ∞) Range (-∞, -6] 48. Vertex (-6, 18) Domain (-∞, ∞) Range [18, ∞)

27 Graphing Calculators are needed!
Using the calculators complete problems on page 406 17. Vertex (7, -57) Domain (-∞, ∞) Range (-∞, -57] 18. Vertex (-3, 685) Domain (-∞, ∞) Range [685, ∞)

28

29 Dimensions for Max Area 16 yds by 16 yds
Example 6 You have 64 yards of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area? Area = Length ∙ Width Area = x ∙ (32 – x) Area = 32x – x2 Now we find the “vertex” Dimensions for Max Area 16 yds by 16 yds Resulting in an area of 256 sq yds Page 314 – 315 Problems 57-76

30 You can always use Graphing Calculator
Problem continued on the next slide

31 Graphing Calculator- continued

32 Quadratic Regression on TI-Nspire
M Kwh 1 8.79 2 8.16 3 6.25 4 5.39 5 6.78 6 8.61 7 8.78 8 10.96 Put the data from the table into the table on your calculator. When you are done, hit /~ to add 5: Data & Statistics page. Click to add your labels on the x- and y-axes. Now select b, 4: Analyze, then 6:Regression. Make sure you select the appropriate regression for the shape of our data. What is your equation?

33 (d) (a) (b) (c) (d)

34 (b) (a) (b) (c) (d) More practice can be found on page Prob


Download ppt "Section 2.2 Quadratic Functions"

Similar presentations


Ads by Google