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GRAPHING PARABOLAS. Pattern Probes 1. Which of the following relationships describes the pattern: 1,-1,-3,-5…? A. t n = -2n B. t n = -2n +3 C. t n = 2n.

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Presentation on theme: "GRAPHING PARABOLAS. Pattern Probes 1. Which of the following relationships describes the pattern: 1,-1,-3,-5…? A. t n = -2n B. t n = -2n +3 C. t n = 2n."— Presentation transcript:

1 GRAPHING PARABOLAS

2 Pattern Probes 1. Which of the following relationships describes the pattern: 1,-1,-3,-5…? A. t n = -2n B. t n = -2n +3 C. t n = 2n – 3 D. t n = 2n

3 Pattern Probes 2. What is the 62th term of the pattern 1,9,17,25,…? A. 489 B. 496 C. 503 D. 207

4 A linear pattern is given by t n = t 1 +(n-1)d Quadratics Patterns Last time we learned that: A quadratic pattern is defined by t n = an 2 + bn +c The CD = 2a for a quadratic A quadratic pattern has its CD on D 2. A linear pattern has a common difference on D 1. Note: Another phrase for linear pattern is an arithmetic sequence.

5 Let’s Graph (Put on your graphing shoes and graph in blue?) Quadratics, as we know, do not increase by the same amount for every increase in x. (How do we know this?) Therefore, their graphs are not straight lines, but curves. In fact, if you’ve ever thrown a ball, watched a frog jump, or taken Math 12, you’ve seen a quadratic graph! Consider the function f(x) = x 2 - the simplest quadratic (Why?) The pattern would be Term #, n12345 Value, t n 1491625 Its table of values would be y = x 2 xy -39 -24 1 00 11 24 39

6 Graphing Cont xy -39 -24 1 00 11 24 39 We can plot these points to get the PARABOLA Vertex Once we have the vertex, the rest of the points follow the pattern: Over 1, up 1 Over 2, up 4 Over 3, up 9 Axis of Symmetry: x = 0 Domain: xε R Range: y ε [0,∞) or y ≥ 0

7 Transformations of y= x 2 That parabola (which represents the simplest quadratic function), can be Stretched vertically Slid horizontally Reflected Slid verticallyThese are called transformations: Ref, VS, HT, VT, These show up (undone) in the transformational form of the equation: The form y = ax 2 + bx + c is called the GENERAL FORM

8 Transformations Ex. Graph the function What are the transformations? Reflection: No (the left side is not multiplied by -1) Vertical Stretch, VS? Yes VS = 2 Vertical Translational, VT? Yes VT = -3 Horizontal Translational, HT? Yes HT = 1 So we can use these to get a MAPPING RULE: (x,y)  (x+1, 2y-3) We now take the old table of values and do what the mapping rule says.

9 Transformations Ex. Graph the function So we can use these to get a MAPPING RULE: (x,y)  (x+1, 2y-3) We now take the old table of values and do what the mapping rule says. y = x 2 xy -39 -24 1 00 11 24 39 ½(y+3) = (x-1) 2 xy -215 5 0 1-3 2 35 415

10 Transformations Ex. Graph the function ½(y+3) = (x-1) 2 xy -215 5 0 1-3 2 35 415 y = x 2 xy -39 -24 1 00 11 24 39 New Vertex (1,-3) that is (HT, VT)

11 Transformations Ex. Graph the function ½(y+3) = (x-1) 2 xy -215 5 0 1-3 2 35 415 New Vertex (1,-3) that is (HT, VT) Once we find the new vertex, the rest of the points follow the pattern Over 1, up 1 x VS Over 2, up 4 x VS Over 3, up 9 x VS Domain: x ε R Range: y ε [-3,∞) Axis of symmetry: x = 1

12  Graph the following. State the axis of symmetry, range, domain and the y-intercept of each.  A)  B)  C)

13 A) Ref: No VS: 2 VT: 3 HT: 2 (x,y)  (x+2,2y+3) ½(y-3) = (x-2) 2 xy 21 011 15 23 3 5 4 521 Domain: x ε R Range: y ε [3,∞) Axis of symmetry: x = 2 Y-int (0,11)

14 B) Ref: Yes VS: 1/2 VT: 0 HT: -1 (x,y)  (x-1,-0.5y) -2y = (x+1) 2 xy -4-4.5 -3-2 -0.5 0 0 -0.5 1-2 2-4.5 Domain: x ε R Range: y ε (-∞,0] Axis of symmetry: x = -1 Y-int (0,-0.5)

15 C) Ref: Yes VS: 1 VT: 4 HT: -6 (x,y)  (x-6,-y+4) -(y-4) = (x+6) 2 xy -9-5 -80 -73 -64 -53 -40 -3-5 Domain: x ε R Range: y ε (∞,4] Axis of symmetry: x = -6 Y –int (0,-32)

16 Going backwards Find the equation of this graph: Ans: 2(y+1)=(x-1) 2

17 Going Backwards a. What quadratic function has a range of y ε(- ∞,6], a VS of 4 and an axis of symmetry of x = -5? b. What quadratic function has a vertex of (2,-12) and passes through the point (1,-9)? a) b)

18 Solving for x using transformational form We can use transformational form to solve for x- values given certain y-values. Example: The parabola passes through the point (x,2). What are the values of x?

19 Solving for x using transformational form Find the x-intercepts of the following quadratic functions:

20 Changing Forms General Form: y = ax 2 +bx + c Transformational Form: Standard Form:

21 What are you good for? General Form: f(x)=y = ax 2 +bx + c Transformational Form: Good for: Getting the mapping rule Graphing the function’s parabola Draw backs: Functions are usually written in the f(x) notation. Getting the equation of the graph Good for: Finding the y-intercept (0,c) Finding the y-intercept (0,c) Finding the x-intercept(s) (soon) Finding the x-intercept(s) (soon) Finding the vertex (soon) Finding the vertex (soon) Standard Form: Good for:Nothing

22 Going To General Form

23

24 Going to General Form Ex. Put the following into general form:


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