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Quadratic Patterns and Function Notation

Quadratic Patterns and Function Notation. Linear Function Probes. Which graph has a slope of 1/3?. Linear Function Probes. How do the slopes and y-intercepts of the graphs below compare?. Starter:. Find the equation of the line that passes through the points (−3 , 5) and (4, − 44 ).

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Quadratic Patterns and Function Notation

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  1. Quadratic Patterns and Function Notation

  2. Linear Function Probes Which graph has a slope of 1/3?

  3. Linear Function Probes How do the slopes and y-intercepts of the graphs below compare?

  4. Starter: Find the equation of the line that passes through the points (−3, 5) and (4, −44). Solution: The equation of a line can be written as y = mx + b. To write the equation for this specific line, we need its slope (m) and y-intercept. The slope is the change in y divided by the change in x: Now use one of the points to find the y-intercept: y = −7x+ b −44 = −7(4) + b5 = −7(−3) +b −44 = -28 + b5 = 21 + b −44 + 28 = b5 −21 = b b= −16b = −16

  5. Starter: So the equation of the line that passes through the points (−3, 5) and (4, −44) is y = −7x−16

  6. Linear Patterns We can set this up using two variables: the term number (n) and the term value (tn). The equations we saw last class represent relationships. These can also be seen as patterns. Ex. 4, 7, 10, 13,… You can predict the next term, and for that matter, every term. You’re using the equation of this pattern whether you know it or not!

  7. Linear Patterns −3−3 −3 We can find the COMMON DIFFERENCE between consecutive terms. This is given the symbol d.

  8. In general: tn = t1 + (n −1)d Linear Patterns The pattern is: n = 1 4 + (0 × 3) n = 2 4 + (1 × 3) n = 3 4 + (2× 3) n = 4 4 + (3× 3) . . n = 100 4 + (99 × 3) So: tn = 4 + (n −1)3 The first term:a or t1 The common difference: CD or d

  9. Linear Patterns Examples What is the equation or general rule for the following pattern? 5, 1,−3, −7,… What is the 100th term of this pattern? Solution: t1 = 5 (or a = 5) CD = −4 (or d = −4) tn = 9 −4n t100 = 9 – 4(100) t100 = –391 tn = 5 + (n−1)(−4) tn = 5 −4n + 4 tn = −4n+ 9

  10. Linear Patterns Examples How many terms are in the following pattern: −9,−6,−3, 0, … , 81? Solution: t1 = −9 CD = 3 tn = 3n – 12 81 = 3n – 12 93 = 3n n = 31 tn = −9 + (n−1)(3) tn = −9 + 3n − 3 tn = 3n−12 There are 31 terms in this pattern.

  11. Linear Patterns Practice • 1) Find the general rule for each pattern. • −8, −10, −12, … • 1, 5, 9, 13, … • 0.5, 1, 1.5, … • 2) Find the 50th term for the patterns above. 3) Which term in the pattern −4, −1, 2, … has a value of 59? 4) Which term in the pattern tn = −2n − 1 has a value of −85? Answers: 1a) tn = -2n – 6 t50 = −106 3) termn = 22 b) tn = 4n – 3 t50 = 197 4) termn = 42 c) tn = 0.5nt50 = 25

  12. Deeper Patterns What’s the pattern? +8 +10+12 +14 NOT COMMON +2 +2+2 COMMON But what does this 2nd level common difference mean? Let’s find thecommon difference

  13. Quadratic Patterns A quadratic pattern is one where the common difference is found on the second level of difference. Every quadratic pattern can be represented by the equation: In general: tn = t1 + (n - 1)d In general: tn = an2 + bn + c (oops…that’s for a linear pattern) But where does the second level common difference fit into this equation?

  14. Quadratic Patterns tn = an2 + bn + c a + b + c 4a +2b + c 9a +3b + c 16a +4b + c 25a +5b + c + 3a+b+ 5a + b +7a + b + 9a + b + 2a + 2a + 2a This shows us that the common difference of ANY quadratic pattern is equal to 2a!! We need to look at the general pattern of a quadratic

  15. Deeper Patterns What’s the pattern? +8 +10+12 +14 NOT COMMON +2 +2+2 COMMON But what does this 2nd level common difference mean? Since the common difference occurs at the 2nd level, we know the relationship is quadratic (y = ax2 + bx + c). Since the common difference is 2, and since we just showed that CD = 2a, we know that a=1 !!!

  16. Quadratic Patterns Example 1 So the equation representing this pattern (the nth term) can be written (so far): tn = an2 + bn + c 1 But what are the values of b and c? To find b and c we need two data points from the pattern. Where we see an n in the equation we’ll put a 2, and where we see the tn we’ll put the 18. Where we see an n in the equation we’ll put a 1, and where we see the tn we’ll put the 10. tn = 1n2 + bn + c 18 = 1(2)2 + b(2) + c 18 = 4 + 2b + c 14 = 2b + c tn = 1n2 + bn + c 10 = 1(1)2 + b(1) + c 10 = 1 + b + c 9 = b + c

  17. Quadratic Patterns Example 1 9 = b + c 14 = 2b + c Let’s subtract these two equations to ELIMINATE the variable c. 9 = b + c 5 = b So we have another piece of the equation: tn= 1n2 + 5n + c We can now use one of these 2 equations to solve for ‘c’. 9 = b + c 9 = (5) + c 4 = c So here’s the equation that represents the pattern: tn= n2 + 5n + 4

  18. Quadratic Patterns Example 1 Let’s test it out! The fourth term in the pattern is 40. Does our equation predict this? In other words,if n = 4, does t4 = 40? Can you find the 10th term? 154 tn = n2 + 5n + 4 t4 = 42 + 5(4) + 4 t4 = 16 + 20 + 4 t4 = 40 Can you find the 11th term? WOW! 180 Can you find which term has a value of 270? Not yet…

  19. Quadratic Patterns Example 2 What is the 50th term of the pattern: 6, 14, 28, 48, 74… 8 14 20 26 6 6 6 If we only knew the type of pattern… It’s quadratic (tn = an2 + bn + c) If only the 2nd level Common Difference of 6 was related to one of the constants… So 6 = 2a, a = 3 CD = 2a If only we knew the equation of this pattern… tn = 3n2 + bn + c If only we could find b and c… (6) = 3(1)2 + b(1) + c (14) = 3(2)2 + b(2) + c 3 = b + c 2 = 2b + c b= −1 and c = 4 (by elimination or substitution) when n = 1 t1 = 6 when n = 2 t2 = 14

  20. Quadratic Patterns Example 2 What is the 50th term of the pattern: 6, 14, 28, 48, 74… So tn = 3n2− 1n+ 4 represents the above pattern. To find the 50th term plug in n = 50 and calculate t50 t50 = 3(50)2− 1(50) + 4 = 7454 BUT… how can we find which term has value of 756? Stay tuned….

  21. Quadratic Patterns Practice Find the 12th term in each quadratic pattern. -306 433

  22. Function Notation The term “function” means a relationship between two variables. One variable is responds to the other. Say that the responding variable (y) depends on the variable x, through the function, f, as in our last example. • This allows us to write things like: • f(50) which means: “What is the y value whenx = 50?” • f(x) = 756 which says: “Find the value of x which gives a y value of 756.” instead of writing: tn = 3n2− 1n+ 4 or y = 3x2− 1x+ 4 we can write: f(x) = 3x2− 1x+ 4 In our last example, we set n = 50 to find t50 . We will use function notation to simplify writing it out.

  23. Function Notation Examples • If f(x) = 4x− 5, then calculate • f(3) • f(0) • x if f(x) = 3 • x if f(x) = 0 • 7 • −5 • x = 2 • x = 5/4 • If f(x) = 3x − 2 and g(x) = 3 −5x, then calculate • g(7) • x if f(x) =7 • x if g(x) = f(x) • d. g(f(x)) • e. f(g(x)) • f. f(g(2)) • −32 • x = 3 • x = 5/8 d. g(f(x)) = 3 – 5(3x − 2) = −15x +13 e. f(g(x)) = 3(3 − 5x) − 2 = −15x +7 f. f(g(2)) = 3(3 − 5(2)) − 2 = −23

  24. More (Fun)ction Notation Examples • 3. If f(x) = 2x2 + 3x – 8, calculate • the first 5 terms of the pattern • f(−4) • f(0) • x if f(x) = −9 • −3, 6, 19, 36, 57 • 12 • −8 • stay tuned..

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