Quadratic Patterns and Function Notation

Linear Function Probes

Which graph has a slope of 1/3?

Linear Function ProbesHow do the slopes and y-intercepts of the graphs

below compare?

Slopes Y-intercepts

A. Q>R>S Q>R>S

B. S>Q>R Q>S>R

C. R>S>Q Q>S>R

D. R>Q>S R>S>Q

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:

77

4943

)44(5

x

ym

Now use one of the points to find the y-intercept:

y = −7x + b

−44 = −7(4) + b 5 = −7(−3) +b −44 = -28 + b 5 = 21 + b−44 + 28 = b 5 − 21 = b b = −16 b = −16

Starter:

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

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!

We can set this up using two variables: the term number (n) and the term value (tn).

term # (n)

1 2 3 4

value(tn)

4 7 10 13

We can find the COMMON DIFFERENCE between consecutive terms. This is given the symbol d.

−3 −3 −3

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 common difference:CD or d

The first term:a or t1

In general:tn = t1 + (n − 1)d

term # (n)

1 2 3 4

value(tn)

4 7 10 13

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 = 5 + (n − 1)(−4)tn = 5 − 4n + 4tn = −4n + 9

tn = 9 − 4nt100 = 9 – 4(100)t100 = –391

How many terms are in the following pattern:−9, −6, −3, 0, … , 81?

There are 31 terms in this pattern.

Solution:

t1 = −9CD = 3

tn = −9 + (n − 1)(3)tn = −9 + 3n − 3tn = 3n − 12

tn = 3n – 1281 = 3n – 1293 = 3n n = 31

1) Find the general rule for each pattern.

a) −8, −10, −12, …b) 1, 5, 9, 13, …c) 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) term n = 22 b) tn = 4n – 3 t50 = 197 4) term n = 42 c) tn = 0.5n t50 = 25

term #(n)

1 2 3 4 5

value(tn)

10 18 28 40 54

Deeper Patterns

NOT COMMON

Let’s find thecommon difference

What’s the pattern?

But what does this 2nd level common difference mean?

+8 +10 +12 +14

+2 +2 +2COMMON

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

(oops…that’s for a linear pattern)

In general:tn = an2 + bn + c

But where does the second level common difference fit into this equation?

We need to look at the general pattern of a quadratic

term # (n) 1 2 3 4 5

value, (tn)

tn = an2 + bn + c

a + b + c 4a +2b + c 9a +3b + c 16a +4b + c 25a +5b + c

This shows us that the common difference of ANY quadratic pattern is equal to 2a!!

+ 3a + b + 5a + b +7a + b + 9a + b

+ 2a + 2a + 2a

term #(n)

1 2 3 4 5

value(tn)

10 18 28 40 54

Deeper Patterns

NOT COMMON

What’s the pattern?

But what does this 2nd level common difference mean?

+8 +10 +12 +14

+2 +2 +2COMMON

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 !!!

tn = 1n2 + bn + c10 = 1(1)2 + b(1) + c10 = 1 + b + c 9 = b + c

So the equation representing this pattern (the nth term) can be written (so far):

tn = an2 + bn + c1

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 1, and where we see the tn we’ll put the 10.

Where we see an n in the equation we’ll put a 2, and where we see the tn we’ll put the 18.

tn = 1n2 + bn + c18 = 1(2)2 + b(2) + c18 = 4 + 2b + c14 = 2b + c

term # (n) 1 2 3 4 5

value (tn) 10 18 28 40 54

9 = b + c 14 = 2b + c9 = b + c

Let’s subtract these two equations to ELIMINATE the variable 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 + c9 = (5) + c4 = c

So here’s the equation that represents the pattern:tn = n2 + 5n + 4

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?

tn = n2 + 5n + 4 t4 = 42 + 5(4) + 4 t4 = 16 + 20 + 4 t4 = 40

WOW!

Can you find the 10th term?

154Can you find the 11th term? 180

Can you find which term has a value of 270? Not yet…

term # (n) 1 2 3 4 5

value (tn) 10 18 28 40 54

What is the 50th term of the pattern: 6, 14, 28, 48, 74…

If only we knew the equation of this pattern…

tn = 3n2 + bn + c

If we only knew the type of pattern…

It’s quadratic (tn = an2 + bn + c)

8 14 20 26 6 6 6

If only the 2nd level Common Difference of 6 was related to one of the constants…

So 6 = 2a, a = 3CD = 2a

If only we could find b and c…

(6) = 3(1)2 + b(1) + c (14) = 3(2)2 + b(2) + c3 = b + c 2 = 2b + c

b = −1 and c = 4 (by elimination or substitution)

when n = 1 t1 = 6 when n = 2 t2 = 14

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….

term # (n) 1 2 3 4 5

value (tn) −9 −16 −27 −42 −61

Find the 12th term in each quadratic pattern.

term # (n) 1 2 3 4 5

value (tn) 4 13 28 49 76

-306

433

In our last example, we set n = 50 to find t50 .

We will use function notation to simplify writing it out.

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.

instead of writing:tn = 3n2 − 1n + 4

ory = 3x2 − 1x + 4

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.”we can write: f(x) = 3x2 − 1x + 4

1. If f(x) = 4x − 5, then calculate

a. f(3)b. f(0)c. x if f(x) = 3d. x if f(x) = 0

a. 7b. −5c. x = 2d. x = 5/4

2. If f(x) = 3x − 2 and g(x) = 3 − 5x, then calculate

a. g(7)b. x if f(x) =7 c. x if g(x) = f(x)d. g(f(x))e. f(g(x))f. f(g(2))

a. −32b. x = 3c. 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

3. If f(x) = 2x2 + 3x – 8, calculate

a. the first 5 terms of the patternb. f(−4) c. f(0)d. x if f(x) = −9

a. −3, 6, 19, 36, 57b. 12c. −8d. stay tuned..