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Graphing Quadratic Functions: Parabolas. Which of the following relationships describes the pattern: 1, −1 , −3 , −5 …? A. t n = −2 n B. t n = −2 n + 3 C. t n = 2 n – 3 D. t n = 2 n. Pattern Probes. 2. What is the 62th term of the pattern: 1, 9, 17, 25, …? A. 489 B. 496 - PowerPoint PPT Presentation

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2. What is the 62th term of the pattern:1, 9, 17, 25, …?

A. 489B. 496C. 503D. 207

1. Which of the following relationships describes the pattern:1, −1, −3, −5 …?

A. tn = −2nB. tn = −2n + 3C. tn = 2n – 3D. tn = 2n

Pattern Probes

1. Which of the following relationships describes the pattern:1, −1, −3, −5 …?

A. tn = −2nB. tn = −2n + 3C. tn = 2n – 3D. tn = 2n

Pattern Probes2. What is the 62th term of

the pattern:1, 9, 17, 25, …?

A. 489B. 496C. 503D. 207

Last time we learned that:

A linear pattern has a common difference on level 1.A linear pattern is given by tn = t1 + (n − 1)d

A quadratic pattern has its common difference on level 2.The CD = 2a for a quadraticThe quadratic pattern is defined by tn = an2 + bn + cThe general form of the quadratic function is defined by

f(x) = ax2 + bx + c

Note: Another phrase for linear pattern is an arithmetic sequence.

Let’s Graph

Quadratics, as we know, do not increase by the same amount for every increase in x (or n). (How do we know this?)Therefore, their graphs are not straight lines,but curves, called parabolas. In fact, if you’veever thrown a ball, or watched a frog jumpyou’ve seen a quadratic graph!

Consider the functionf(x) = x2

a = 1b = 0c = 0

y = x2

x y–3 9

–2 4

–1 1

0 0

1 1

2 4

3 9

Its table of values would be:

Let’s Graph

Quadratics, as we know, do not increase by the same amount for every increase in x (or n). (How do we know this?)Therefore, their graphs are not straight lines,but curves, called parabolas. In fact, if you’veever thrown a ball, or watched a frog jumpyou’ve seen a quadratic graph!

Consider the functionf(x) = x2

The pattern would be

term #, (n) 1 2 3 4 5value, (tn) 1 4 9 16 25

Its table of values would be:

a = 1b = 0c = 0

y = x2

x y–3 9

–2 4

–1 1

0 0

1 1

2 4

3 9

Let’s Graph

Once we have the vertex, the rest of the points follow the pattern:over 1 (from the vertex), up 1 (from the vertex)

over 2 (from the vertex), up 4 (from the vertex)

over 3 (from the vertex), up 9 (from the vertex)

over 7 (from the vertex), up 49 (from the vertex)

y = x2

x y–3 9–2 4–1 10 01 1 2 43 9

Properties of the graph y = x2

axis of symmetry:the line x = 0

vertex:the point (0, 0)

domain:{x R}

range:{y ≥ 0, y R}

Transformations of y = x2

This simplest parabola can be:

We give these transformations short forms: VS, HT, and VT,

These show up (undone) in the transformational form of the equation:

2HTVTVS1

xy

TransformationsEx. Graph the function:(FYI, its general formis y = 2x2 – 4x – 1)

2)1(321

xy

What are the transformations that have been applied to the base graph of y = x2?

Vertical StretchVS = 2 (positive, so no reflection)

Vertical TranslationalVT = –3

Horizontal TranslationalHT = 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.

old x and y values, from y = x2

new x and y values

TransformationsEx. Graph the function:(FYI, its general formis y = 2x2 – 4x – 1)

2)1(321

xy

Apply the mapping rule:(x, y)→ (x + 1, 2y – 3)

y = x2

x y–3 9–2 4–1 10 01 1 2 43 9

“old” x and y values(from the base parabola)

½(y + 3) = (x – 1)2

x y–2 15–1 50 –11 –32 –13 54 15

New x and y values for this transformed parabola

TransformationsEx. Graph the function: 2)1(3

21

xy

½(y + 3) = (x – 1)2

x y–2 15–1 50 –11 –32 –13 54 15

Transformationson y = x2:VS = 2VT = –3HT = 1

TransformationsEx. Graph the function: 2)1(3

21

xy

Short-cut:Once we have the vertex,

the rest of the points follow the pattern:

over 1, up 1 × 2over 2, up 4 × 2over 3, up 9 × 2

.

.

.over #, up # × VS

(again, always from the vertex)

TransformationsEx. Graph the function: 2)1(3

21

xy

vertex (1, –3)(that is (HT, VT))

axis of symmetry x = 1(that is the line x = HT)

domain {x R}(unchanged)

range {y ≥ –3, y R}(that is {y ≥ VT, y R })

Graphing Parabolas: PracticeGraph the following. State the vertex, axis of symmetry, range,

domain and the y-intercept of each.

A) (or y = 2x2 – 8x + 11, in general form)

B) (or y = – ½x2 – x – ½ , in general form)

C) (or y = –x2 – 12x – 32, in general form)

2)2(321

xy

2)1(2 xy

2)6()4( xy

2)2(321

xy (or y = 2x2 – 8x + 11 in general form)

½(y – 3) = (x – 2)2

x y

-1 21

0 11

1 5

2 3

3 5

4 11

5 21

VS = 2HT = 2VT = 3

(x, y) → (x + 2, 2y + 3)

domain: {x R}range: {y ≥ 3, y R}axis of symmetry: x = 2y-intercept (0,11)

A)

(or y = -½x2 - x - ½ , in general form)

VS = –½HT = –1VT = 0

(x, y) → (x – 1, – ½y)

domain: {x R}range: {y ≤ 0, y R}axis of symmetry: x = –1y-intercept (0, – ½)

2)1(2 xy

-2y = (x + 1)2

x y

-4 -4.5

-3 -2

-2 -0.5

-1 0

0 -0.5

1 -2

2 -4.5

B)

(or y = –x2 – 12x – 32 in general form)

½(y – 3) = (x – 2)2

x y

-1 21

0 11

1 5

2 3

3 5

4 11

5 21

VS = –1 HT = –6VT = 4

(x, y) → (x – 6, – y + 4)

domain: {x R}range: {y ≤ 4, y R}axis of symmetry: x = –6y-intercept (0, –32)

2)6()4( xyC)

Going backwards: Finding the Equation

Find the equation of this graph:

Give the general form of this equation:

23

21 2 xxy

More on this to come….

a. What quadratic function has a range of {y ≥ 6, y R}, 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, –15)?

a) 2)5()6(41

xy b) 2)2()12(31

xy

Going backwards: Finding the Equation

Example:The parabola

2)3()6(21

xy

passes through the points (x, 2). What are the values of x?

We can use transformational form to solve for the x-values given a certain y-value.(Note – there will (almost) always be TWO such x-values!)

1,53234

34

3821

36221

)3()6(21

2

2

2

2

xxxx

x

x

x

xy

Solving for x given yusing transformational form

2)8()5(4 xy 2)1()2(21

xy

x

x

x

x

x

528

208

820

820

85042

2

11

11

)1(1

12021

2

2

x

x

x

x

Solving for x given yusing transformational form

Find the x-intercepts (roots) of the following quadratic functions:

2)1()2(21

xy

422

)1(22

)1(221

1)0(221

2

2

2

yy

y

y

y

Solving for y given xusing transformational form

FYI, this is much easier from standard form: 442 2 xxy

Find the y-intercept of the following quadratic functions:

Comparing FormsGeneral Form:

Transformational Form:

Standard Form:

Factored Form:

2)HT()VT(VS1

xy

khxay 2)(

c bx axy 2

21VS rxrxy coming soon…

not coming soon…

What are you good for?Transformational form is good for:• finding the vertex

(HT, VT)• finding the range

{y ≤ or ≥ VT, y R}• finding the axis of symmetry

x = HT• finding the max/min value

VT• getting the mapping rule

(x, y) → (x + HT, VSy + VT)• graphing the function• getting the equation from the graph

Drawbacks:• Functions are usually

written in the f(x) notation

2)HT()VT(VS1

xy

What are you good for?General form is good for:• finding the y-intercept

(0, c)• finding the x-intercepts

…coming soon….• finding the vertex

…coming soon….• graphing the parabola

…coming soon…• finding the roots

…coming soon…

c bx axy 2

Drawbacks:• Some work is involved to

obtain this form when only graph, or only transformations are given

What are you good for?

Standard form is good for:• nothing

khxay 2)(

What are you good for?

Factored form is good for:• finding roots

r1 and r2

21VS rxrxy

can be factored

Going Between Forms

general → transformational• 4 steps: includes completing

the square• …coming soon….

transformational → general• 2 steps: FOIL then solve for x

2)1(321

xy 142 2 x – – x y 4 steps

2 steps

Transformational to General Form

FOILirstutsidensideast

2)3( x

)3)(3( xx

2x x3 x3 9

96)3( 22 xxx

Ex. Put the following into general form: 2)4()3(21

xy

)4)(4()3(21

xxy

1644)3(21 2 xxxy

)168(2)3(212 2

xxy

321623 2 xxy

29162 2 xxy

Transformational to General Form

168)3(21 2 xxy