Daily CheckGive the transformations for each of the following functions?

1) f(x) = (x - 2)2 + 4

2) f(x) = -3x2

3) f(x) = ½ (x+3)2

Write the equation in vertex form for the following graph.

Warm-upMultiply out each of the following functions.

1) y = (x – 1)2 + 8

2) y = 2(x + 3)2 – 5

3) y = -(x – 4)2 + 3

4) y = 2(x + 1)2 – 2

This is how you convert from vertex form to standard form.

CCGPS GeometryDay 39 (10-1-13)

UNIT QUESTION: How are real life scenarios represented by quadratic functions?

Today’s Question:How do we change from standard form to vertex form of a quadratic?Standard: MCC9-12.A.SSE.3b, F.IF.8

Summary of Day One Findings

Parabolas

Vertex Form

khxay 2)(Vertex: (h, k)

Axis: x = h

Rate: a (+ up; – down)

• What's the pattern?

(x + 6)2

+

+

x 6

x

6

x2 6x

6x 36

x2 + 12x + 36

• How about these?

x2 + 4x ______ (x _____ )2

x2 + 10x ______ (x _____ )2

x2 – 14x ______ (x _____ )2

+ 4 + 2

+ 25 + 5

+ 49 – 7

COMPLETING THE SQUARE

• Converting from standard form to vertex form can be easy…

x2 + 6x + 9 (x + 3)2

326 932

x2 – 2x + 1 =

x2 + 8x + 16 =

x2 + 20x + 100 =

(x – 1)2

(x + 4)2

(x + 10)2

… but we're not always so lucky

COMPLETING THE SQUARE

• The following equation requires a bit of work to get it into vertex form.

y = x2 + 8x + 10y = (x2 + 8x ) + 10

428

1642

+ 16 – 16

16 is added to complete the square. 16 is sub-tracted to maintain the balance of the equation.

y = (x + 4)2 – 6

The vertex of this parabola is located at ( –4, –6 ).

COMPLETING THE SQUARE

• Lets do another. This time the x2 term is negative.

y = –x2 + 12x – 5

y = –(x2 – 12x ) – 5

6212

36)6( 2

+ 36 + 36

The 36 in parentheses becomes negative so we must add 36 to keep the equation balanced.

y = – (x – 6)2 + 31The vertex of this parabola is located at ( 6, 31 ).

y = –(x2 – 12x ) – 5

Un-distribute a negative so that when can complete the squarey = (–x2 + 12x ) – 5

COMPLETING THE SQUARE

COMPLETING THE SQUARE Find the value to add to the trinomial to create a

perfect square trinomial: (Half of “b”)2

[A] cxx 102[B] cxx 52

[C] cxx 82 2[D] cxx 93 2

Example 1 Type 1: a = 1Write in vertex form. Identify the vertex and axis of symmetry.

[A] 862 xxy [B] 342 xxy

Write in standard form. Identify the vertex and axis of symmetry.

[A] 50243 2 xxy [B] 322 xxy

Example 2 Type 1: a≠1

Method #2: SHORTCUT1. Find the AXIS of SYMMETRY :

2. Find VERTEX (h, k)h = x k is found by substituting “x”

3. “a” – value for vertex form should be the same coefficient of x2 in standard form. Check by using another point (intercept)

a

bx

2

khxay 2)(

Method #2 ExampleGiven f(x) = x2 + 8x + 10

1) Find a, b, and c.

2) Find the line of symmetry or “h” using x = -b/2a

3) Find the y value of the vertex, or “k” by substituting

“x” into the equation.

So, the vertex is at (-4, 6).

4) Write the equation in vertex form using the “h” and “k” found. “a” will be the same thing as in Step 1.

84

2 2 1

bx

a

( 4) 16 32 10 6f

2( 4) 6y x

1, 8, 10a b c

[1] 182 xxy

PRACTICE METHOD #2:Write in vertex form. Find vertex and axis of symmetry.

[2] 20102 xxy

[3] 563 2 xxy

PRACTICE METHOD #2:Write in vertex form. Find vertex and axis of symmetry.

[4] 32162 2 xxy