4-2 Quadratic Functions:

Standard Form

Today’s Objective:I can graph a quadratic function in standard form

The function models the height h of the soccer ball as it travels distance x. What is the maximum height of the ball? Explain.

h=− 0.01 ( 45 )2+0.9(45)¿20.25 𝑓𝑡

Warm Up

x

y

Quadratic Function: Vertex Form𝑓 (𝑥)=±𝑎(𝑥− h)2+𝑘 Attributes:

• Opens up (a > 0) or down (a < 0)• Vertex is maximum or minimum• Vertex: (h, k)• Axis of symmetry:

x

y

(h ,𝑘)

𝑥=h

Quadratic Function: Standard FormAttributes:• Opens up (a > 0) or down (a < 0)• Vertex is maximum or minimum• y-intercept: (0, c)

Can be determined with a little work• Axis of symmetry: • Vertex:

𝑓 (𝑥)=𝑎𝑥2+𝑏𝑥+𝑐

x

y

x

y

x

y

x

y

(0, c)

𝑥=−𝑏2𝑎

(−𝑏2𝑎 , 𝑓 (−𝑏

2𝑎 ))Evaluate f(x) at

Graphing a Quadratic Function: Standard form1. Plot the vertex2. Find and plot two points

to the right of vertex. 3. Plot the point across axis

of symmetry.4. Sketch the curve.

Vertex:

Axis of Symmetry:Domain:

Range:All Real Numbers

𝑦=𝑥2+2𝑥+3

Units right of vertex

x

Units up from

vertex

12

14𝑦 ≥ 2

𝑥2

𝑥=−1

𝑥=−𝑏2𝑎¿

−22(1)¿−1

𝑦=(−1)2+2 (−1 )+3¿2

(−𝟏 ,𝟐)

-5 5

-2

8

x

y

x

y

x

y

x

y

Graphing a Quadratic Function: Standard form1. Plot the vertex2. Find and plot two points

to the right of vertex. 3. Plot the point across axis

of symmetry.4. Sketch the curve.

Vertex:

Axis of Symmetry:Domain:

Range:All Real Numbers

𝑦=2 𝑥2− 4 𝑥−5

Units right of vertex

x

Units up from

vertex

12

28𝑦 ≥ −7

2

𝑥=1

𝑥=−𝑏2𝑎¿

42(2)¿1

𝑦=2(1)2 − 4 (1 ) −5¿−7

(𝟏 ,−𝟕)

-5 5

-8

2

x

y

x

y

x

y

x

y

Graphing a Quadratic Function: Standard formVertex:

Axis of Symmetry:Domain:

Range:All Real Numbers

𝑦=− 0.5𝑥2+2 𝑥−3

Units right of vertex

x

Units up from

vertex

12

− 0.5−2

𝑦 ≤ −1

-0.5

𝑥=2𝑥=

−𝑏2𝑎¿

− 22(− .5)¿2

𝑦=− 0.5 (2 )2+2 (2 ) −3¿−1

(𝟐 ,−𝟏)

-5 5

-8

2

x

y

x

y

x

y

x

y

Vertex on Calculator:[2nd], [trace]Choose minimum or maximumMove curser left of vertex, [enter]Move curser right of vertex, [enter][enter]

Vertex Form of a Quadratic Equation

• Vertex form y = a·(x – h)2 + k allows us to find vertex of the parabola without graphing or creating a x-y table.

y = (x – 2)2 + 5 a = 1vertex at (2, 5)

y = 4(x – 6)2 – 3 a = 4vertex at (6, –3)y = 4(x – 6)2 + –3

y = –0.5(x + 1)2 + 9 a = –0.5vertex at (–1, 9)y = –0.5(x – –1)2 + 9

Vertex Form of a Quadratic Equation

• Check your understanding… 1. What are the vertex coordinates of the parabolas with

the following equations?a. y = (x – 4)2 + 1b. y = 2(x + 7)2 + 3c. y = –3(x – 5)2 – 12

vertex at (4, 1)vertex at (–7, 3)vertex at (5, –12)

2. Create a quadratic equation in vertex form for a "wide" parabola with vertex at (–1, 8).

y = 0.2(x + 1)2 + 8

Vertex Form of a Quadratic Equation

• Finding the a value (cont'd)• If we know the coordinates of the vertex and some other

point on the parabola, then we can find the a value.• For example,

What is the a value in the equation for a parabola that has a vertex at (3, 4) and an x-intercept at (7, 0)?

y = a·(x – h)2 + k0 = a·(7 – 3)2 + 40 = a·(4)2 + 40 = a·16 + 4-4 = a·16

-0.25 = a

substitutesimplifysimplifysubtract 4divide by 16y = -0.25·(x – 3)2 + 4

-8 2

-8

2

x

y

Writing a Quadratic function: vertex form

Identify the Vertex:

Finding stretch factor:• Choose another known point and

solve for a.

(− 2 ,−7 )𝑦=𝑎(𝑥+2)2−7

−5=𝑎(− 1+2)2− 72=𝑎

𝑦=2(𝑥+2)2− 7

-8 2

-8

2

x

y

-8 2

-8

2

x

y

(-2, -7)

(-1, -5)

Standard form to Vertex form𝑦=±𝑎(𝑥− h)2+𝑘𝑦=𝑎𝑥2+𝑏𝑥+𝑐

• a value is the same• Find the vertex

𝑦=2 𝑥2+10 𝑥+7

𝑥=−102(2)¿− 2.5

𝑦=2(−2.5)2+10 (−2.5 )+7¿−5.5

𝑦=2(𝑥+2.5)2 −5.5

𝑦=−𝑥2+4 𝑥−5𝑥=

− 42(−1)¿2

𝑦=− (2 )2+4 (2 )− 5¿−1

𝑦=−(𝑥− 2)2 −1

Bungee JumpingYou can model the arch of this bridge with the function How high above the river is the arch?

x

y

x

y Maximum

𝑥=− 0.847

2(− 0.000498)¿850

𝑦=− 0.000498 (850 )2+0.847(850)¿360

(850,360)Arch height:516+360¿876 𝑓𝑡

p.206:8-30 evens