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

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 (−𝑏2𝑎 ))

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

1

2

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

1

2

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

1

2

− 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]

Standard form to Vertex form

𝑦=±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−5

𝑥=− 4

2(−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