1. Warm Up – copy the problem into your notes and solve. Show your work!!
The function ℎ=−0.01 𝑥 𝑥 models the height h of the soccer ball as it travels distance x. What is the maximum height of the ball? Explain.
ℎ=− (45)
=20.25 𝑓𝑡

2. 4-2 Quadratic Functions: Standard Form
Today’s Objective:
I can graph a quadratic function in standard form

3. Quadratic Function: Vertex Form
𝑓(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=ℎ
(ℎ,𝑘)
𝑥=ℎ

4. Quadratic Function: Standard Form
𝑓(𝑥)=𝑎 𝑥 2 +𝑏𝑥+𝑐
Attributes:
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: 𝑥= −𝑏 2𝑎
Vertex: −𝑏 2𝑎 ,𝑓 −𝑏 2𝑎
(0, c)
−𝑏 2𝑎 ,𝑓 −𝑏 2𝑎
𝑥= −𝑏 2𝑎
Evaluate f(x) at −𝑏 2𝑎

5. Graphing a Quadratic Function: Standard form
𝑦= 𝑥 2 +2𝑥+3
Vertex:
(−𝟏, 𝟐)
Plot the vertex
Find and plot two points to the right of vertex.
Plot the point across axis of symmetry.
Sketch the curve.
𝑥= −𝑏 2𝑎
= −2 2(1)
=−1
𝑦= (−1) 2 +2 −1 +3 =2
Units right of vertex x
Units up from vertex 1 2
Axis of Symmetry:
Domain:
Range:
𝑥=−1
All Real Numbers
𝑥 2 1
𝑦≥2 4

6. Graphing a Quadratic Function: Standard form
𝑦= 2𝑥 2 −4𝑥−5
Vertex:
(𝟏, −𝟕)
Plot the vertex
Find and plot two points to the right of vertex.
Plot the point across axis of symmetry.
Sketch the curve.
𝑥= −𝑏 2𝑎
= 4 2(2) =1
𝑦=2 (1) 2 −4 1 −5
=−7
Units right of vertex x
Units up from vertex 1 2
Axis of Symmetry:
Domain:
Range:
𝑥=1
All Real Numbers
2 𝑥 2 2
𝑦≥−7 8

7. Graphing a Quadratic Function: Standard form
𝑦= −0.5𝑥 2 +2𝑥−3
Vertex:
(𝟐, −𝟏)
Axis of Symmetry:
Domain:
Range:
𝑥=2
W.S. Finding the Vertex of Quadratic Equations
𝑥= −𝑏 2𝑎
= −2 2(−.5) =2
All Real Numbers
𝑦=− −3
𝑦≤−1
=−1
Vertex on Calculator:
[2nd], [trace]
Choose minimum or maximum
Move curser left of vertex, [enter]
Move curser right of vertex, [enter]
[enter]
Units right of vertex x
Units up from vertex 1 2
-0.5 𝑥 2
−0.5 −2

8. Standard form to Vertex form
a value is the same
Find the vertex
−𝑏 2𝑎 ,𝑓 −𝑏 2𝑎
𝑦=𝑎 𝑥 2 +𝑏𝑥+𝑐
𝑦=±𝑎 (𝑥−ℎ) 2 +𝑘
𝑦=2 𝑥 2 +10𝑥+7
𝑦=− 𝑥 2 +4𝑥−5
𝑥= −4 2(−1)
𝑥= −10 2(2) =2
=−2.5
𝑦=− −5
𝑦=2 (−2.5) −2.5 +7
=−1
=−5.5
𝑦=− (𝑥−2) 2 −1
𝑦=2 (𝑥+2.5) 2 −5.5

9. 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 an x-y table.
y = (x – 2)2 + 5
a = 1
vertex at (2, 5)
y = 4(x – 6)2 – 3
a = 4
vertex at (6, –3)
y = 4(x – 6)2 + –3
y = –0.5(x + 1)2 + 9
a = –0.5
vertex at (–1, 9)
y = –0.5(x – –1)2 + 9

10. 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 + 1
vertex at (4, 1)
vertex at (–7, 3)
b. y = 2(x + 7)2 + 3
c. y = –3(x – 5)2 – 12
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

11. 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 + k
substitute
0 = a·(7 – 3)2 + 4
simplify
0 = a·(4)2 + 4
simplify
0 = a·16 + 4
subtract 4
-4 = a·16
divide by 16
-0.25 = a
y = -0.25·(x – 3)2 + 4

12. Bungee Jumping
You can model the arch of this bridge with the function 𝑦=− 𝑥 𝑥
How high above the river is the arch?
𝑥= − (− )
=850
𝑦=− (850)
=360
Maximum
Arch height:
=876 𝑓𝑡
(850,360)