### 5.1 & 5.2 - Souderton Math

```5.1 – Introduction to Quadratic
Functions
Objectives: Define, identify, and graph quadratic functions.
Multiply linear binomials to produce a quadratic
expression.
Standard: 2.8.11.E. Use equations to represent curves.
The Standard Form of a Quadratic Function is:
f ( x )  ax  bx  c
2
(A quadratic function is any function that can be written
in the form f(x)= ax2 + bx + c, where a ≠ 0.)
A Quadratic function is any function that can be written
in the form f(x)= ax2 + bx + c, where a ≠ 0.
Ex 1a. Let f(x) = (2x – 5)(x - 2). Show that f represents
a quadratic function. Identify a, b, and c when the
function is written in the form f(x) = ax2 + bx + c.
FOIL  First – Outer – Inner – Last
OR
Distribute each term in the first set of parentheses to
each term in the second set of parentheses!
(2x – 5)(x – 2) =
a = 2, b = -9, c = 10
2x2 – 4x – 5x + 10
2x2 – 9x + 10
p.278: #13-19 ODD
The graph of a quadratic function is
called a parabola.
• Each parabola has an axis of symmetry, a line that
divides the parabola into two parts that are mirror
images of each other.
• The vertex of a parabola is either the lowest point on
the graph or the highest point on the graph.
Axis of
Symmetry
Vertex
Ex 2a. Identify whether f(x) = -2x2 - 4x + 1 has a maximum value or a
minimum value at the vertex. Then give the approximate coordinates
of the vertex.
• First, graph the function:
• Next, find the maximum value of the parabola (2nd, Trace):
• Finally, max(-1, 3).
III. Minimum and Maximum Values
• Let f(x) = ax2 + bx + c, where a ≠ 0. The graph
of f is a parabola.
– If a > 0, the parabola opens up and the vertex is
the lowest point. The y-coordinate of the vertex is
the minimum value of f.
– If a < 0, the parabola opens down and the vertex
is the highest point. The y-coordinate of the vertex
is the maximum value of f.
a. f(x) = x2 + x – 6
b. g(x) = 5 + 4x – x2
c. f(x) = 2x2 - 5x + 2
d. g(x) = 7 - 6x - 2x2
Opens up, has minimum value
Opens down, has maximum value
Opens up, has minimum value
Opens down, has maximum value
(an introduction)
Solve 5x2 – 19 = 231. Give exact solutions. Then give
approximate solutions to the nearest hundredth.
5 x  19  231
2
5 x  250
2
x  50
2
x   50
x  7 . 07
x   7 . 07
Ex 2a. Solve 4(x+2)2 = 49
49
( x  2) 
2
4
( x  2)
2
 
49
4
x2 
7
2
x 
x 
7
2
or
x  
7
2
2
3
11
2
or
x  
2
2
A rescue helicopter hovering 68 feet above a boat in distress drops a
life raft. The height in feet of the raft above the water can be
modeled by h(t) = -16t2 + 68, where t is the time in seconds after it
is dropped. After how many seconds will the raft dropped from
the helicopter hit the water?
When the raft hits the water, the
height will = 0, so h(t) = 0:
0   16t  68
2
 16t   68
2
68
t 
2
68 ft
16
t 
2
17

4
t
2
 
17
4
t  
17
2
t   2 .1 seco n d s
Since only positive values of time make
sense, the answer is 2.1 seconds.
If ∆ABC is a right triangle with the right angle
at C, then a2+ b2 = c2.
When you apply the Pythagorean Theorem,
use the principal square root because
distance and length cannot be negative.
c
b
a
Homework:
p 278 #20-26even, 27-32,
38-44even, 49
p 287 #24-30even,
36-42even, 50
```