3.1 Quadratic Functions and Models, or …
Everything you always wanted to know
about parabolas,
but were afraid to ask
The “geometry” of a quadratic function: a parabola
The algebra of a quadratic function
Quadratic functions have two algebraic forms:
 general form: f(x) = ax2 + bx + c
  b   b 
,f 
vertex (using the vertex formula) = 
 2a  2a  
 vertex form: f(x) = a(x - h)2 + k
vertex = (h, k)
The role of a (a is referred to as the leading coefficient):
if a > 0, parabola opens upward
if a < 0, parabola opens downward
3.1-1
What’s a quadratic function and what’s not
A quadratic function:
1. must have a quadratic term (ax2)
2. may or may not have a linear term (bx)
3. may or may not have a constant term (c)
4. may have no x-expression to powers other than 0, 1, 2
5. may have no x-expression in a denominator
f(x) = 3x2 - 2x + 1
f(x) = -x2 - 2
f(x) = x -2 + 1
1
f(x) =
+ x2
2
(1  x)
f(x) = 2x + 1
(yes, a = 3, b = -2, c = 1)
(yes, a = -1, b = 0, c = -2)
(no, violates 4)
(no, violates 5)
(no, violates 1)
Finding the vertex
Example:
f(x) = (x + 1)2 – 3 vertex form
vertex: (-1, -3)
_______________________________________
Example:
f(x) = x2 + 2x – 2 general form
Using the vertex formula:
x-coordinate of vertex = -b/(2a) = -2/2 = -1
y-coordinate of vertex = f( -1 ) = -3
vertex: (-1, -3)
3.1-2
Sketching the graph of a parabola: f(x) = ax2 + bx + c
1. if a > 0, it opens up
if a < 0, it opens down
  b   b 
, f    this is a key point!
2. plot vertex: 
 2a  2a  
3. plot x-intercepts (if any): solve f(x) = 0 to get them
4. plot y-intercept: (f(0) = c so y-int. = (0,c))
5. plot more points on either side of the vertex as
necessary to round out picture
Ex. f(x) = -x2 + 2x + 3
1. a < 0 so
-b
-2
2. 2a = 2(-1) = 1
f( 1 ) = 4 , so
3. solve -x2 + 2x + 3 = 0:
(-x + 3)(x + 1) = 0
x = 3, -1
4. c = 3, so
5. f(2) = 3
(a = -1, b = 2, c = 3)
opens down
vertex: (1,4)
x-intercepts: (3, 0), (-1, 0)
y-intercept: (0, 3)
additional point: (2,3)
3.1-3
Finding the equation of a parabola
Given that the vertex of a parabola is (-1, 3) and it passes
through the point (-2,7), find its equation.
vertex form: y = a(x + 1)2 + 3 But what is a?
(-2, 7) must satisfy the above equation, so:
7 = a(-2 + 1)2 + 3  a = 4
vertex form of equation:
general form of equation:
y = 4(x + 1)2 + 3
y = 4x2 + 8x + 7
Quadratic regression (omit)
3.1-4
Maximum and minimum of a quadratic function
Here’s your familiar parabola (graph of a quadratic function):
Maximum:
 refers to the largest value that f(x) can ever have
 for this example, it is 4
 maximum is second coordinate of the vertex point
 we say that "f reaches its maximum of 4 for x = 1"
 f has a maximum value because it opens down
The following parabola opens up, therefore doesn't have a
maximum, but rather a minimum:
Minimum:
 refers to the smallest value that f(x) can ever have
 for this example, it is -3 (reaches it for x = 5)
3.1-5
Finding maximum or minimum for a quadratic
function
 really quite easy
 just finding the vertex tells you everything you need to
know about the maximum (or minimum)
 don't need to graph
 don't need to do much of anything
JUST FIND THE VERTEX!!
Example: When a cannon is fired at a certain angle, the distance
of the shell above the ground h (in meters) at time t (in seconds) is
given by the formula
2
h(t) = -4.9t + 24t + 5
Find the maximum height attained by the shell.
Notes:
 h(t) is a quadratic function (parabola)  has a vertex!
 for this parabola, a = - 4.9 and b = 24
 a < 0  the parabola opens downward  has a max value
THE CALCULATION:
-b
-24
=
2.45
2a
2(-4.9) =
h(2.45) = -4.9(2.45)2 + 24(2.45) + 5 = 34.4
vertex = (2.45, 34.4)
vertex
THE ANSWER: The shell reaches its maximum height of 34.4
meters 2.45 seconds after the cannon is fired.
3.1-6