1-4 Quadratic Functions and Their Graphs
Quadratic function: has a squared term, but none of
higher degree
standard form of the quadratic function:
f(x) = ax2 + bx + c
(a  0)
  b   b 
,f 
vertex = 
 2a  2a  
vertex form of the quadratic function:
f(x) = a(x - h)2 + k
vertex = (h, k)
parabola: the graph of a quadratic function
The anatomy of a parabola
The role of a:
if a > 0, parabola opens upward
if a < 0, parabola opens downward
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p. 1
Finding the vertex
Example:
f(x) = (x + 1)2 - 3
It is already in vertex form f(x)= (x - h)2 + k
so vertex = (h, k) = (-1, -3)
Example:
f(x) = x2 + 2x - 2
x-coordinate of vertex = -b/2a = -2/2 = -1
y-coordinate of vertex = f( -1 ) = (-1)2 + 2(-1) –2 = -3
so vertex = (-1, -3)
Note: this is not the way shown in the book (i.e. by
completing the square), but is far superior to it
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p. 2
Sketching the graph of a quadratic function
Example: R(x) = x(2000 – 60x) 1 ≤ x ≤ 25
(1) write it in standard form: R(x) = 2000x – 60x2
(2) find the vertex:
x-coordinate = -b/2a = -2000/-120 = 50/3 = 16.67
y-coordinate = 50/3(2000 – 60(50/3)) = 16667
(3) find the points at extreme left and right of range
R(1) = 1940  point on graph is (1, 1940)
R(25) = 12500  point on graph is (25,12500)
(4) graph points and draw
R(x)




1-4




x

p. 3
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 attains 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:
this parabola attains its minimum of -3 for x = 5
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p. 4
Finding maximum or minimum for a quadratic
function
JUST FIND THE VERTEX!!
Example:
When a cannon is fired at a certain angle, the distance h (in
meters) of the shell above the ground t seconds after firing
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
=
2a
2(-4.9) =
2.45
vertex
h(2.45) = -4.9(2.45)2 + 24(2.45) + 5 = 34.4
vertex = (2.45, 34.4)
THE ANSWER: The shell reaches its maximum height of 34.4
meters 2.45 seconds after the cannon is fired.
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p. 5
Break-even analysis
Example:
R(x) = x(2000 – 60x)
C(x) = 5000 + 500x
1 ≤ x ≤ 25
1 ≤ x ≤ 25
Graphing them on the same system, we see:
R(x), C(x)








x

A break-even point is a production level in response to demand
(x) for which
Revenue = Cost
Graphically, this means x-values where the graphs intersect.
Algebraically, we solve the equation Revenue = Cost
x(2000 – 60x) = 5000 + 500x
to get x = 4 or 21 as production levels where we break even.
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p. 6