
Mathematical Investigations III
Name:
Mathematical Investigations III- A View of the World
Quadratics and Completing the Square:
One of the most effective tools for understanding quadratic expressions is completing the square. By
completing the square, one can easily find solutions to quadratic equations and the vertex of a parabola
(that is, the highest or lowest point on the parabola).
Below, we demonstrate the process of completing the square and use it to find solutions and a vertex.
Solve 3x 2 10x  8  0
Find the vertex of the parabola:
2
y = 3x 10x  8
Solution:
Group the x 2 and x terms:
Solution:
Based on our solution to the left, we already
know how to complete the square on the righthand side:
(3x 2  10 x)  8  0
Factor out the coefficient on x 2 :
2
10 

3 x 2  x   8  0
3 

This is the key step. Look at the coefficient of
x in the parentheses and divide it by 2 to get a
number c . Add and subtract c 2 :
10
25 25 

3 x 2  x 
 8  0
3
9
9 

The first three terms in the parentheses factor
2
as  x  c  :
2

5
25 
3  x      8  0

3
9 

2
Multiply out, leaving the  x  c  in factored
form:
2
5
25

3 x   
8  0
3
3

Solve algebraically:
2
5
49

x   
3
9

5
7

3
3
2
Therefore x  4 or x  .
3
5
49

y  3 x   
3
3

The vertex of the parabola is its highest or lowest
point. If the coefficient of x 2 is positive, the vertex
will be the lowest point. Otherwise, if the
coefficient of x 2 is negative, the vertex will be the
highest point.
Since we want to know the lowest point on the
parabola, the question we now ask is “When
is y the smallest it can be?” The answer is “when
2
5

 x   is as small as it can be.” Since squares
3

cannot be negative, the smallest that quantity can
5
be is 0, which happens when x 
.
3
5
 49
Plugging in x 
, we get that y 
.
3
3
x
5 49
Therefore the vertex is at 
 3 , 3 
Poly 1.1
Rev. S11
Mathematical Investigations III
Name:
For each of the following:
a. Solve f ( x)  0 by completing the square.
b. Find the vertex of the parabola y  f (x) by completing the square.
c. Sketch a quick graph of y  f (x) , labeling the vertex and zeroes.
1. f ( x)  x 2  8 x  48
a.
b.
c.
2. f ( x)  4 x 2  8 x  5
a.
b.
c.
3. f ( x)  2 x 2  10 x  9
a.
b.
c.
Poly 1.2
Rev. S11
Mathematical Investigations III
Name:
4. f ( x)  x 2  6 x  13
a.
General formulas:
b.
c.
The Quadratic Formula and the Vertex of a Parabola.
Now that you are familiar with completing the square in specific examples, try it in a more general setting
to derive the quadratic formula and a formula for the vertex of a parabola.
5. Given the quadratic function f ( x)  ax 2  bx  c :
a. Complete the square on the expression on the right-hand side.
b. Use your answer to Part (a) to solve f ( x)  0 .
c. Use your answer to Part (a) to find the vertex of the parabola y  f (x) .
a.
b.
c.
Poly 1.3
Rev. S11
Mathematical Investigations III
Name:
In Question 5, you should have observed the following:
For a quadratic function f ( x)  ax 2  bx  c :
(1) (The Quadratic Formula) The solutions to f ( x)  0 are x 
 b  b 2  4ac
.
2a
  b  b 2  4ac 
 .
(2) The vertex of the parabola y  f (x) is at the point 
,
4a
 2a

Extra Observation:
Notice the relationship between the two results above. Look at the
graph of the parabola at the left, with vertex C and zeros at A and B.

C
A
B
D
2
y  1 x  5  9
x
Find the coordinates of C, A, B, and D:
Find the distances AD and DB.
What is the relationship of these to the quadratic formula
values for A and B ?
2
 b  b  4ac
2a
2a
Poly 1.4
Rev. S11