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Mathematical Investigations III
Name:
Mathematical Investigations III- A View of the World
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
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