1. Summary of 2.1 y = -x2 graph of y = x2 is reflected in the x- axis
Note: Negative in front of x2 makes parabola “frown”.
Reflections:
y = (-x)2 graph of y = x2 is reflected in the y- axis
Note: If –x is in ( ), the parabola stays “smiling”.
y = Ax2 where A > 1, graph of y = x2 stretches vertically by a factor of A
Note: Parabola gets “narrower”
Vertical Change:
y = Ax2 where 0< A < 1, graph of y = x2 shrinks vertically by a factor of A
Note: Parabola gets “wider”

2. Summary of 2.1
y = (x - B)2 graph of y = x2 is shifted B units to the right.
Horizontal Shifts:
y = (x + B)2 graph of y = x2 is shifted B units to the left.
Note: Think opposite direction of the sign of B.
y = x2 + C graph of y = x2 is shifted C units UP.
Vertical Shifts:
y = x2 - C graph of y = x2 is shifted C units DOWN.
Note: Follow the direction of the sign of C.

3. 2.2 Graphing Quadratic Functions by completing the square
10/15/12

4. When a is positive, the parabola opens up. The vertex is minimum.
Vertex Form:
a quadratic equation written in the form
Y = a(x –h)2 + k
(h, k) is the vertex
When a is positive, the parabola opens up. The vertex is minimum.
When a is negative, the parabola opens down. The vertex is maximum.
Graph of
y = (x – 2)2 - 9:
x - intercept:
Where parabola crosses the x-axis.
To find the x-intercept, set y = 0 and solve for x.
y - intercept:
Where parabola crosses the y-axis.
To find the y-intercept, set x = 0 and solve for y.
If equation is in standard form y = ax2 + bx + c
then c is the y-intercept.

5. Steps for completing the square
Standard form : y = x2 + bx + c
y = x2 + bx
( )
+ c
y = (x2 + bx + 𝑏 )
+ c - 𝑏
y = (x 𝑏 2 ) d
Put ( ) around x2 + bx and move c outside ( )
Take half of b and square it. Add it to the ( ) and subtract it from c.
3. Rewrite what’s in the ( ) as (x + 𝑏 2 )2 and simplify c - 𝑏 2 2
Note: if y = x2 – bx + c
Then y = (x - 𝑏 2 )2 + d

6. Example 1 y = x2 + 6x ( ) + 5 y = (x2 + 6x + 9) + 5 - 9
Rewrite y = x2 + 6x + 5 in Vertex Form and describe the transformation.
Determine the x-int and y int. (if it exists) and vertex.
Determine if the vertex is a maximum or a minimum.
a.) Rewrite in vertex form and describe the transformation.
y = x2 + 6x
( )
+ 5
Put ( ) around x2 + 6x and move +5 outside ( )
Take half of 6 and square it. Add 9 to the ( ) and subtract 9 from 5.
3. Rewrite what’s in the ( ) as (x + 3)2
y = (x2 + 6x + 9)
y = (x + 3)
Graph of x2 shifts 3 units to the left and 4 units down.

7. Example 1 Since a is positive 1, the vertex is minimum.
b.) Determine the x and y intercepts and vertex
x –int. set y = 0 and solve for x.
y –int. set x = 0 and solve for y.
0 = (x + 3)
4 = (x + 3)2
±2 = x + 3
2 = x + 3 and -2 = x + 3
-1 = x = x
y = ( )
y = (3)2 - 4
y = 9 - 4
y = 5
Or since the original problem is written in standard form, y –int is the c term which is 5.
y = x2 +6x + 5
Vertex (h, k) = (-3, -4)
c.) Minimum or Maximum?
Since a is positive 1, the vertex is minimum.

8. Example 2 y = x2 - 6x ( ) + 10 y = (x2 - 6x + 9) + 10 - 9
Rewrite y = x2 - 6x + 10 in Vertex Form and describe the transformation.
Determine the x-int and y int. (if it exists) and vertex.
Determine if the vertex is a maximum or a minimum.
a.) Rewrite in Vertex Form and describe the transformation.
y = x2 - 6x
( )
+ 10
Put ( ) around x2 - 6x and move +10 outside ( )
Take half of 6 and square it. Add 9 to the ( ) and subtract 9 from 10.
3. Rewrite what’s in the ( ) as (x - 3)2
y = (x2 - 6x + 9)
y = (x - 3)
Graph of x2 shifts 3 units to the right and 1 units up.

9. Example 2 Since a is positive 1, the vertex is a minimum. y –int.
b.) Determine the x and y intercepts and vertex
x –int. set y = 0 and solve for x.
y –int.
y = x2 -6x + 10
Since the original problem is written in standard form, y –int is the c term which is 10.
0 = (x - 3)
-1 = (x - 3)2
Since you cannot take the square root of a negative number, there is no x- intercept
Vertex (h, k) = (3, 1)
c.) Minimum or Maximum
Since a is positive 1, the vertex is a minimum.

10. Example 3 y = 2x2 - 4x + 6 2 2 2 2 𝑦 2 = x2 - 2x ( ) + 3
Rewrite y = 2x2 - 4x + 6 in Vertex Form and describe the transformation.
Determine the x-int and y int. (if it exists) and vertex.
Determine if the vertex is a maximum or a minum.
a.) Rewrite y = x2 - 6x + 10 in Vertex Form and describe the transformation.
y = 2x2 - 4x + 6
Divide everything by 2.
Put ( ) around x2 - 2x and move +3 outside ( )
Take half of 2 and square it. Add 1 to the ( ) and subtract 1 from 3.
4. Rewrite what’s in the ( ) as (x - 1)2
5. Multiply both sides by 2.
𝑦 2 = x2 - 2x
( )
+ 3
𝑦 2 = (x2 - 2x +1)
𝑦 2 = (x - 1)
𝑦 = 2(x - 1)
Graph of x2 stretches vertically by factor of 2 and shifts 1 unit to the right and 4 units up.

11. Example 3 Since a is positive 2, the vertex is a minimum. y –int.
b.) Determine the x and y intercepts and vertex
x –int. set y = 0 and solve for x.
y –int.
y = 2x2 -4x + 6
Since the original problem is written in standard form, y –int is the c term which is 6.
0 = 2(x - 1)
-4 = 2(x - 1) Divide by 2
-2 = (x - 1)2
Since you cannot take the square root of a negative number, there is no x- intercept
Vertex (h, k) = (1, 4)
c.) Minimum or Maximum
Since a is positive 2, the vertex is a minimum.