# Math 112 Notes - De Anza College ```Quadratic Functions
Vertex Form
The vertex form of a Quadratic Function is


y  a( x  h)2  k where
( h, k ) is the vertex (lowest or highest point)
a tells you whether the Quadratic Function opens up or down
a  0; opens up
and gives you an idea how wide the function is
a  0; opens down
x  h is the axis of symmetry

To go from standard form
the Square.
Example 1:
( y  ax 2  bx  c) to vertex form, use a process called Completing
y  x2  2x 1
y  1  x2  2x
1

2
  2  1;1  1
2

Put the
x 2 and 2 x term on one side of the equation
Calculate &frac12; of the coefficient of the
x term, square it
and add the result to both sides of the equation.
y 11  x  2x 1
y  2  ( x  1)( x  1)
2
y  2  ( x  1) 2
Simplify the left side and factor the right side.
y  ( x  1)2  2
Solve for y.
Notice that if we compare our equation to the format
y  ( x  1) 2  2
y  a ( x  h)  k
a  1, h  1, k  2
vertex  (1, 2) and x  1, axis of symmetry
y  ( x  1)2  2 quickly by graphing the axis of symmetry and the vertex
and a few points like the pair (0, 1) and ( 2, 1) and the pair (1, 2) and ( 3, 2) . Notice that,
We can graph, by hand,
for each pair, the y-coordinates are the same.
Example 2:
y  2 x2  5x  1
y  1  2 x2  5x
y  1  2( x 2 
5
x)
2
Notice that the leading coefficient is 2.
Put the
2 x 2 and 5x term on one side of the equation
2
The leading coefficient for x is not 1 so factor out the 2
to make it 1.
 1 5 5  5  2 25 
   ;    
 2 2 4  4  16 
Calculate &frac12; of the coefficient of the
x term,
25
to both sides of the
16
25
equation. Notice that the term
is inside the
16
square it and add 2 
parentheses on the right side.
25
5
25
 2( x 2  x  )
16
2
16
5
5
 2( x  )( x  )
4
4
5
 2( x  ) 2
4
y 1 2 
66
16
66
y
16
y
5
66
y  2( x  ) 2 
4
16
Simplify the left side and factor the right side.
Solve for y.
Notice that if we compare our equation to the format
5
66
y  2( x  ) 2 
4
16
y  a ( x  h)  k
5
66
a  2, h   , k  
4
16
5
 5 66 
vertex    ,   and x   , axis of symmetry
4
 4 16 
```