- The standard form of a quadratic equation is y = ax2 + bx + c. A quadratic equation graphs as a parabola. - The vertex of a parabola is the highest/lowest point, which occurs at x = -b/2a. To find the vertex, substitute -b/2a into the original equation to solve for y. - For the equation y=-x2 + 4x –1, the vertex occurs at point (2,3). Additional points are graphed to sketch the parabola.

2. −
b
a2
• The standard form of a quadratic equation is y = ax2
+ bx + c.
• The graph of a quadratic equation is a parabola.
• When a is positive, the graph opens up.
• When a is negative, the graph opens down.
• Every parabola has a vertex. For graphs opening up, the vertex is
a minimum (low point). For graphs opening down, the vertex is a
maximum (high point).
• The x-coordinate of the vertex is equal to .

3. To find the x coordinate of the vertex, use the equation
Then substitute the value of x back into the equation of the
parabola and solve for y.
x
b
a
= −
2
You are given the equation y=-x2
+ 4x –1. Find the coordinates
of the vertex.
x
b
a
= −
2
x = 2
−= =a b1 4,
( −
x = −
4
2 1)( )
2
y x x= − + −4 1
y = − + −4 8 1
y = 3
y = − + −2
2 4 2 1( ) ( )
The coordinates of the vertex are (2,3)

4. Choose two values of x that are to the right or left of the x-
coordinate of the vertex.
Substitute those values in the equation and solve for y.
Graph the points.
Since a parabola is symmetric, you can plot those same y-values on
the other side of the parabola, the same distance away from the
vertex.
x y = -x2
+ 4x -1 y
1
-1
y = -(1)2
+ 4(1) - 1
y = -1 +4 - 1
2
y = -(-1)2
+ 4(-1) -1
y = -1 -4 -1
-6

5. x
y
Plot the vertex and the points from
your table of values: (2,3), (1,2),
(-1,-6).
Since a parabola is symmetric and
has a vertical line of symmetry
through the vertex of the parabola,
points which are the same distance
to the right of the vertex will also
have the same y-coordinate as
those to the left of the vertex.
Therefore, we can plot points at
(3,2) and (5,-6).
Sketch in the parabola.

6. Find the vertex of the following quadratic equations. Make a
table of values and graph the parabola.
1 42
. y x x= −
2 2 32
. y x= − +
3 6 42
. y = x x− +

7. y x x= −2
4
a b= = −1 4
x
b
a
x
x
= −
= −
−
=
2
4
2 1
2
( )
y x x
y
y
y
= −
= −
= −
= −
2
2
4
2 4 2
4 8
4
( )
The vertex is at (2,-4)
x y = x 2
- 4 x y
1 y = 1 2
- 4 ( 1 ) - 3
0 y = 0 2
- 4 ( 0 ) 0
x
y

8. y x= − +2 32
a b= − =2 0
x
b
a
x
x
= −
= −
−
=
2
0
2 2
0
( )
y x
y
y
= − +
= − +
=
2 3
2 0 3
3
2
2
( )
The vertex is at (0,3)
x y = - 2 x 2
+ 3 y
- 1 y = - 2 ( - 1 ) 2
+ 3 1
- 2 y = - 2 ( - 2 ) 2
+ 3 - 5
x
y

9. y x x= − +2
6 4
a b= = −1 6
x
b
a
x
x
= −
= −
−
=
2
6
2 1
3
( )
y x x
y
y
= − +
= − +
= −
2
2
6 4
3 6 3 4
5
( )
The vertex is at (3,-5)
x y = x 2 - 6 x + 4 y
1 y = 1 2 - 6 ( 1 ) + 4 - 1
0 y = 0 2
- 6 ( 0 ) + 4 4
x
y

10. y x x= − +2
6 4
a b= = −1 6
x
b
a
x
x
= −
= −
−
=
2
6
2 1
3
( )
y x x
y
y
= − +
= − +
= −
2
2
6 4
3 6 3 4
5
( )
The vertex is at (3,-5)
x y = x 2 - 6 x + 4 y
1 y = 1 2 - 6 ( 1 ) + 4 - 1
0 y = 0 2
- 6 ( 0 ) + 4 4
x
y