SECTION 5.1 –
QUADRATIC FUNCTIONS
y
f(x)=x^2
5
4
3
2
1
x
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
-1
f ( x)  x
2
-2
-3
-4
-5
2
3
4
5
6
7
8
9
The Graph of a Quadratic Function
•
The vertex of the parabola is the point where the graph does one
of the following:
– Stops increasing and starts decreasing, or
– Stops decreasing and starts increasing
9
y
f ( x)  ax  bx  c
2
f(x)=x^2 + 3
f(x)=- x^2 + 1
8
7
6
• If a parabola is concave up, the vertex is a minimum point.
5
4
3
2
1
-9
-8
-7
-6
-5
-4
-3
-2
-1
-1
-2
-3
-4
-5
-6
-7
-8
-9
x
1
2
3
4
5
6
7
8
9
• If a parabola is concave down, the vertex is a maximum point.
Axis of Symmetry
The vertical line through the vertex is called the axis of symmetry. This line divides the graph into
two halves that are reflections of each other.
9
y
f(x)=(x - 4)^2
8
7
6
5
4
3
2
1
-9
-8
-7
-6
-5
-4
-3
-2
-1
-1
-2
-3
-4
-5
-6
-7
-8
-9
x
1
2
3
4
5
6
7
8
9
The axis of symmetry is:
x4
Vertex Form
y  a ( x  h)  k
2
If a > 0, the graph is concave up – vertex is minimum
If a < 0, the graph is concave down – vertex is maximum
If 0 < |a| < 1, the parabola is shrunk vertically – looks wider than x2.
If |a| > 1, the graph is stretched vertically – looks more narrow than x2.
Vertex: (ℎ, 𝑘)
Axis of Symmetry: x = h
VERTEX FORM
1. Determine (a) the vertex, (b) whether it is a minimum or a maximum, (c) axis of symmetry, (d) where
the graph is increasing/decreasing and (e) domain and range.
y  2( x  3)  4
2
𝑎 > 0 ⇒ 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝
𝑎 = 2 > 1 ⇒ 𝑠𝑡𝑟𝑒𝑡𝑐ℎ𝑒𝑑/𝑛𝑎𝑟𝑟𝑜𝑤
Vertex is a minimum
Vertex: (3, 4)
Axis of symmetry: 𝑥 = 3
Increasing : (3, )
Decreasing : (,3)
Domain : (, )
Range : [4, )
VERTEX FORM
2. Determine (a) the vertex, (b) whether it is a minimum or a maximum, (c) axis of symmetry, (d) where
the graph is increasing/decreasing and (e) domain and range.
1
y   ( x  1) 2  4
2
𝑎 < 0 ⇒ 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑑𝑜𝑤𝑛
1
𝑎 = < 1 ⇒ 𝑠ℎ𝑟𝑢𝑛𝑘/𝑤𝑖𝑑𝑒
2
Vertex is a maximum
Vertex: (−1, −4)
Axis of symmetry: 𝑥 = −1
Increasing : (,1)
Decreasing : (1, )
Domain : (, )
Range : (,  4]
Standard Form
f ( x)  ax  bx  c
2
If a > 0, the graph is concave up – vertex is minimum
If a < 0, the graph is concave down – vertex is maximum
If 0 < |a| < 1, the parabola is shrunk vertically – looks wider than x2.
If |a| > 1, the graph is stretched vertically – looks more narrow than x2.
Vertex : (h, k )
y - intercept : (0, c)
The vertex of the parabola can be found by
• Completing the Square or
𝑏
• Using the formula: ℎ = − and then substituting 𝑥 = ℎ to find 𝑘 .
2𝑎
STANDARD FORM
3. Determine (a) the vertex by completing the square, and (b) whether it is a minimum or a maximum.
y  x2  4x  8
a > 0  concave up
|a| = 1 not wide, not skinny
Vertex is a minimum
Vertex: ( 2,4)
Axis of symmetry: 𝑥 = 2
2
1

2
  (4)    2   4
2

y  x2  4x  4  4  8
y  ( x 2  4 x  4)  4
y  ( x  2) 2  4
STANDARD FORM
4. Determine (a) the vertex by using the formula, and (b) whether it is a minimum or a maximum.
y  3x 2  6 x  8
a < 0  concave down
|a| = 3 stretched/skinny
Vertex is a maximum
Vertex: (1,5)
y  3( x  1) 2  5
b
2a
 ( 6)

2(3)
6

6
1
h
k  3(1) 2  6(1)  8
 3  6  8
 5
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 AND 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠
5. Find the 𝑥 − and 𝑦 −intercepts.
y  3x 2  5 x  2
To find 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠, let 𝑦 = 0.
To find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, let 𝑥 = 0.
3x  5 x  2  0
y  3(0) 2  5(0)  2  2
(3 x  1)( x  2)  0
0,  2
2
3x  1  0 x  2  0
1
x
x  2
3
1 
 , 0 ,  2, 0 
3 
DETERMINE THE FORMULA
6. Determine the formula for the quadratic function pictured below.
y  a ( x  h)  k
2
vertex : (2,  3)
another point : (4,1)
y  a ( x  (2)) 2  (3)
 1  4a  3
4a  2
1
a
2
y  a ( x  2) 2  3
 1  a (4  2) 2  3
 1  a (2) 2  3
1
2
y  ( x  2)  3
2
HEIGHT OF A ROCKET
Height of a Rocket. A model rocket is launched with an initial velocity of 150 ft/sec from a height
of 40 ft. The height of the rocket t seconds after it has been launched is given by the function
𝑠 𝑡 = −16𝑡 2 + 150𝑡 + 40. Determine the time at which the rocket reaches its maximum height
and find the maximum height.
Since a < 0, we know that the maximum height occurs at the vertex.
b
2a
 150

 32
75

16
 4.6875
h
k  16(4.6875) 2  150(4.6875)  40
 351.563  703.125  40
 391.5625
(time, height )
(4.687, 391.5625)
The rocket reaches its maximum height
of 391.5625 feet after 4.6875 seconds.