1. Graphs of Quadratic Functions
10.2
Graphs of Quadratic Functions
The graph shown below is a graph of the simplest quadratic function, defined by y = x2. This graph is called a parabola.
The point (0, 0), the lowest point on the curve, is the vertex.

2. The vertical line through the vertex is the axis of the parabola, here x = 0.
A parabola is symmetric about its axis.

3. Parabolas do not need to have their vertices at the origin.
The graph of
F(x) = x2 + k
is shifted, or translated k units vertically compared to f(x) = x2.

4. Quadratic Function
A function that can be written in the form
f(x) = ax2 + bx + c
for real numbers a, b, and c, with a ≠ 0, is a quadratic function.
The graph of any quadratic function is a parabola with a vertical axis.

5. General Principles
1. The graph of the quadratic function defined by
F(x) = a(x – h)2 + k, a ≠ 0, is a parabola with vertex (h, k) and the vertical line x = h as axis.
2. The graph opens up if a is positive and down if a is negative.
3. The graph is wider than that of f(x) = a2 if 0 < |a| < 1. The graph is narrower than that of f(x) = x2 if |a| > 1.

6. Reflections in the Axes
Transformations
Horizontal Shifts
Vertical Shifts
Up: f(x) + k
Down: f(x) – k
Right: f(x - h)
Left: f(x + h)
Reflections in the Axes
-f(x)
Over x
f(-x)
Over y

7. Vertical Stretch or Shrink
Transformations
Vertical Stretch or Shrink
a f(x)
If |a| > 1, then vertical stretch by factor a
If 0 < |a| < 1, then vertical shrink by factor a
When graphing, graph the parent function and then follow order of operations when performing the shifts.

8. Vertex
You can find this by using , then find y, write as an order pair. Or You can use max / min in the calculator

9. Parent Function

10. EXAMPLE 1
Graph f(x) = x Describe the transformation. Give the vertex, domain, and range.
The graph has the same shape as f(x) = x2, but shifted up 3 units. x
x2 + 3 2 7 1 4 3 1 2
vertex (0, 3)
domain: (, )
range: [3, )

11. EXAMPLE 2
Graph f(x) = (x  2) Describe the transformation. Give the vertex, axis, domain, and range.
The graph has the same shape as f(x) = x2, but shifted 2 units to the right and 1 unit up. x
f(x) 5 1 2 3 4
vertex (2, 1)
axis x = 2
domain: (, )
range: [1, )

12. EXAMPLE 3
Graph f(x) = 2x2  3. Give the vertex, axis, domain, and range.
The coefficient (2) affects the shape of the graph; the 2 makes the parabola narrower.
The negative sign makes the parabola open down.
The graph is shifted down 3 units.

13. continued Graph f(x) = 2x2  3. vertex (0, 3) axis x = 0
11 1 5 3 1  2
vertex (0, 3)
axis x = 0
domain: (, )
range: (, 3]

14. EXAMPLE 4
The number of higher order multiple births in the United States is rising. Let x represent the number of years since 1970 and y represent the rate of higher-order multiples born per 100,000 births since The data are shown in the following table.
Use the data points (1, 29.1), (6, 35), (26, 152.6) to find a quadratic equation model for the data.
Use ax2 + bx + c
(1, 29.1) = a(1)2 + b(1) + c = 29.1
(6, 35) = a(6)2 + b(6) + c = 35
(26, 152.6) = a(26)2 + b(26) + c = 152.6

15. continued Simplify the system: a + b + c = 29.1 (1)
To eliminate c, multiply equation (1) by –1 and add the result to equation (2).
a  b  c =  1  (1)
36a + 6b + c = (2)
35a + 5b = (4)

16. continued
To eliminate c again, multiply equation (2) by –1 and add the result to equation (3).
36a  6b  c =  1  (2)
676a + 26b + c = (3)
640a + 20b = (5)
To eliminate b, multiply equation (4) by –4 and add the result to equation (5).
140a  20b =  4 (4)
640 a + 20b = (5)
500a = 94
a = 94/500 = 0.188

17. continued To find b, substitute 0.188 for a in equation (4).
35a + 5b = 5.9
35(0.188) + 5b = 5.9
b = 5.9
5b = –0.68
b = – 0.136
To find c, use equation (1). a + b + c = (1)
c = 29.1 – a – b
c = 29.1 – – (–0.136) =
The quadratic model using the three points is:
y = 0.188x2 – 0.136x