Parabolas 4'1

1
Parabolas 4.1
2
f(x) x2 x 1

• x x2 x 1 f(x)
• 0 0 0 1 0
• 1 1 1 1 3
• 2 4 2 1 7
• 3 9 3 1 13
• 4 16 4 1 21

3
f(x) x2 x 1

• x x2 x 1 f(x)
• 10 100 10 1 111
• 100 10000 100 1 10,101
• 1000 1,000,000 1000 1 1,001,001

4
The First Term
x2
..carries the most weight and determines the
overall shape (for large numbers of x ).
x
1
5
ax2 bx c a gt 0
opens up
0x2 bx c a 0
opens down
ax2 bx c a lt 0
6
x -b vb2 - 4ac 2a
7
axis of symmetry
x -b/2a
x intercept
x intercept
vertex
8
y a(x h)2 kwhere (h, k) is the vertex
There are 5 holes in this equation. We can now
fill 4 of them. Use the vertex y a(x-1)2
(-5) y a(x-1)2 5 Use the second point -3
a(0-1)2 5 -3 1a 5 2 a
9
y a(x h)2 kwhere (h, k) is the vertex
y 2(x-h)2 k y 2(x-1)2 (-5) y 2(x-1)2
5 y 2(x 2 2x 1) 5 y 2x 2 4x 2
5 y 2x 2 4x 3
10
The min or max is just f(-b/2a) !
11
What is the max area you can enclose in a
rectangular with 50 feet of fence?
25-x
x
Area x(25-x)
12

• Area x(25-x)
• 25x x2
• -x2 25x

What is a? So it opens down and the vertex is a
maximum. At what x is the maximum? -b/2a
-25/-2 12.5 So a rectangle 12.5 ft. wide holds
the max area.
13
A(x) -x2 25x
0 12.5 25
14
Homework

• Page 306
• 11-18, 35-66