Transcript for Lesson 4-3

Start-Up Day 13
Solve each system of equations.
1.
3a + b = –5
2a – 6b = 30
2.
9a + 3b = 24
a+b=6
a = 0, b = –5
a = 1, b = 5
HOME LEARNING QUESTIONS?
OBJECTIVE:
Students will be able to use quadratic
functions to model data.
ESSENTIAL QUESTION:
How can you find three coefficients a, b, and c, of
f(x) = ax2 + bx + c which create a quadratic model
for 3 points on the curve?
HOME LEARNING:
p. 212 # 7, 13, 16, 17, 19,20, 24 & 26+
MathXL Unit 2 Test Review (Lessons 4.14.3)
Objectives
Use quadratic functions to model data.
Use quadratic models to analyze and
predict.
Vocabulary
quadratic model
Just as two points define a linear function, three noncollinear points define a quadratic function. You can
find three coefficients a, b, and c,
of f(x) = ax2 + bx + c
by using a system of three equations, one for each
point.
The points do not need to have equally spaced xvalues.
http://www.virtualnerd.com/tutorials/?id=Alg2_05_01_0020
Reading Math
Collinear points lie on the same line. Noncollinear
points do not all lie on the same line.
Example 1: Writing a Quadratic Function from Data
Write a quadratic function that fits the points
(1, –5), (3, 5) and (4, 16).
Use each point to write a system of equations to
find a, b, and c in f(x) = ax2 + bx + c.
(x, y)
(1, –5)
f(x) = ax2 + bx + c
–5 = a(1)2 + b(1) + c
System in a, b, c
a + b + c = –5
(3, 5)
5 = a(3)2 + b(3) + c
9a + 3b + c = 5
(4, 16)
16 = a(4)2 + b(4) + c
16a + 4b + c = 16
1
Example 1 Continued
Add equation #1 to the
negative of equation #2
to get #4 .
2
1
4
9a + 3b + c = 5
-a - b - c = 5
8a + 2b + 0c = 10
Add equation #1 to the
negative of equation #3
to get #5 .
16a + 4b + c = 16
-a - b - c = 5
3
1
5
15a + 3b + 0c = 21
Example 1 Continued
Solve equation #4 and equation #5 for a and b
using elimination.
5
4
2(15a + 3b = 21)
–3(8a + 2b = 10)
30a + 6b = 42
– 24a – 6b = –30
6a + 0b = 12
a =2
Multiply by 2.
Multiply by –3.
ADD.
Solve for a.
Example 1 Continued
Substitute 2 for a into equation #4 or equation
#5 to get b.
8(2) +2b = 10
15(2) +3b = 21
2b = –6
b = –3
3b = –9
b = –3
Example 1 Continued
Substitute a = 2 and b = –3 into equation #1 to
solve for c.
(2) +(–3) + c = –5
–1 + c = –5
c = –4
Write the function using a = 2, b = –3 and c = –4.
f(x) = ax2 + bx + c
f(x)= 2x2 – 3x – 4
Example 1 Continued
f(x) = ax2 + bx + c
f(x)= 2x2 – 3x – 4
Check Substitute your 3 given points, (1, –5), (3, 5),
and (4, 16), into your model to verify that
they actually satisfy the function rule.
Got it? Example 1
Write a quadratic function that fits the points
(0, –3), (1, 0) and (2, 1).
Use each point to write a system of equations to
find a, b, and c in f(x) = ax2 + bx + c.
(x,y)
(0, –3)
f(x) = ax2 + bx + c
–3 = a(0)2 + b(0) + c
System in a, b, c
1
c = –3
(1, 0)
0 = a(1)2 + b(1) + c
a+b+c=0
(2, 1)
1 = a(2)2 + b(2) + c
4a + 2b + c = 1
1
2
Got It? Example 1 Continued
Substitute c = –3 from equation
equation 2 and equation 3 .
2
a+b+c=0
a+b–3=0
a+b=3
3
4
1
into both
4a + 2b + c = 1
4a + 2b – 3 = 1
4a + 2b = 4
5
Got It? Example 1 Continued
Solve equation
elimination.
4
5
4
and equation
-4(a + b) = -4(3)
4a + 2b = 4
5
for b using
-4a - 4b = -12
4a + 2b = 4
Multiply by- 4.
0a - 2b = -8
b=4
Solve for b.
ADD
Got It? Example 1 Continued
Substitute 4 for b into equation
to find a.
4
a+b=3
a+4=3
a = –1
or equation
4
5
5
4a + 2b = 4
4a + 2(4) = 4
4a = –4
a = –1
Write the function using a = –1, b = 4, and c = –3.
f(x) = ax2 + bx + c
f(x)= –x2 + 4x – 3
Got It? Example 1 Continued
f(x) = ax2 + bx + c
f(x)= –x2 + 4x – 3
Check Substitute to verify that (0, –3), (1, 0),
and (2, 1) satisfy the function rule.
Recall that the
maximum
value will occur
at the vertex,
so we will need
to find
the vertex for
each .
f(x) = ax2 + bx + c
f(x)= –16t2 + 150t + 1
RECALL: The x value of the vertex of the parabola
can be found by computing
-b
x=
2a
-b -150
x value of vertex =
=
= 4.6875
2a -32
f(x) = ax2 + bx + c
f(x)= –16t2 + 150t + 1
RECALL: The y value of the vertex of the parabola
can be found by computing
æ -b ö
fç ÷
è 2a ø
æ -b ö
y value of vertex = f ç ÷
è 2a ø
= -16 ( 4.6875) +150 ( 4.6875) +1
2
= 352.5625
Conclusion: Rocket 2 went higher by more than 227 feet!
YOUR TURN: f(x) = ax2 + bx + c
f(x)= –16t2 + 72t + 5
An arrow shot into the air is modeled by the f(x).
Determine the maximum height of its flight path.
-b
x=
2a
æ -b ö
y= fç ÷
è 2a ø
Mini Quiz
Write a quadratic function that fits the points
(2, 0), (3, –2), and (5, –12).
f(x) = –x2 + 3x – 2