start-up day 13 solve each system of equations. a = 0, b = –5 1. 2. 2a – 6b = 30 3a + b = –5 a...
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Start-Up Day 13Solve each system of equations.
a = 0, b = –5 1.
2.
2a – 6b = 30
3a + b = –5
a + b = 6
9a + 3b = 24a = 1, b = 5
Students will be able to use quadratic functions to model data.
Objective:
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 4.1-4.3)
Just as two points define a linear function, three non-collinear 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 x-values.
Collinear points lie on the same line. Noncollinear points do not all lie on the same line.
Reading Math
http://www.virtualnerd.com/tutorials/?id=Alg2_05_01_0020
Write a quadratic function that fits the points (1, –5), (3, 5) and (4, 16).
Example 1: Writing a Quadratic Function from Data
Use each point to write a system of equations to find a, b, and c in f(x) = ax2 + bx + c.
(x, y) f(x) = ax2 + bx + c System in a, b, c
(1, –5) –5 = a(1)2 + b(1) + c a + b + c = –5 1
9a + 3b + c = 5
16a + 4b + c = 16
(3, 5) 5 = a(3)2 + b(3) + c
(4, 16) 16 = a(4)2 + b(4) + c
Example 1 Continued
Add equation #1 to the negative of equation #2 to get #4 .
Add equation #1 to the negative of equation #3 to get #5 .
2 9a + 3b + c = 51 - a - b - c = 5
4 8a + 2b + 0c = 10
3 16a + 4b + c = 161 - a - b - c = 5
5 15a + 3b + 0c = 21
Solve equation #4 and equation #5 for a and b using elimination.
4
5 2(15a + 3b = 21) –3(8a + 2b = 10)
30a + 6b = 42 – 24a – 6b = –30
6a + 0b = 12
Multiply by –3.
Multiply by 2.
ADD.
Solve for a. a = 2
Example 1 Continued
Substitute 2 for a into equation #4 or equation #5 to get b.
8(2) +2b = 10
2b = –6
15(2) +3b = 21
3b = –9
b = –3b = –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.
Example 1 Continued
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.
Example 1 Continued
f(x) = ax2 + bx + c f(x)= 2x2 – 3x – 4
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) f(x) = ax2 + bx + c System in a, b, c
(0, –3) –3 = a(0)2 + b(0) + c c = –3 1
a + b + c = 0
4a + 2b + c = 1
(1, 0) 0 = a(1)2 + b(1) + c
(2, 1) 1 = a(2)2 + b(2) + c
1
2
Got it? Example 1
Substitute c = –3 from equation into both equation and equation .2
32
3
a + b + c = 0 4a + 2b + c = 1
Got It? Example 1 Continued
1
a + b – 3 = 0 4a + 2b – 3 = 1
a + b = 3 4a + 2b = 4 54
Solve equation and equation for b using elimination.
4
5
4
5
-4(a + b) = -4(3)4a + 2b = 4
-4a - 4b = -12 4a + 2b = 4
0a - 2b = -8
Got It? Example 1 Continued
b = 4
Multiply by- 4.
ADD
Solve for b.
Substitute 4 for b into equation or equation to find a.
a + b = 3
a + 4 = 3 4a + 2b = 4
4a + 2(4) = 4
4a = –4 a = –1
Write the function using a = –1, b = 4, and c = –3.
4 5
a = –1
Got It? Example 1 Continued
4 5
Check Substitute to verify that (0, –3), (1, 0), and (2, 1) satisfy the function rule.
Got It? Example 1 Continued
f(x) = ax2 + bx + c f(x)= –x2 + 4x – 3
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 + 1RECALL: The x value of the vertex of the parabola can be found by computing
f(x) = ax2 + bx + c
f(x)= –16t2 + 150t + 1RECALL: The y value of the vertex of the parabola can be found by computing
Conclusion: Rocket 2 went higher by more than 227 feet!
YOUR TURN: f(x) = ax2 + bx + c
f(x)= –16t2 + 72t + 5An arrow shot into the air is modeled by the f(x). Determine the maximum height of its flight path.
Mini Quiz
Write a quadratic function that fits the points
(2, 0), (3, –2), and (5, –12).
f(x) = –x2 + 3x – 2