modeling with quadratic functions
DESCRIPTION
Modeling with Quadratic Functions. Depending on what information we are given will determine the form that is the most convenient to use. Graphing y=x 2. 1. Write a quadratic function for the parabola shown. Use vertex form because the vertex is given. y = a ( x – h ) 2 + k. - PowerPoint PPT PresentationTRANSCRIPT
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Modeling with Quadratic Functions
Depending on what information we are given will determine the form that is the most convenient to use.
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Graphing y=x2
Standard Vertex
y = ax2+bx+c y= a (x - h)2+k
axis: axis: x = h
vertex ( x , y ) vertex ( h , k )
C is y-intercept
a>0 U shaped, Minimum a>0 U shaped, Minimum
a<0 ∩ shaped, Maximum a<0 ∩ shaped, Maximum
a
bx
2
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1. Write a quadratic function for the parabola shown.
Use vertex form because the vertex is given.
y = a(x – h)2 + k Vertex form
y = a(x – 1)2 – 2 Substitute 1 for h and –2 for k.
Use the other given point, (3, 2), to find a.2 = a(3 – 1)2 – 2 Substitute 3 for x and 2 for y.
2 = 4a – 2 Simplify coefficient of a.
1 = a Solve for a.
A quadratic function for the parabola is y = 1(x – 1)2 – 2.
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2. Write a quadratic function whose graph has the given characteristics.
vertex: (4, –5)
passes through: (2, –1)
Use vertex form because the vertex is given.y = a(x – h)2 + k Vertex form
y = a(x – 4)2 – 5 Substitute 4 for h and –5 for k.
Use the other given point, (2,–1), to find a.–1 = a(2 – 4)2 – 5 Substitute 2 for x and –1 for y.
–1 = 4a – 5 Simplify coefficient of x.
1 = a Solve for a.
A quadratic function for the parabola is y = 1(x – 4)2 – 5.
ANSWER
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Use vertex form because the vertex is given.
y = a(x – h)2 + k Vertex form
y = a(x + 3)2 + 1 Substitute –3 for h and 1 for k.
Use the other given point, (0,–8), to find a.–8 = a(0 + 3)2 + 1 Substitute 2 for x and –8 for y.
–8 = 9a + 1 Simplify coefficient of x.
–1 = a Solve for a.
A quadratic function for the parabola is y = 1(x + 3)2 + 1.
ANSWER
3. vertex: (–3, 1)
passes through: (0, –8)
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Steps for solving in 3 variablesSteps for solving in 3 variables
1.1. Using the 1Using the 1stst 2 equations, cancel one of the variables. 2 equations, cancel one of the variables.
2.2. Using the last 2 equations, cancel the same variable Using the last 2 equations, cancel the same variable from step 1.from step 1.
3.3. Use the results of steps 1 & 2 to solve for the 2 Use the results of steps 1 & 2 to solve for the 2 remaining variables.remaining variables.
4.4. Plug the results from step 3 into one of the original 3 Plug the results from step 3 into one of the original 3 equations and solve for the 3equations and solve for the 3rdrd remaining variable. remaining variable.
5.5. Write the quadratic equation in standard form.Write the quadratic equation in standard form.
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4. Write a quadratic function in standard form for the parabola that passes through the points
(–1, –3), (0, – 4), and (2, 6).
STEP 1 Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.
–3 = a(–1)2 + b(–1) + c Substitute –1 for x and -3 for y.
–3 = a – b + c Equation 1
–4 = a(0)2 + b(0) + c Substitute 0 for x and – 4 for y.
– 4 = c Equation 2
6 = a(2)2 + b(2) + c Substitute 2 for x and 6 for y.
6 = 4a + 2b + c Equation 3
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Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 4 for c in Equations 1 and 3.
STEP 2
a – b + c = – 3 Equation 1
a – b – 4 = – 3 Substitute – 4 for c.
a – b = 1 Revised Equation 1
4a + 2b + c = 6 Equation 3
4a + 2b - 4 = 6 Substitute – 4 for c.
4a + 2b = 10 Revised Equation 3
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a – b = 1 2a – 2b = 2
4a + 2b = 10 4a + 2b = 10
6a = 12
a = 2
So 2 – b = 1, which means b = 1.
The solution is a = 2, b = 1, and c = – 4.
A quadratic function for the parabola is y = 2x2 + x – 4.
ANSWER
STEP 3Solve the system consisting of revised Equations 1 and 3. Use the elimination method.
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5. Write a quadratic function in standard form for the parabola that passes through the given points.
(–1, 5), (0, –1), (2, 11)
STEP 1 Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.
5 = a(–1)2 + b(–1) + c Substitute –1 for x and 5 for y.
5 = a – b + c Equation 1
–1 = a(0)2 + b(0) + c Substitute 0 for x and – 1 for y.
– 1 = c Equation 2
11 = a(2)2 + b(2) + c Substitute 2 for x and 11 for y.
11 = 4a + 2b + c Equation 3
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a – b + c = 5 Equation 1
a – b – 1 = 5 Substitute – 1 for c.
a – b = 6 Revised Equation 1
4a + 2b + c = 11 Equation 3
4a + 2b – 1 = 11 Substitute – 1 for c.
4a + 2b = 12 Revised Equation 3
Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 1 for c in Equations 1 and 3.
STEP 2
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a – b = 6 2a – 2b = 12
4a + 2b = 10 4a + 2b = 12
6a = 24
a = 4
So, 4 – b = 6, which means b = – 2
A quadratic function for the parabola is
4x2 – 2x – 1y =
ANSWER
STEP 3Solve the system consisting of revised
Equations 1 and 3. Use the elimination method.
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6. (1, 0), (2, -3), (3, -10)
STEP 1 Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.
0 = a(1)2 + b(1) + c Substitute 1 for x and 0 for y.
0= 1a + 1b + c Equation 1
-3 = a(2)2 + b(2) + c Substitute 2 for x and- 3 for y.
-3 = 4a +2b + c Equation 2
-10 = a(3)2 + b(3) + c Substitute 3 for x and -10 for y.
-10 = 9a + 3b + c Equation 3
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Equation 1
Equation 2
Equation 2
Equation 3
Rewrite the system of three equations in Step 1 as a system of two equations. Use Elimination to simplify one set of equations then use substitution & elimination to find the
STEP 2
0 = 1a + 1b + c
-3 = 4a +2b + c
-10 = 9a + 3b + c
-1 0 = -1a - 1b - c
-3 = 4a +2b + c -3 = 3a +1b
PART 1
PART 1
-3 = 4a +2b + c
PART 2
-1 3 = -4a -2b - c
-10 = 9a + 3b + c
-7 = 5a + 1b PART 2
PART 1
PART 2
-3 = 3a +1b
-7 = 5a + 1b
-1 3 = -3a - 1b-7 = 5a + 1b
-4 = 2a
-2 = a
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STEP 3Solve for the remaining variables by substitution.
0 = 1a + 1b + c
-10 = 9a + 3b + c
-3 = 4a +2b + c
Original equations
-2 = a
-7 = 5a + 1b
-3 = 3a +1b PART 1
PART 2
-3 = 3a +1b -3 = 3(-2) +1b
-3 = -6 +1b
3 = b
0 = 1a + 1b + c 0 = 1(-2) + 1(3) + c
0 = -2 + 3 + c
0 = 1 + c -1 = c
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-2 = a 3 = b -1 = c
y = ax2 + bx + c
y = -2x2 + 3x -1
Substitute
FINAL ANSWER!!
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QUIZ Wednesday, February 15, 2012
One question each similar to the following examples (in this power point):
Examples: 2, 5, & 6
GOOD LUCK!!