```Modeling with Quadratic Functions
Depending on what information we are given will
determine the form that is the most convenient to use.
Graphing
2
y=x
Standard
Vertex
y = ax2+bx+c
y= a (x - h)2+k
axis:
x
b
2a
vertex ( x , y )
axis: x = h
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
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.
Substitute 3 for x and 2 for y.
2 = a(3 – 1)2 – 2
Simplify coefficient of a.
2 = 4a – 2
Solve for a.
1=a
A quadratic function for the parabola is y = 1(x – 1)2 – 2.
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.
Vertex form
y = a(x – h)2 + k
y = a(x – 4)2 – 5
Substitute 4 for h and –5 for k.
Use the other given point, (2,–1), to find a.
Substitute 2 for x and –1 for y.
–1 = a(2 – 4)2 – 5
Simplify coefficient of x.
–1 = 4a – 5
Solve for a.
1=a
A quadratic function for the parabola is y = 1(x – 4)2 – 5.
3. vertex: (–3, 1)
passes through: (0, –8)
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.
Substitute 2 for x and –8 for y.
–8 = a(0 + 3)2 + 1
Simplify coefficient of x.
–8 = 9a + 1
Solve for a.
–1 = a
A quadratic function for the parabola is y = 1(x + 3)2 + 1.
Steps for solving in 3 variables
1. Using the 1st 2 equations, cancel one of the variables.
2. Using the last 2 equations, cancel the same variable
from step 1.
3. Use the results of steps 1 & 2 to solve for the 2
remaining variables.
4. Plug the results from step 3 into one of the original 3
equations and solve for the 3rd remaining variable.
5. Write the quadratic equation in standard form.
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
–4=c
6 = a(2)2 + b(2) + c
Substitute 0 for x and – 4 for y.
6 = 4a + 2b + c
Equation 2
Substitute 2 for x and 6 for y.
Equation 3
STEP 2
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.
a–b+c=–3
a–b–4=–3
a–b=1
4a + 2b + c = 6
4a + 2b - 4 = 6
4a + 2b = 10
Equation 1
Substitute – 4 for c.
Revised Equation 1
Equation 3
Substitute – 4 for c.
Revised Equation 3
STEP 3
Solve the system consisting of revised Equations 1
and 3. Use the elimination method.
a–b=1
4a + 2b = 10
2a – 2b = 2
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.
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
–1=c
Substitute 0 for x and – 1 for y.
11 = a(2)2 + b(2) + c
Substitute 2 for x and 11 for y.
11 = 4a + 2b + c
Equation 3
Equation 2
STEP 2
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.
a–b+c=5
a–b–1=5
a–b=6
4a + 2b + c = 11
4a + 2b – 1 = 11
4a + 2b = 12
Equation 1
Substitute – 1 for c.
Revised Equation 1
Equation 3
Substitute – 1 for c.
Revised Equation 3
STEP 3 Solve the system consisting of revised
Equations 1 and 3. Use the elimination method.
a–b=6
4a + 2b = 10
2a – 2b = 12
4a + 2b = 12
6a = 24
a= 4
So, 4 – b = 6, which means b = – 2
A quadratic function for the parabola is
y = 4x2 – 2x – 1
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
0= 1a + 1b + c
Substitute 1 for x and 0 for y.
Equation 1
-3 = a(2)2 + b(2) + c
-3 = 4a +2b + c
Substitute 2 for x and- 3 for y.
-10 = a(3)2 + b(3) + c
Substitute 3 for x and -10 for y.
-10 = 9a + 3b + c
Equation 3
Equation 2
STEP 2 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
PART 1
Equation 1
Equation 2
0 = 1a + 1b + c
-1
0 = -1a - 1b - c
-3 = 4a +2b + c
-3 = 3a +1b PART 1
-1
3 = -4a -2b - c
-10 = 9a + 3b + c
-7 = 5a + 1b PART 2
-3 = 4a +2b + c
PART 2
Equation 2
-3 = 4a +2b + c
Equation 3
-10 = 9a + 3b + c
PART 1
PART 2
-3 = 3a +1b
-7 = 5a + 1b
-1
3 = -3a - 1b
-7 = 5a + 1b
-4 = 2a
-2 = a
STEP 3
Solve for the remaining variables by substitution.
Original equations
0 = 1a + 1b + c
-3 = 4a +2b + c
-10 = 9a + 3b + c
-3 = 3a +1b
-3 = 3a +1b PART 1
-7 = 5a + 1b
PART 2
-2 = a
-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
Substitute
-2 = a
3=b
-1 = c
y = ax2 + bx + c
y=
2
-2x
+ 3x -1
FINAL
QUIZ
Wednesday, February 15, 2012
One question each similar to the following
examples (in this power point):
Examples: 2, 5, & 6
GOOD LUCK!!
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