Modeling with Quadratic Functions

Depending on what information we are given will determine the form that is the most convenient to use.

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

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.

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

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)

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.

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

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

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.

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

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

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.

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

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

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

-2 = a 3 = b -1 = c

y = ax2 + bx + c

y = -2x2 + 3x -1

Substitute

FINAL ANSWER!!

QUIZ Wednesday, February 15, 2012

One question each similar to the following examples (in this power point):

Examples: 2, 5, & 6

GOOD LUCK!!