84

Sect 3.1 – Quadratic Functions and Models

Objective 1: Polynomial Function

In modeling, the most common function used is a polynomial function. A polynomial function has the property that the powers of the variables are whole numbers (0, 1, 2, 3, ...). There can be no variables in the denominator or under a radical sign.

Polynomial Function A polynomial function f of degree n is in the form: f(x) = anxn + an – 1xn – 1 + … + a1x + a0 where n is a whole number, a0 through an are real numbers, and an ≠ 0. The term anxn with the highest power is called the leading term. The coefficient an of the term with the highest power is called the leading coefficient. Determine if the following is a polynomial. If so, state leading term and leading coefficient: Ex. 1a x5 –

89 x4 +

56 x3 Ex. 1b x2/3 – 8x – 11

Ex. 1c –

€

7x4

3 Ex. 1d

€

3511x2

Ex. 1e 4.95x2 – 9 Solution: a) The powers of the variables are 5, 4, and 3 which are all whole numbers. So, the expression is a polynomial. The leading term is x5 and the leading coefficient is 1.

b) Since 2/3 is a not a whole number, then the expression is not a polynomial.

c) The powers of the variables are 3, and 1 which are all whole numbers. So, the expression is a polynomial. The

leading term is –

€

7x4

3 and the leading coefficient is –

€

73

.

d) Since x2 is in the denominator, then the expression is not a polynomial.

e) The powers of the variable is 2 which is a whole number. So, the expression is a polynomial. The leading term is 4.95x2 and the leading coefficient is 4.95.

85

Objective 2: Quadratic Functions

A quadratic function is a second-degree polynomial in one variable. Quadratic Function A quadratic function is a function of the form: f(x) = ax2 + bx + c where a, b, and c are real numbers and a ≠ 0. This is usually referred to as standard form. The graph of a quadratic function is called a parabola. The vertex is always either the lowest point or the highest point on the graph and the vertical line through the vertex is called the axis of symmetry. It is where we can fold the graph paper so that both halves of the parabola coincide. The coordinates of the vertex is (h, k) and the axis of symmetry is x = h: Graphs of f(x) = a(x – h)2 + k 1) The vertex is the point (h, k). 2) The axis of symmetry is the line x = h. 3) If a > 0, the graph opens upward (smile), and k is the minimum value of the function. 4) If a < 0, the graph opens downward (frown), and k is the maximum value of the function. a > 0 a < 0

Previously, we graphed functions using transformations. In order to do so, our quadratic functions needs to be rewritten from standard form into graphing form (f(x) = a(x – h)2 + k).

Vertex (h, k) Minimum Point

Axis of Symmetry x = h

Axis of Symmetry x = h

Maximum Point Vertex (h, k)

86

Objective 3: Using Completing the Square and Transformations to Graph a Quadratic Function. To get a quadratic function into graphing form, we need to complete the square. To complete the square, we follow this procedure: Procedure for converting from f(x) = ax2 + bx + c to f(x) = a(x – h)2 + k 1) If the coefficient of the squared term is not one, factor it out from both the squared term and the linear term. The coefficient of the linear term inside the parenthesis will be b divided by a. 2) Divide the coefficient of the linear term by 2 (or multiply it by a half) to get p. Square p and write plus the result (p2) and minus the result (p2) inside of the set of parenthesis. 3) Move the minus p2 out of the parenthesis by distributing a to that value. Combine the result with c. 4) Rewrite the perfect square in the parenthesis as (x + p)2. The quadratic function should be in graphing form. Given p(x) = 4x2 – 16x + 13 Ex. 2 a) Write the function in graphing form. b) Find the vertex, axis of symmetry, and the maximum or minimum value. c) Find the x- and y-intercepts. d) Sketch the graph. Solution: a) p(x) = 4x2 – 16x + 13 (factor out 4 from 4x2 + 16x) p(x) = 4(x2 – 4x) + 13 (p = – 4 ÷ 2 = – 2. Add and subtract (– 2)2 = 4) p(x) = 4(x2 – 4x + 4 – 4) + 13 (move the – 4 out and times it by 4) p(x) = 4(x2 – 4x + 4) – 16 + 13 (combine – 16 and 13) p(x) = 4(x2 – 4x + 4) – 3 (rewrite in the form (x + p)2) p(x) = 4(x – 2)2 – 3 a = 4, h = 2, and k = – 3 b) Vertex: (2, – 3) Axis of symmetry: x = 2 Since a > 0, the graph opens upward, so p(x) has a minimum value of – 3 at x = 2. c) x-intercepts. Let p(x) = 0: 0 = 4x2 – 16x + 13

87

-4

-3

-2

-1

0

1

2

3

4

-2 -1 0 1 2 3 4 5 6

a = 4, b = – 16, and c = 13

x =

€

−b± b2 −4ac2a

=

€

−(−16)± (−16)2 −4(4)(13)2(4)

=

€

16± 256−2088

=

€

16± 488

=

€

16± 42•38

=

€

16±4 38

=

€

4(4± 3 )8

=

€

4± 32

The x-intercepts are (

€

4− 32

, 0) and (

€

4+ 32

, 0) y-intercepts. Let x = 0: p(0) = 4(0)2 – 16(0) + 13 = 13 The y-intercept is (0, 13).

d) Let's go through the steps of our general strategy : i) Since | a | = 4, the graph is stretched by a factor of 4. ii) Since a is positive, the graph is not reflected across the x-axis. iii) Since k is – 3 and h is 2, the graph is shifted down by 3 units and to the right by 2 unit.

Objective 4: Deriving a Formula for the Vertex and Axis of Symmetry. Given a quadratic function in the form f(x) = ax2 + bx + c where a ≠ 0, we want to derive a formula for getting the function into graphing form much in the same way as we derived the quadratic formula. To write the function in graphing form (f(x) = a(x – h)2 + k), we need to find a, h, and k. The value of a is the same value of a in the standard form of the quadratic equation. If we know h, we can find k by evaluating the function at x = h since f(h) = a((h) – h)2 + k = a(0)2 + k = k. Thus, k = f(h). So, let’s find a formula for h. We will also do a numerical example to see how the steps work: f(x) = 3x2 + 5x – 7 f(x) = ax2 + bx + c Factor out a from the variable terms. f(x) = 3(x2 +

€

53

x) – 7 f(x) = a(x2 +

€

ba

x) + c

p =

€

12•

€

53

=

€

56

. (

€

56 )

2 =

€

2536

p =

€

12•

€

ba

=

€

b2a

. (

€

b2a )

2 =

€

b2

4a2

88

Add and subtract the result inside the parenthesis.

f(x) = 3(x2 +

€

53

x +

€

2536

–

€

2536

) – 7 f(x) = a(x2 +

€

ba

x +

€

b2

4a2 –

€

b2

4a2 ) + c

Move the negative constant term out and times it by a.

f(x) = 3(x2 +

€

53

x +

€

2536

) – 3(

€

2536

) – 7 f(x) = a(x2 +

€

ba

x +

€

b2

4a2 ) – a(

€

b2

4a2 ) + c

f(x) = 3(x2 +

€

53

x +

€

2536

) –

€

2512

–

€

8412

f(x) = a(x2 +

€

ba

x +

€

b2

4a2 ) –

€

b2

4a +

€

4ac4a

Combine the constant terms outside of the parenthesis.

f(x) = 3(x2 +

€

53

x +

€

2536

) –

€

10912

f(x) = a(x2 +

€

ba

x +

€

b2

4a2 ) +

€

4ac−b2

4a

Rewrite in the perfect square trinomial in the form (x + p)2.

f(x) = 3(x +

€

56

)2 –

€

10912

f(x) = a(x +

€

b2a

)2 +

€

4ac−b2

4a

f(x) = 3(x – (–

€

56

))2 –

€

10912

f(x) = a(x – (–

€

b2a

))2 +

€

4ac−b2

4a

Thus, h = –

€

56

Thus, h = –

€

b2a

Incidentally, we do have a formula for k: k =

€

4ac−b2

4a, but it is usually easier

to use k = f(h) to find k. Notice that the formula h = –

€

b2a

looks like the quadratic formula without the ± radical. Properties of the Graph of a Quadratic Function Given f(x) = ax2 + bx + c where a ≠ 0, 1) The vertex (h, k) is (–

€

b2a

, f(–

€

b2a

)).

2) The axis of symmetry is the line x = h = –

€

b2a

3) If a > 0, the vertex is a minimum point. If a < 0, the vertex is a maximum point. Objective 5: Graphing Quadratic Functions. For the following quadratic functions: a) Find the vertex, the axis of symmetry, and the minimum or maximum value. b) Find the x- and y-intercepts. c) Write the function in graphing form. d) Sketch the graph.

89

-2

0

2

4

6

8

10

-6 -4 -2 0 2 4 6

Ex. 3 f(x) = 2x2 + 12x + 17 Solution: a) f(x) = 2x2 + 12x + 17, a = 2, b = 12, & c = 17 (vertex formula) h = –

€

b2a

= –

€

(12)2(2)

= – 3

k = f(– 3) = 2(– 3)2 + 12(– 3) + 17 = 18 – 36 + 17 = – 1 Thus, the vertex is (– 3, – 1). The axis of symmetry is x = – 3. Since a > 0, f has a minimum value of – 1 at x = – 3.

b) x-intercepts. Let f(x) = 0: 0 = 2x2 + 12x + 17, a = 2, b = 12, & c = 17 (quadratic formula)

x =

€

−b± b2 −4ac2a

=

€

−(12)± (12)2 −4(2)(17)2(2)

=

€

− 12± 144−1364

=

€

− 12± 84

=

€

− 12± 22•24

=

€

− 12 ± 2 24

=

€

2(− 6± 2 )4

=

€

− 6 ± 22

. y - intercepts. Let x = 0: f(0) = 2(0)2 + 12(0) + 17 = 17

So, the x-intercepts are (

€

− 6− 22

, 0) and (

€

− 6+ 22

, 0) and the y-intercept is (0, 17).

c) Since a = 2, h = – 3, and k = – 1, then the graphing form is f(x) = 2(x + 3)2 – 1

d) Let's go through the steps of our general strategy: i) Since | a | = 2, the graph is stretched by a factor of 2. ii) Since a is positive, the graph is not reflected across the x-axis. iii) Since k is – 1 and h is – 3, the graph is shifted down by 1 unit and to the left by 3 unit. Ex. 4 g(x) = –

€

13

x2 + 4x – 14

90

-16

-14

-12

-10

-8

-6

-4

-2

0

2

-2 0 2 4 6 8 10 12 14 16

Solution: a) g(x) = –

€

13

x2 + 4x – 14, a = –

€

13

, b = 4, & c = – 14 (vertex formula) h = –

€

b2a

= –

€

(4)2(− 1

3) = – 4 ÷ (–

€

23 ) = –

€

41•(–

€

32 ) = 6

k = g(6) = –

€

13 (6)2 + 4(6) – 14 = – 12 + 24 – 14 = – 2

Thus, the vertex is (6, – 2). The axis of symmetry is x = 6. Since a < 0, f has a maximum value of – 2 at x = 6.

b) x-intercepts. Let f(x) = 0: 0 = –

€

13

x2 + 4x – 14, a = –

€

13

, b = 4, & c = – 14 (quadratic formula)

x =

€

−b± b2 −4ac2a

=

€

−(4) ± (4)2 −4(− 13

)(−14)

2(− 13

) =

€

−4± 16− 563

− 23

=

€

−4± 483− 56

3

− 23

=

€

−4± − 83

− 23

, but this is not a real number.

y - intercepts. Let x = 0: f(0) = –

€

13

(0)2 + 4(0) – 14 = – 14 So, there no x-intercepts. y-intercept is (0, – 14). c) Since a = –

€

13

, h = 6, and k = – 2, then the graphing form is

f(x) = –

€

13

(x – 6)2 – 2

d) Let's go through the steps of our general strategy: i) Since | a | =

€

13

, the graph

is shrunk by a factor of

€

13

. ii) Since a is negative, the graph is reflected across the x-axis. iii) Since k is – 2 and h is 6, the graph is shifted down by 2 units and to the right by 6 units.

91

Notice that the discrimnant b2 – 4ac under the radical in the quadratic formula determine the number of x-intercepts the quadratic function has: 1) If b2 – 4ac > 0, then the quadratic function has two distinct x-intercepts. 2) If b2 – 4ac = 0, then the quadratic function has one x-intercept. 3) If b2 – 4ac < 0, then the quadratic function has no x-intercepts. b2 – 4ac > 0 b2 – 4ac = 0 b2 – 4ac < 0

Objective 6: Finding the Quadratic Function Given its Vertex and One Other Point. If we have the vertex and one other point on the graph of the parabola, we can plug into the graphing form of the quadratic function to find the function. Use the information to find the Quadratic Function in standard form: Ex. 5 The vertex of the parabola is (4, – 5) and the parabola also passes through the point (2, – 7). Solution: First, plug the vertex into f(x) = a(x – h)2 + k: f(x) = a(x – 4)2 – 5 Since y = f(2) = – 7, plug in the point (2, – 7) in for x and f(x) respectively: – 7 = a(2 – 4)2 – 5 – 7 = a(– 2)2 – 5 – 7 = 4a – 5 – 2 = 4a – 0.5 = a So, f(x) = – 0.5(x – 4)2 – 5. Now expand and combine like terms: – 0.5(x – 4)2 – 5 = – 0.5(x2 – 8x + 16) – 5 = – 0.5x2 + 4x – 8 – 5 = – 0.5x2 + 4x – 13. Hence, f(x) = – 0.5x2 + 4x – 13

-5 -4 -3 -2 -1 0 1 2 3 4

-5 -4 -3 -2 -1 0 1 2 3 4

-5

-4

-3

-2

-1

0

1

-3 -2 -1 0 1 2 3

-1

0

1

2

3

4

5

-3 -2 -1 0 1 2 3

92

Objective 7: Quadratic Models The maximum and minimum value of a quadratic function occurs at the vertex. The sign of a will determine if we have a maximum or minimum value: Solve the following: Ex. 6 A manufacturer of wide screen TV sets determines that their profits (p(x) is dollars) depends on the number of TV sets, x, that they produce and sell and is given by: p(x) = – 0.15x2 + 45x – 1450 where x ≥ 0 a) Find the y-intercept and interpret its meaning. b) How many TV sets must be produced and sold for the manufacturer to break even? c) How many TV sets must be produced and sold to maximize the profit? What is the maximum profit? Solution: a) y-intercept. Let x = 0: p(0) = – 0.15(0)2 + 45(0) – 1450 = – 1450 So, the y-intercept is (0, – 1450). This means that is the manufacturer produces no TV sets, then the manufacture will lose $1450.

b) To break even means the profit is 0. Setting the p(x) = 0 and solving yields: – 0.15x2 + 45x – 1450 = 0, a = – 0.15, b = 45, & c = – 1450

x =

€

−b± b2 −4ac2a

=

€

−(45)± (45)2 −4(−0.15)(−1450)2(−0.15)

=

€

−45± 2025−870−0.3

=

€

−45± 1155−0.3

.

But, x =

€

−45− 1155−0.3

≈ 263 or x =

€

−45+ 1155−0.3

≈ 37

Either they can produce and sell 37 TV sets or 263 TV sets.

c) Since a < 0, the maximum profit will occur at the vertex. h = –

€

b2a

= –

€

(45)2(−0.15)

= 150

k = p(150) = – 0.15(150)2 + 45(150) – 1450 = – 3375 + 6750 – 1450 = 1925 Thus, the vertex is (150, 1925). The manufacturer will have a maximum profit of $1925 when 150 TV sets are produced and sold.

93

Ex. 7 A farmer can get $30 per 50-lb carton of potatoes on July 1st. After that, the price drops by 30 cents per carton per day. A farmer has 80 cartons in the field on July 1st and estimates that the crop is increasing at a rate of one carton per day. When should the farmer harvest the potatoes to maximize revenue and what is the maximum revenue? Solution: Let x = the number of days after July 1st

Since the price is decreasing by 30 cents per day, then, after x days, the price is 30 – 0.3x. Since the number of cartons is increasing by one carton per day, the number of cartons after x days is 80 + x. We can then find the revenue function: R(x) = (price)(number of cartons)

= (30 – 0.3x)(80 + x) = – 0.3x2 + 6x + 2400 The domain of the function is [0, ∞) since x could equal 0. The maximum revenue will occur at the vertex since a < 0, so we need to calculate the vertex: h = –

€

b2a

= –

€

62(−0.3)

= 10

k = R(10) = – 0.3(10)2 + 6(10) + 2400 = 2430 Since x = 10 corresponds to ten days after July 1st, the farmer should harvest on July 11 to maximize the total revenue. The maximum revenue is $2430.