2.3 first and second degree functions

Quadratic functions or second degree functions are

functions of the form y = f(x) = ax2 + bx + c, a = 0.

Second Degree Functions




Below is the graph of y = f(x) = –x2.

x y

–4

–3

–2

–1

0

1

2

3

4









x y

–4

–3

–2

–1

0 0

1 –1

2 –4

3 –9

4 –16





x y

–4 –16

–3 –9

–2 –4

–1 –1

0 0

1 –1

2 –4

3 –9

4 –16




Below is the graph of y = f(x) = –x2 .

x y

–4 –16

–3 –9

–2 –4

–1 –1

0 0

1 –1

2 –4

3 –9

4 –16

The graphs of 2nd (quadratic)

equations are called parabolas.


The graphs of 2nd (quadratic)

equations are called parabolas.

Parabolas describe the paths of

thrown objects (or the upside–

down paths).



Properties of Parabolas:



I. A parabola is symmetric to a center line.




II. The highest or lowest point of the parabola sits on

the center line.





the center line. This point is called the vertex.






III. If the positions of the vertex and another point on

the parabola are known, the parabola is determined.


The vertex position is given by the following formula.





III. If the positions of the vertex and another point on

the parabola are known, the parabola is determined.

Vertex Formula: The vertex of the

parabola y = ax2 + bx + c is at x = –b2a





Moreover, if a > 0, the parabola opens up,

if a < 0 the parabola opens down.



To graph a parabola y = ax2 + bx + c:







1. Set x = in the equation to find the vertex.–b2a







1. Set x = in the equation to find the vertex.

2. Find another point.

–b2a








2. Find another point. (Use x=0 to get the y intercept

if its not the vertex.)

–b2a










3. Locate the reflection across the center line, these

three points form the tip of the parabola.

–b2a











three points form the tip of the parabola. Trace the

parabola.

–b2a












parabola.

4. Set y=0 to find the x intercept.

–b2a












parabola.

4. Set y=0 to find the x intercept. If no real solution

exists, there is no x intercept.

–b2a




Example C. Graph y = x2 – 4x – 12



Vertex: set x = –(–4)

2(1)





Vertex: set x = = 2–(–4)

2(1)




Second Degree FunctionsExample C. Graph y = x2 – 4x – 12

Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16, –(–4)

2(1)





2(1)so v = (2, –16).





2(1)so v = (2, –16).



(2, –16)



2(1)so v = (2, –16).




(Use x=0 or the y intercept if possible.)

–b2a

(2, –16)

Another point: Set x = 0



2(1)so v = (2, –16).





–b2a

(2, –16)

Another point: Set x = 0 then y = –12 or (0, –12)



2(1)so v = (2, –16).





–b2a

(2, –16)



(2, –16)

(0, –12)



2(1)so v = (2, –16).





–b2a



(2, –16)

(0, –12)



2(1)so v = (2, –16).





3. Locate the reflection across the center line,

these three points form the tip of the parabola.

–b2a

(2, –16)

(0, –12) (4, –12)



It's reflection across the mid–line is (4, –12)



2(1)so v = (2, –16).







–b2a

(2, –16)

(0, –12) (4, –12)






2(1)so v = (2, –16).







Trace the parabola.

–b2a

(2, –16)

(0, –12) (4, –12)




Set y = 0 and get x–int:x2 – 4x – 12 = 0



2(1)so v = (2, –16).

(2, –16)

(0, –12) (4, –12)




Set y = 0 and get x–int:x2 – 4x – 12 = 0(x + 2)(x – 6) = 0x = –2, x = 6



2(1)so v = (2, –16).



2(1)so v = (2, –16).

(2, –16)

(0, –12) (4, –12)


Following is an example of maximization via the vertex.



Set y = 0 and get x–int:x2 – 4x – 12 = 0(x + 2)(x – 6) = 0x = –2, x = 6

Example D. The "Crazy Chicken" can sell 120 whole

roasted chicken for $8 in one day. For every $0.50

increases in the price, 4 less chickens are sold. Find

the price that will maximize the revenue (the sale).







Set x to be the number of $0.50 increments above the

base price of $8.00.








So the price is (8 + 0.5x) or (8 + x/2).









The number of chicken sold at this price is (120 – 4x).










Hence the revenue, depending on the number of

times of $0.50 price hikes x, is

R(x) = (8 + x/2)(120 – 4x)












R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.












R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.

This a 2nd degree equation whose graph is a

parabola that opens downward.












R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.

This a 2nd degree equation whose graph is a

parabola that opens downward. The vertex is the

highest point on the graph where R is the largest.


x

R

R = 960 + 28x – 2x2.


x

RThe vertex of this

parabola is at

x = = 7. –(28)2(–2)

R = 960 + 28x – 2x2.


x

RThe vertex of this

parabola is at

x = = 7.

So R(7) = $1058 gives

the maximum revenue.

–(28)2(–2)

R = 960 + 28x – 2x2.

v = (7, 1058)


x

RThe vertex of this

parabola is at

x = = 7.

So R(7) = $1058 gives


–(28)2(–2)

R = 960 + 28x – 2x2.

v = (7, 1058)

More precisely, raise

the price x = 7 times to

8 + 7(0.50) = 11.50 per

chicken, chicken,


x

RThe vertex of this

parabola is at

x = = 7.

So R(7) = $1058 gives


–(28)2(–2)

R = 960 + 28x – 2x2.

v = (7, 1058)



8 + 7(0.50) = 11.50 per

chicken, chicken, and we can sell 120 – 4(7) = 92 chickens

per day with the revenue of 11.5(92) = 1058.


x

RThe vertex of this

parabola is at

x = = 7.

So R(7) = $1058 gives


–(28)2(–2)

R = 960 + 28x – 2x2.

v = (7, 1058)



8 + 7(0.50) = 11.50 per

chicken, chicken, and we can sell 120 – 4(7) = 92 chickens

per day with the revenue of 11.5(92) = 1058.

Any inverse square law in science is 2nd degree.

Laws related to distance or area in mathematics are

2nd degree. That’s why 2nd equations are important.

Finally, we make the observation that given y = ax2 + …,

if a > 0, then the parabola opens upward.

Graphs of Quadratic Equations

if a < 0, then the parabola opens downward.

Exercise A. 1. Practice drawing the parabolas y = ax2 + bx + c.


a > 0

a < 0

Graphs of Quadratic EquationsA. Graph the parabolas by plotting the vertex and using

the symmetry. Find the x and y intercepts, if any.

4. y = x2 – 4 5. y = x2 + 4

2. y = x2 3. y = –x2

6. y = –x2 + 4 7. y = –x2 – 4

8. y = x2 – 2x – 3 9. y = x2 + 2x + 3

10. y = –x2 + 2x – 3 11. y = –x2 – 2x + 3

12. y = x2 – 2x – 8 13. y = x2 + 2x – 8

14. y = –x2 + 2x + 8 15. y = –x2 – 2x + 8

16. a. y = x2 – 4x – 5 b. y = –x2 + 4x + 5

17. a. y = x2 + 4x – 5 b. y = –x2 – 4x + 5

19. y = x2 + 4x – 21 20. y = x2 – 4x – 45

21. y = x2 – 6x – 27 22. y = –x2 – 6x + 27

Graphs of Quadratic EquationsEx B. Graph the following parabolas by plotting the vertex

point, the y–intercept and its reflection.

Verify that there are no x intercepts

(i.e. they have complex roots).

1. y = x2 – 2x + 8 2. y = –x2 + 2x – 5

3. y = x2 + 2x + 3 4. y = –x2 – 3x – 4

5. y = 2x2 + 3x + 4 6. y = x2 – 4x + 32

Graphs of Quadratic EquationsEx C. Carry out the following steps to find the answer for 1 to 3.

a. Let x be the number of times we would raise the price at

each time from the base price of

so the new price would be .

b. Each price hike drop out so the number of customers is .

c. The revenue R(x) is the product of the formulas in a and b which is

2nd degree formula. Find it’s vertex which gives the maximum revenue.

1. Mom’s Kitchen can sell 60 Mom’s Famous apple pies for $12 each

everyday. For every fifty-cent increase in the price, the number of pie sold

drops by 1 pie per day. Use this information to maximize the revenue.

What should you charge?

2. Popi’s Dinner can sell 72 Popi’s Special meal at $24 per person.

For each dollar rise in the price, 2 less Popi’s Special meals would be sold.

Use this information to maximize the revenue.


3. Jr. Eatery can sell 480 fried chickens at $8 each everyday.

For every 25 cents increase in the price, four less chickens are sold.

Use this information to maximize the revenue.



Carry out the following steps to find the answer for 4 to 6.

a. Let (x, y) be any point on the line, solving y in terms of x,

the coordinates may be expressed as .

b. Using the distance formula, the distance between P and a

point on the line is.

c. The minimum distance happens at the vertex of the 2nd

degree equation in the square-root (why?).

5. The distance from the origin to the line y = 2x + 1. Draw.

4. The distance from the point (2, 3) to the line

y = –2x + 3. Draw.

6. The distance from the point (–2, 1) to the line

y = –2x + 2. Draw.

(Answers to odd problems) Exercise A.5. y = –x2 + 4

(0,0)

(0, 4)

7. y = –x2 – 4

(0, –4)

3. y = -x2

11. y = –x2 – 2x + 3 13. y = x2 + 2x – 89. y = x2+2x+3

(-1, 2)

(0, 3)

(-1, 4)

(0, 3)

(-3, 0) (1, 0)(-1, -9)

(0, -8)

(2, 0)(-4, 0)

(2, 0)(–4 , 0)

(–1, 9)

15. y = –x2 – 2x + 8 17. a. y = x2 + 4x – 5 b. y = –x2 – 4x + 5

(1, 0)(–5, 0)

(–2 , –9)

(0, –5) (1, 0)(–5, 0)

(–2 , 9)

(0, 5)


19. y = x2 + 4x – 21 21. y = x2 – 6x – 27

(3, 0)(–7, 0)

(–2 , –25)(0, –21)

(9, 0)(–3, 0)

(0 , –27) (0, –21)

(3 , –36)

Exercise B.

1. y = x2 – 2x + 8 3. y = x2 + 2x + 3 5. y = 2x2 + 3x + 4

(1, 7)

(0, 8)

(-1, 7)

(0, 8)

(-3/4, 23/8)

(0, 4)

Exercise C.

1. The maximum

revenue is

R(18)=$882. The

price should be

$21 per pie.

3. The maximum

revenue is is

R(44)=5776. The

price should be

$19 per chicken.

5. The minimum distance

happens at (-2/3, 1/3)

(2, 8) (-2, 8)(-1.5, 4)

(-2/3, 1/3)

2.3 first and second degree functions

Technology

Transcript of 2.3 first and second degree functions