Linear and Quadratic Functions and Modeling

67

Chapter 2 Polynomial, Power, and Rational Functions

■ Section 2.1 Linear and Quadratic Functions and Modeling

Exploration 1

1. –$2000 per year 2. The equation will have the form v(t)=mt+b. The value of the building after 0 year is v(0)=m(0)+b=b=50,000. The slope m is the rate of change, which is –2000 (dollars per year). So an equation for the value of the building (in dollars) as a function of the time (in years) is v(t)=–2000t+50,000. 3. v(0)=50,000 and v(16)=–2000(16)+50,000=18,000 dollars 4. The equation v(t)=39,000 becomes –2000t+50,000=39,000 –2000t=–11,000 t=5.5 years

6. (x-4)2=(x-4)(x-4)=x2-4x-4x+16 =x2-8x+16 7. 3(x-6)2=3(x-6)(x-6)=(3x-18)(x-6) =3x2-18x-18x+108=3x2-36x+108 8. –3(x+7)2=–3(x+7)(x+7) =(–3x-21)(x+7)=–3x2-21x-21x-147 =–3x2-42x-147 9. 2x2-4x+2=2(x2-2x+1)=2(x-1)(x-1) =2(x-1)2 10. 3x2+12x+12=3(x2+4x+4)=3(x+2)(x+2) =3(x+2)2

Section 2.1 Exercises

1. Not a polynomial function because of the exponent –5 2. Polynomial of degree 1 with leading coefficient 2 3. Polynomial of degree 5 with leading coefficient 2 4. Polynomial of degree 0 with leading coefficient 13 5. Not a polynomial function because of cube root 6. Polynomial of degree 2 with leading coefficient –5 5 5 5 18 7. m= so y-4= (x-2) ⇒ f(x)= x+ 7 7 7 7 y 5 (2, 4) x

Quick Review 2.1

1. y=8x+3.6 2. y=–1.8x-2 3 3. y-4=– (x+2), or y=–0.6x+2.8 5 y 7

3 ( 2, 4) (3, 1) 5 x (–5, –1)

7 7 7 8 8. m= - so y-5= - (x+3) ⇒ f(x)=– x+ 9 9 9 3 8 8 7 4. y-5= (x-1), or y= x+ 3 3 3 y 6 (1, 5) 10 (6, –2) 5 ( 2, 3) x (–3, 5) x y 10

5. (x+3)2=(x+3)(x+3)=x2+3x+3x+9 =x2+6x+9

68

Chapter 2

Polynomial, Power, and Rational Functions

16. (f)—the vertex is in Quadrant I, at (1, 12), meaning it must be either (b) or (f). Since f(0)=10, it cannot be (b): if the vertex in (b) is (1, 12), then the intersection with the y-axis occurs considerably lower than (0, 10). It must be (f). 17. (e)—the vertex is at (1, –3) in Quadrant IV, so it must be (e). 10 x

2 4 4 4 9. m= - so y-6= - (x+4) ⇒ f(x)= - x+ 3 3 3 3

y 10 (–4, 6) (–1, 2)

18. (c)—the vertex is at (–1, 12) in Quadrant II and the parabola opens down, so it must be (c). 19. Translate the graph of f(x)=x2 3 units right to obtain the graph of h(x)=(x-3)2, and translate this graph 2 units down to obtain the graph of g(x)=(x-3)2-2. y

5 5 5 3 10. m= so y-2= (x-1) ⇒ f(x)= x+ 4 4 4 4

y 10 (5, 7) (1, 2) 10 x

10

10

x

11. m=–1 so y-3=–1(x-0) ⇒ f(x)=–x+3

y 5 (0, 3) (3, 0) x 5

20. Vertically shrink the graph of f(x)=x2 by a factor of 1 1 to obtain the graph of g(x)= x2, and translate this 4 4 1 graph 1 unit down to obtain the graph of h(x)= x2-1. 4 y 10

10

x

1 1 1 12. m= so y-2= (x-0) ⇒ f(x)= x+2 2 2 2

y 10

(0, 2) (–4, 0) 10 x

21. Translate the graph of f(x)=x2 2 units left to obtain the graph of h(x)=(x+2)2, vertically shrink this graph by 1 1 a factor of to obtain the graph of k(x)= (x+2)2, 2 2 and translate this graph 3 units down to obtain the graph 1 of g(x)= (x+2)2-3. 2 y 10

13. (a)—the vertex is at (–1, –3), in Quadrant III, eliminating all but (a) and (d). Since f(0)=–1, it must be (a). 14. (d)—the vertex is at (–2, –7), in Quadrant III, eliminating all but (a) and (d). Since f(0)=5, it must be (d). 15. (b)—the vertex is in Quadrant I, at (1, 4), meaning it must be either (b) or (f). Since f(0)=1, it cannot be (f): if the vertex in (f) is (1, 4), then the intersection with the y-axis would be about (0, 3). It must be (b).

10

x

Section 2.1

22. Vertically stretch the graph of f(x)=x2 by a factor of 3 to obtain the graph of g(x)=3x2, reflect this graph across the x-axis to obtain the graph of k(x)=–3x2, and translate this graph up 2 units to obtain the graph of h(x)=–3x2+2.1 y 10

Linear and Quadratic Functions and Modeling

69

32. h(x)=–2 a x2 +

10

x

7 x b -4 2 7 49 49 =–2 a x2 + 2 # x + b -4+ 4 16 8 7 2 17 =–2 a x + b + 4 8 7 17 7 Vertex: a - , b ; axis: x = 4 8 4 33....