Find the first, fourth, and eighth terms of the sequence. A(n) = –3 • 2n–1

–6; –48; –768

–3; –24; –384

1; –216; –279,936

–12; –96; –1,536
2.
Find the first, fourth, and eighth terms of the sequence. A(n) = –2 • 5n–1

–50; –6,250; –3,906,250

–10; –1,250; –781,250

1; –1,000; –10,000,000

–2; –250; –156,250
3.
You drop a ball from a height of 1.5 meters. Each curved path has 71% of the height of the previous path. a. Write a rule for the sequence using centimeters. The initial height is given by the term n = 1. b. What height will the ball be at the top of the sixth path?

A(n) = 1.5 • (0.71)n–1 ; 0.27 cm

A(n) = 150 • (71)n–1 ; 270,634,402,650 cm

A(n) = 150 • (0.71)n–1 ; 27.06 cm

A(n) = 0.71 • (1.5)n–1 ; 5.39 cm
4.
Find the first, fourth, and eighth terms of the sequence. A(n) = –5 • 3n–1

–15; –405; –32,805

1; –3,375; –170,859,375

–45; –1,215; –98,415

–5; –135; –10,935
5.
Find the first, fourth, and eighth terms of the sequence. A(n) = –5 • 2n–1

–5; –40; –640

–20; –160; –2,560

1; –1,000; –10,000,000

–10; –80; –1,280
6.
You drop a ball from a height of 1.5 meters. Each curved path has 75% of the height of the previous path. a. Write a rule for the sequence using centimeters. The initial height is given by the term n = 1. b. What height will the ball be at the top of the sixth path?

A(n) = 150 • (75)n–1 ; 355,957,031,250 cm

A(n) = 150 • (0.75)n–1 ; 35.6 cm

A(n) = 0.75 • (1.5)n–1 ; 5.7 cm

A(n) = 1.5 • (0.75)n–1 ; 0.36 cm
7.
Find the common ratio of the sequence. 3, 9, 27, 81, . . .

3

–6

6

8.
You drop a ball from a height of 0.5 meter. Each curved path has 52% of the height of the previous path. a. Write a rule for the sequence using centimeters. The initial height is given by the term n = 1. b. What height will the ball be at the top of the third path?

A(n) = 0.52 • (0.5)n–1 ; 0.13 cm

A(n) = 50 • (0.52)n–1 ; 13.52 cm

A(n) = 0.5 • (0.52)n–1 ; 0.14 cm

A(n) = 50 • (52)n–1 ; 135,200 cm
9.
Find the common ratio of the sequence. –4, 8, –16, 32, . . .

–12

–2

12
10.
Find the first, fourth, and eighth terms of the sequence. A(n) = –2 • 2n–1

–4; –32; –512

–2; –16; –256

1; –64; –16,384

–8; –64; –1,024

1. f(x)=x²+10x+16

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

so, vertex = (-5, -9)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

2. f(x)=−(x−3)(x+1)

by multiplying the factors, the general form is f(x)= -x²+2x+3.

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

so, vertex = (1, 4)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

3. f(x)= −x²+4

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

so, vertex = (0, 4)

now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

4. f(x)=2x²+16x+30

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

so, vertex = (-4, -2)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

5. y=(x+2)²+4

the general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a> 0 parabola is opened upward.

if a< 0 parabola is opened downward.

compare the given equation with general form of parabola.

-h=2 ⇒h=-2

k=4

so, vertex= (-2, 4)

as, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.

hope this : )

the answer to this question is 3/19

step-by-step explanation: