Answer:
C. g(x) = 4x²
Step-by-step explanation:
The general equation for a parabola is
y = ax² + bx +c
Since ƒ(x) = x², a=1, b =0, c =0
For g(x), the vertex is still at the origin, so
g(x) = ax²
The graph passes through (1,4).
Insert the coordinates of the point.
4 = a(1)²
a = 4
g(x) = 4x²
The figure below shows that the graph of g(x) = 4x² passes through the point (1, 4).
Answer:
1) Domain: { 1,2,3,4,5,6}
Range: {12,24,36,48,60,72}
2) First four terms are: 2,5,8,11
3) First four terms are: 8,13,20,29
4) First four terms are: 0,0,2,6
5) First four terms are: 0,1,1.41,1.73
Step-by-step explanation:
1.
n 1 2 3 4 5 6
f(n) 12 24 36 48 60 72
Domain: { 1,2,3,4,5,6}
Range: {12,24,36,48,60,72}
Write first four terms of each sequence
2) f(n)=3n-1
Put n = 1, 3(1)-1 = 3-1 = 2
Put n = 2, 3(2)-1 = 6-1 = 5
Put n = 3, 3(3)-1 = 9-1 = 8
Put n = 4, 3(4)-1 = 12-1 = 11
So, First four terms are: 2,5,8,11
3) f(n)=n^2+2n+5
Put n = 1, (1)^2+2(1)+5= 1+2+5 = 8
Put n = 2, (2)^2+2(2)+5 = 4+4+5= 13
Put n = 3, (3)^2+2(3)+5 = 9+6+5 = 20
Put n = 4, (4)^2+2(4)+5 = 16+8+5 = 29
So, First four terms are: 8,13,20,29
4) f(n)=(n-1)(n-2)
Put n = 1, (1-1)(1-2) = 0(-1) = 0
Put n = 2, (2-1)(2-2) = 1(0) = 0
Put n = 3, (3-1)(3-2) = 2(1) = 2
Put n = 4, (4-1)(4-2) = 3(2) = 6
So, First four terms are: 0,0,2,6
5) f(n) =
Put n=1
Put n=2
Put n=3
Put n=4
So, First four terms are: 0,1,1.41,1.73