Given values are

The difference table is

x₀ + nh = x ⇒ 1 + n = x ⇒ n = x - 1
∴ By Newton's forward interpolation formula,
y= y₀\(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)

[∵ Δ³y₀ = 0]
⇒ y = -1 + 0 + x² - 3x + 2
⇒ y = x² - 3x + 1
Hence, the required second degree polynomial is  y = x² - 3x + 1

Scale:
In x-axis 1 cm = 100 units
In y-axis 1 cm = 1000 units
Plot the points (200, 2500), (300, 2800) (400, 3100) (640; 3820), (540,3220) and (580, 3640).

At x = 350, draw a vertical line and from the intersecting point on the curve, draw a horizontal line.
From the graph, we find that when x = 350, y = 2900.
Hence, the expense at a level of 350 units is Rs. 2900

Scale:
In x axis 1 cm = 10 units
In y axis 1 cm = 2 units
Plot the points (40, 6.2), (50, 7.2), (60, 9.1) and (70, 12). At x = 48, draw a vertical line to the graph and from the intersecting point, draw a horizontal line to meet the y-axis
From the graph, we find that when x = 48, the value of Y is equal to 6.8.

Given f(x) = x²+ 3h
LHS = Δ f(x)
= f(x + h) - f(x)
= [(x + h)² + 3 (x + h)] - [x² + 3x]

= h² + 2xh + 3h
when h = 1,
LHS = 1² + 2x(1) + 3(1)
= 1+2x+3
= 2x+4= RHS
Hence proved.

Given h = 1
LHS = (E⁻¹Δ) x³
= Δ(E^-1(x³))
= Δ(x - h)³  [∵ E^-1f(x) = f(x - nh)]
= Δ(1 - h)³ [∵ h = 1]
= (x - 1+ 1)³ - (x - 1)³ [∵ Δf(x) =f(x + h) - f(x)]
= x³ - (x - 1)³
= x³ - (x³ - 3x² + 3x - 1)
[∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
= x³ - x³ + 3x² - 3x + 1
= 3x² - 3x + 1
= RHS
Hence proved