Question

1. a) find the equation of the polynomial that passes through the points: (1,-10),(2,-1),(3,32), (4,101),(5,218), (6,395) b) plot the points c) find iroc at x = 3.5

Explanation:

Step1: Assume cubic polynomial form

Let the polynomial be $y = ax^3 + bx^2 + cx + d$ (since 6 points suggest degree ≤5, but finite differences show cubic).

Step2: Set up equations from points

For $(1,-10)$: $a + b + c + d = -10$
For $(2,-1)$: $8a + 4b + 2c + d = -1$
For $(3,32)$: $27a + 9b + 3c + d = 32$
For $(4,101)$: $64a + 16b + 4c + d = 101$

Step3: Solve the system of equations

Subtract Eq1 from Eq2: $7a + 3b + c = 9$
Subtract Eq2 from Eq3: $19a + 5b + c = 33$
Subtract Eq3 from Eq4: $37a + 7b + c = 69$

Subtract first new eq from second: $12a + 2b = 24 \implies 6a + b = 12$
Subtract second new eq from third: $18a + 2b = 36 \implies 9a + b = 18$

Subtract these two: $3a = 6 \implies a=2$
Then $6(2)+b=12 \implies b=0$
Then $7(2)+3(0)+c=9 \implies c=9-14=-5$
Then $2+0-5+d=-10 \implies d=-7$

Polynomial: $y=2x^3 -5x -7$ (verify with (5,218): $2(125)-25-7=250-32=218$, correct; (6,395): $2(216)-30-7=432-37=395$, correct)

Step4: Find derivative for IROC

Instantaneous rate of change (IROC) is the derivative: $y' = 6x^2 -5$

Step5: Calculate IROC at x=3.5

Substitute $x=3.5$: $y'(3.5)=6(3.5)^2 -5 = 6(12.25)-5 = 73.5 -5 = 68.5$

---

Answer:

a) The polynomial equation is $\boldsymbol{y=2x^3 -5x -7}$
b) [Plot instructions: Plot the points (1,-10), (2,-1), (3,32), (4,101), (5,218), (6,395) on a coordinate plane, with x-axis from 1 to 6 and y-axis scaled to fit the y-values, then sketch the cubic curve $y=2x^3 -5x -7$ through them]
c) The IROC at $x=3.5$ is $\boldsymbol{68.5}$