how do I draw cubic function as a dashed curve using the points (-4,3),(0,0),(2-6), and (5,5)

Question

draw cubic function as a dashed curve using the points (-4,3),(0,0),(2-6), and (5,5).

Answer 1

Newtons form :

The coordinates are (-4, 3), (0, 0), (2, -6), (5, 5)

Write the n = 4 points (-4, 3), (0, 0), (2, -6), (5, 5) we can write

P(x) = C0  + C1 (x + 4) + C2 (x + 4)(x) +  C3 (x + 4)(x)(x - 2) ---> (1)

This polynomial must go through the four points, so

C0 = 3

C0 + C1(0 + 4) = 0

C0 + C1(2 + 4) + C2(2 + 4)(2) = -6

C0 + C1(5 + 4) + C2(5 + 4)(5) + C3(5 + 4)(5)(5 - 2) = 5

C0 = 3

 4C1 = -3

C1 = -3/4

C0 + 6 C1 +12 C2 = -6

3 + 6(-3/4) + 12(C2) = -6

12(C2) = -6 -3 + 9/2

12C2 = -9 + 4.5

12C2 = -4.5

C2 = -4.5/12

C2 = - 0.375

C0 + C1(5 + 4)+C2(5 + 4)(5) + C3(5 + 4)(5)(5 - 2) = 5

3 + 9C1 + 45C2 + 165C3 = 5

3 +9(-3/4) + 45(-0.375) + 165C3 = 5

3 - 6.75 - 16.875 + 165C3 = 5

-20.625 + 165C3 = 5

165C3 = 5 + 20.625

C3 = 25.625/165

C3 = 0.15

MW = b

W = [C0 C1 C2 C3]

Substitute the C0, C1, C2, C3 values in equation (1)

P(x) = C0  + C1 (x + 4) + C2 (x + 4)(x) +  C3 (x + 4)(x )(x - 2)

P(x) = 3 + (-3/4) (x + 4) + (-0.375) (x^2 + 4x) +(0.15) (x^3 + 2 x^2 - 8x)

P(x) = 3 - 0.75 x - 3 -0.375 x^2 - 1.5x + 0.15x^3 + 0.3x^2 - 1.2x

After simplifying required polynomial is P(x) = 0.15x ^3 - 0.075x ^2 - 3.45x

Graph

 

Answer 2

The points are (- 4, 3), (0, 0), (2, -6) and (5, 5).

The general form of third degree polynomial function is y = ax³ + bx² + cx + d.

The above function passes through the points (- 4, 3), (0, 0), (2, -6) and (5, 5).

If (x, y) = (0, 0) then (3) = a(0)³ + b(0)² + c(0) + d -------> 0 = d

If (x, y) = (- 4, 3) then (3) = a(- 4)³ + b(- 4)² + c(- 4) + d -------> 3 = - 64a + 16b - 4c + d -------> 3 = - 64a + 16b - 4c ----->(1).

If (x, y) = (2, - 6) then (- 6) = a(2)³ + b(2)² + c(2) + d -------> - 6 = 8a + 4b + 2c + d -------> - 3 = 4a + 2b + c.----->(2)

If (x, y) = (5, 5) then (5) = a(5)³ + b(5)² + c(5) + d -------> 5 = 125a + 25b + 5c + d -------> 1 = 25a + 5b + c.----->(3)

Solve (1) and (2) to eliminate c - variable and obtain two variable equation.

eq(1) + 4·eq(2) ----> 3 + 4(-3) = (- 64a + 16b - 4c) + 4(4a + 2b + c)

                            ----> 3 - 12 = - 64a + 16b - 4c + 16a + 8b + 4c

                            ----> - 9 = - 48a + 24b -------> (4)

Solve (2) and (3) to eliminate c - variable and obtain two variable equation.

eq(2) - eq(3) ----> - 3 - 1 = (4a + 2b + c) - (25a + 5b + c)

                            ----> - 4 = 4a + 2b + c - 25a - 5b - c

                            ----> - 4 = - 21a - 3b. -------> (5)

Solve (4) and (5) to eliminate b - variable and obtain one variable equation.

eq(4) + 8·eq(5) ----> (- 9) + 8(- 4) = (- 48a + 24b) + 8(- 21a - 3b)

                            ----> - 9 - 32 = - 48a + 24b - 168a - 24b

                            ----> - 41 = - 216a  -------> a = 41/216.

Substitute a = 41/216 in equation 5 : - 4 = - 21a - 3b.

- 4 = - 21(41/216) - 3b -------> - 3b = - 4 + 861/216 = (- 864 + 861)/216 = - 3/216 -------> b = 1/216.

Substitute a = 41/216 and b = 1/216 in equation 2 : - 3 = 4a + 2b + c.

- 3 = 4(41/216) + 2(1/216) + c ------> - 3 - 164/216 - 2/216 = c -----> (- 648 - 164 - 2)/216 = c ------> c = - 814/216 = - 407/108.

Substitute a = 41/216, b = 1/216, c = - 407/108 and d = 0 in the original function : y = ax³ + bx² + cx + d.

y = (41/216)x³ + (1/216)x² + (- 407/108)x + (0)

y = 0.1898x³ + 0.0046x² - 3.7658x.

Plot the points (- 4, 3), (0, 0), (2, -6) and (5, 5) on the coordinate points.

Some additional points are obtained when substitute the some values of x in the original function.

If x = 2.5 then y = 0.1898(2.5)³ + 0.0046(2.5)² - 3.7658(2.5) = 6.45 (approximately)

If x = - 2.5 then y = 0.1898(2.5)³ + 0.0046(2.5)² - 3.7658(2.5) = - 6.45 (approximately)

And plot  these points also.

Connected the above points with a smooth curve as shown figure.