FROM THE GIVEN INFORMATION WRITE THE CUBIC FUNCTION

To find cubic equation from the given zeroes, x-intercepts, solutions or values of x, we use the formula given below.

y = k(x - a) (x - b) (x - c)

Here a, b and c are x-intercepts.

Difference between touches and crosses :

Here the curve crosses the x-axis at three different points.

The curve crosses x-axis at -2 and touches the x-axis at 2. So, we have to take the factor (x - 2) twice. So, the equation would be y = (x + 2)(x - 2)²

Find the equation of the cubic whose graph :

Problem 1 :

Cuts the x - axis at 3, 1, -2 and passes through (2, -4)

Solution :

x = 3, x = 1, x = -2

Factors are :

(x - 3) (x - 1) (x + 2)

P(x) = a (x - 3) (x - 1) (x + 2) --- (1)

Then solve for a. substitute passes through the points (2, -4).

-4 = a (2 - 3) (2 - 1) (2 + 2)

-4 = a (-1) (1) (4)

-4 = -4a

1 = a

a = 1 substitute the equation (1).

P(x) = (x - 3) (x - 1) (x + 2) 

Problem 2 :

Cuts the x - axis at -2, 0 and 1/2 and passes through (-3, -21)

Solution :

x = -2, x = 0, x = 1/2

Factors are :

(x + 2) (x - 0) (x - 1/2)

P(x) = a (x + 2) (x - 0) (x - 1/2)   --- (1)

Then solve for a. substitute passes through the points (-3, -21).

-21 = a (-3 + 2) (-3 - 0) (-3-(1/2))

-21 = a (-1) (-3) (-7/2)

-21 = -21a/2

(-21) (-2/21) = a

2 = a

a = 2 substitute the equation (1).

P(x) = 2(x + 2) (x) (x - 1/2)

Problem 3 :

Touches the x - axis at 1, cuts the x- axis at -2 and passes through (4, 54)

Solution :

x = 1 and x = -2

(x - 1) and (x + 2)

The factor (x - 1) should be repeated twice.

P(x) = a(x - 1)² (x + 2) --- (1)

Then solve for a. substitute passes through the points (4, 54).

54 = a(4 - 1)² (4 + 2)

54 = a(3)² (6)

54 = 54a

1 = a

a = 1 substitute the equation (1).

P(x) = (x - 1)² (x + 2)

Problem 4 :

Touches the x - axis at -2/3, cuts the x- axis at 4 and passes through (-1, -5)

Solution :

x = -2/3 and x = 4

(x + 2/3) and (x - 4)

P(x) = a(x + 2/3)² (x - 4)  --- (1)

Then solve for a. substitute passes through the points (-1, -5).

-5 = a(-1/3)² (-5)

-5 = (a/9) (-5)

-5 = -5a/9

(-5) (-9/5) = a

9 = a 

a = 9 substitute the equation (1).

P(x) = 9[(3x + 2)/3]² (x - 4)

= (9) (1/9) [3x + 2]² (x - 4)

= (3x + 2)² (x - 4)

Find the equation of a real cubic polynomial which cuts :

Problem 5 :

The x - axis at 1/2 and -3, cuts the y - axis at 30 and passes through (1, -20)

Solution :

From the given x-intercepts, we get (x - 1/2) (x + 3). Let another zero of the cubic polynomial be c.

(x - 1/2)  (x + 3) and (x - c)

P(x) = k(x - 1/2) (x + 3) (x - c)  -----(1)

y-intercept, when x = 0

P(0) = k(0 - 1/2) (0 + 3) (0 - c) 

30 = k(-1/2) (-3c)

20 = ck -------(1)

The curve passes through the point (1, -20)

-20 = k(1 - 1/2) (1 + 3) (1 - c)

-20 = k(1/2) (4) (1 - c)

-10 = k(1 - c)

From (1), k = 20/c

-10 = (20/c)(1-c)

-10c = 20(1 - c) 

-c = 2(1 - c)

c = 2, k = 10

= 10 (1/2) (2x - 1) (x + 3) (x - 2)

P (x) = 5(2x - 1) (x + 3) (x - 2)

Problem 6 :

The x  - axis at 1, touches the x - axis at -2 and cuts the y - axis at (0, 8)

Solution :

x = 1 and x = -2

(x - 1) (x + 2)

P(x) = k (x - 1) (x + 2)²

Then solve for a. substitute passes through the points (0, 8).

8 = k (0 - 1) (0 + 2)²

8 = k(-1) (2)²

-8/4 = k

a = -2

P(x) = -2 (x - 1) (x + 2)²

Problem 7 :

The x - axis at 2, the y- axis at -4 and passes through (1, -1) and (-1, -21)

Solution :

Cubic equation will be the form k(x - a) (x - b) (x - c). Here a, b and c are x-intercepts. 

With the given x-intercept, we create a linear factor (x - 2)

y axis = (0, -4)

Since the graph passes through the points (1, -1) and (-1, -21).

P(x) = (x - 2) (ax² + bx + c) --- (1)

(0, -4) substitute the equation (1).

-4 = (0 - 2) (a(0)² + b(0) + c)

-4 = -2c 

2 = c 

c = 2 and (1, -1) substitute the equation (1).

-1 = (1 - 2) (a(1)² + b(1) + 2)

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

a + b = 1 - 2

a + b = -1

The curve passes through the point (-1, -21).

-21 = (-1 - 2) (a(-1)² + b(-1) + 2)

7 = a - b + 2

a - b = 5  ---(2)

(1) + (2)

2a = 4, then a = 2

By applying the value of a in (1), we get

2 + b = -1

b = -3

Applying the values, we get

P(x) = (x - 2) (2x² - 3x + 2)