16.1k views
5 votes
Which polynomial function has x intercepts -1,0, and 2 and passes through the point (1,-6)?

User Almad
by
8.5k points

1 Answer

5 votes

Answer:

f(x) =
3x^3 - 3x^2 - 6x

Explanation:

Which polynomial function has x intercepts -1,0, and 2 and passes through the point (1,-6)?

There are 3 distinct and real roots given in the question, which means that the function must be a third degree polynomial. The roots are -1, 0, and 2. This means that f(x) = 0 at these points. The general form of the cubic equation is given by:

f(x) = ax^3 + bx^2 + cx + d; where a, b, c, and d are arbitrary constants.

From the given data:

f(-1)=0 implies a*(-1)^3 + b*(-1)^2 + c(-1) + d = -a + b - c + d = 0. (Equation 1).

f(0)=0 implies a*(0)^3 + b*(0)^2 + c(0) + d = 0a + 0b + 0c + d = 0. (Equation 2).

f(2)=0 implies a*(2)^3 + b*(2)^2 + c(2) + d = 8a + 4b + 2c + d = 0. (Equation 3).

f(1)=0 implies a*(1)^3 + b*(1)^2 + c(1) + d = a + b + c + d = -6. (Equation 4).

Equation 2 shows that d = 0. So rest of the equations become:

-a + b - c = 0;

8a + 4b + 2c = 0; (Divide 2 on both sides of the equation to simplify).

a + b + c = -6

This system of equation can be solved using the Gaussian Elimination Method. Converting the system into the augmented matrix form:

• 1 1 1 | -6

• -1 1 -1 | 0

• 4 2 1 | 0

Adding row 1 into row 3:

• 1 1 1 | -6

• 0 2 0 | -6

• 4 2 1 | 0

Dividing row 2 with 2 and multiplying row 1 with -4 and add it into row 3:

• 1 1 1 | -6

• 0 1 0 | -3

• 0 -2 -3 | 24

Multiplying row 2 with 2 and add it into row 3:

• 1 1 1 | -6

• 0 1 0 | -3

• 0 0 -3 | 18

It can be seen that when this updated augmented matrix is converted into a system, it comes out to be:

• a + b + c = -6

• b = -3

• -3c = 18 (This implies that c = -6.)

Put c = -6 and b = -3 in equation 1:

• a + (-3) + (-6) = -6

• a = -6 + 3 + 6

• a = 3.

So f(x) =
3x^3 - 3x^2 - 6x (All conditions are being satisfied)!!!

User Taskmaster
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.