Precal Help?

Question

Express the polynomial f(x) in the form 

a_nxⁿ + a_n − 1x^n − 1 + ··· + a₁x + a₀.

 

Find a quadratic function that has zeros −6 and 8 and a graph that passes through the point (3, 5).
f(x) =

Answer 1

The quadratic function form is f(x) = a₂ x² + a₁ x + a₀.

Zeros of the quadratic function are - 6 and 8 ; and function passes through the point (3, 5).

So, f(- 6) = 0, f(8) = 0 and f(3) = 5.

a₂ (- 6)² + a₁ (- 6) + a₀ = 0 ⇒ 36a₂ - 6a₁ + a₀ = 0     ------->  Equation 1.

a₂ (8)² + a₁ (8) + a₀ = 0 ⇒ 64a₂ + 8a₁ + a₀ = 0         ------->  Equation 2.

a₂ (3)² + a₁ (3) + a₀ = 5 ⇒ 9a₂ + 3a₁ + a₀ = 5           ------->  Equation 3.

To eliminate the a₀ - variable subtract equation 1 and equation 2.

         36a₂ - 6a₁ + a₀ = 0

( - )  64a₂ + 8a₁ + a₀ = 0

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      -28a₂ - 14a₁ = 0 ⇒ 2a₂ + a₁ = 0    ------------>    Equation 4.

To eliminate the a₀ - variable subtract equation 2 and equation 3.

        64a₂ + 8a₁ + a₀ = 0

( - )    9a₂ + 3a₁ + a₀ = 5

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        55a₂ + 5a₁ = - 5 ⇒ 11a₂ + a₁ = - 1   ------------>    Equation 5.

To eliminate the a₁ - variable subtract equation 4 and equation 5.

           2a₂ + a₁ = 0

( - )  11a₂ + a₁ = - 1

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    - 9a₂ = 1 ⇒ a₂ = - 1/9.

To find the value of a₁, substitute a₂ = - 1/9 in the equation 4.

2(- 1/9) + a₁ = 0 ⇒ a₁ = 2/9.

To find the value of a₀, substitute a₂ = - 1/9 and a₁ = 2/9 in the equation 1.

36(- 1/9) - 6(2/9) + a₀ = 0

(- 36 -12)/9 + a₀ = 0

- 48/9 + a₀ = 0

a₀ = 48/9.

The quadratic function f(x) = (- 1/9) x² + (2/9) x + 48/9.