Create the equation of a quadratic polynomial, in standard form, that has zeroes of -3, 4, 7 and which passe, through the point (3, -12).

Answer 1

Zeroes of -3, 4, and 7 ⇒ x = -3, 4, and 7

Factors: x + 3, x - 4, and x - 7

Multiply the factors using the box method.

(x + 3)(x - 4)(x - 7) = (x + 3)(x² - 7x - 4x + 28)

= (x + 3)(x² - 11x + 28)

= x³ + 3x² - 11x² - 33x + 28x + 84

= x³ - 8x² - 5x + 84

Our quadratic equation is y = a(x + 3)(x - 4)(x - 7). We need to find 'a' by plugging in x = 3 and y = -12 to the quadratic equation.

a(3 + 3)(3 - 4)(3 - 7) = -12 ⇒ a(6)(-1)(-4) = -12 ⇒ 24a = -12 ⇒ a = -12/24 = -1/2

y = -1/2(x³ - 8x² - 5x + 84) ⇒ y = (-1/2)x³ + 4x² + (5/2)x - 42