Questions: Si une fonction quadratique s'écrit de la même façon sous sa forme canonique que sous sa forme factorisée, que pouvons-nous affirmer à propos de ses zéros. Cette fonction possède deux zéros distincts. Cette fonction possède un zéro unique. Cette fonction possède deux zéros distincts qui sont l'opposé l'un de l'autre. Cette fonction ne possède pas de zéro.

Solution Steps

To determine the nature of the zeros of a quadratic function that is expressed in both its canonical (vertex) form and its factored form, we need to analyze the conditions under which these forms are equivalent. A quadratic function can be written in the canonical form as \( f(x) = a(x-h)^2 + k \) and in the factored form as \( f(x) = a(x-x_1)(x-x_2) \). If these forms are equivalent, it implies that the vertex form has a perfect square, meaning the discriminant of the quadratic is zero, indicating a unique zero.

1. Recognize that if the canonical form and factored form are the same, the quadratic must have a perfect square trinomial.
2. This implies the discriminant (\(b^2 - 4ac\)) is zero, leading to a single, repeated root.

The quadratic function is given in both canonical form \( f(x) = a(x-h)^2 + k \) and factored form \( f(x) = a(x-x_1)(x-x_2) \). For these forms to be equivalent, the quadratic must be a perfect square trinomial.

Expanding the canonical form, we have: \[ f(x) = a(x-h)^2 + k = a(x^2 - 2hx + h^2) + k = a x^2 - 2ahx + ah^2 + k \]

From the expanded form, the coefficients are:

• \( a_{\text{coeff}} = a \)
• \( b_{\text{coeff}} = -2ah \)
• \( c_{\text{coeff}} = ah^2 + k \)

The discriminant \(\Delta\) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \Delta = b^2 - 4ac \] Substituting the coefficients, we get: \[ \Delta = (-2ah)^2 - 4a(ah^2 + k) = 4a^2h^2 - 4a(ah^2 + k) \]

Simplifying the expression for the discriminant: \[ \Delta = 4a^2h^2 - 4a^2h^2 - 4ak = -4ak \]

For the quadratic to have a unique zero, the discriminant must be zero: \[ -4ak = 0 \] This implies: \[ ak = 0 \]

Final Answer