For \( f(x) = 3x^2 – 6x + p \), the discriminant of the quadratic equation \( ax^2 + bx + c = 0 \) is \( \Delta = b^2 – 4ac \). Here, \( a = 3 \), \( b = -6 \), \( c = p \). Since the equation \( f(x) = 0 \) has two equal roots, the discriminant is zero:

\[ \Delta = (-6)^2 – 4 \cdot 3 \cdot p = 36 – 12p \]

\[ \Delta = 0 \implies 36 – 12p = 0 \]

Answer: The discriminant is \( 0 \).

From part (a)(i):

\[ 36 – 12p = 0 \]

\[ 12p = 36 \implies p = \frac{36}{12} = 3 \]

Answer: \( p = 3 \)

Find the vertex of \( f(x) = 3x^2 – 6x + 3 \), which lies on the \( x \)-axis.

Method 1: Axis of Symmetry

The vertex of a quadratic \( ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). Here, \( a = 3 \), \( b = -6 \):

\[ x = -\frac{-6}{2 \cdot 3} = \frac{6}{6} = 1 \]

Evaluate \( f(x) \) at \( x = 1 \):

\[ f(1) = 3(1)^2 – 6 \cdot 1 + 3 = 3 – 6 + 3 = 0 \]

Vertex: \( (1, 0) \).

Method 2: Factorization/Completing the Square

Rewrite \( f(x) \):

\[ f(x) = 3x^2 – 6x + 3 = 3(x^2 – 2x + 1) = 3(x – 1)^2 \]

The vertex form \( f(x) = 3(x – 1)^2 \) shows the vertex at \( x = 1 \), where:

\[ f(1) = 3(1 – 1)^2 = 0 \]

Vertex: \( (1, 0) \).

Method 3: Derivative

Find the vertex by setting the derivative to zero:

\[ f'(x) = 6x – 6 \]

\[ 6x – 6 = 0 \implies x = 1 \]

Evaluate \( f(1) \):

\[ f(1) = 3(1)^2 – 6 \cdot 1 + 3 = 0 \]

Vertex: \( (1, 0) \).

Answer: \( (1, 0) \)

Since \( f(x) = 0 \) has two equal roots and the vertex is at \( (1, 0) \), the root is:

\[ f(x) = 3(x – 1)^2 = 0 \implies x = 1 \]

Answer: \( x = 1 \)

In the form \( f(x) = a(x – h)^2 + k \), compare with \( f(x) = 3x^2 – 6x + 3 = 3(x – 1)^2 \). The coefficient of \( x^2 \) is \( a \):

\[ a = 3 \]

Answer: \( a = 3 \)

In the vertex form \( f(x) = 3(x – 1)^2 \), the x-coordinate of the vertex is \( h \):

\[ h = 1 \]

Answer: \( h = 1 \)

In the vertex form \( f(x) = 3(x – 1)^2 \), the y-coordinate of the vertex is \( k \). Since \( f(1) = 0 \):

\[ k = 0 \]

Answer: \( k = 0 \)

The function \( g(x) \) is obtained by reflecting \( f(x) = 3x^2 – 6x + 3 = 3(x – 1)^2 \) in the \( x \)-axis, then translating by \( \begin{pmatrix} 0 \\ 6 \end{pmatrix} \).

Reflection in the x-axis: Replace \( f(x) \) with \( -f(x) \):

\[ -f(x) = -3(x – 1)^2 \]

Translation by \( \begin{pmatrix} 0 \\ 6 \end{pmatrix} \): Add 6:

\[ g(x) = -3(x – 1)^2 + 6 \]

Expand to \( Ax^2 + Bx + C \):

\[ -3(x – 1)^2 = -3(x^2 – 2x + 1) = -3x^2 + 6x – 3 \]

\[ g(x) = -3x^2 + 6x – 3 + 6 = -3x^2 + 6x + 3 \]

Thus, \( A = -3 \), \( B = 6 \), \( C = 3 \).

Answer: \( g(x) = -3x^2 + 6x + 3 \)