The quadratic function \( f(x) = ax^2 + bx + c \) has a vertex at \( (-2, 10) \). The axis of symmetry is \( x = -2 \). Since \( f(x) = k \) has two solutions and one is \( x = 2 \), the other solution is symmetric about the axis of symmetry. The distance from \( x = 2 \) to \( x = -2 \) is \( 2 – (-2) = 4 \). Thus, the other solution is:

\[ x = -2 – (2 – (-2)) = -2 – 4 = -6 \]

Answer: \( x = -6 \)

Determine the signs of \( a \), \( b \), and \( c \).

For \( a \): The graph is a parabola with a vertex at \( (-2, 10) \). Since the vertex is a maximum (as implied by the downward-opening parabola, given \( a \) is negative), \( a < 0 \).

For \( b \): The vertex of \( f(x) = ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). Given the vertex at \( x = -2 \):

\[ -\frac{b}{2a} = -2 \implies \frac{b}{2a} = 2 \implies b = 4a \]

Since \( a < 0 \), \( b = 4a < 0 \).

For \( c \): The table indicates \( c \) is positive.

Answer:

A quadratic function \( f(x) = ax^2 + bx + c \) with \( a < 0 \) (downward-opening parabola) is decreasing where the derivative \( f'(x) < 0 \). Compute the derivative:

\[ f'(x) = 2ax + b \]

Since \( b = 4a \) and \( a < 0 \), solve:

\[ 2ax + b < 0 \implies 2ax + 4a < 0 \implies 2a(x + 2) < 0 \]

Since \( a < 0 \), divide by \( 2a \) (reversing the inequality):

\[ x + 2 > 0 \implies x > -2 \]

Alternatively, the function decreases to the right of the vertex \( x = -2 \) for a downward-opening parabola. Thus:

Answer: \( x > -2 \)