Question

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Explanation:

Step1: Find vertex of Function 1

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex is \( x = -\frac{b}{2a} \). For \( f(x)=-2x^{2}-12x - 9 \), \( a=-2 \), \( b = - 12 \). So \( x=-\frac{-12}{2\times(-2)}=-\frac{12}{4}=-3 \). Substitute \( x = - 3 \) into the function: \( f(-3)=-2\times(-3)^{2}-12\times(-3)-9=-2\times9 + 36-9=-18 + 36-9 = 9 \). So the vertex of Function 1 is \( (-3,9) \).

Step2: Find vertex of Function 2

From the graph of Function 2, the vertex is the highest point (since it's a downward - opening parabola). By looking at the graph, the vertex has coordinates \( (4,7) \) (we can see that the peak of the parabola is at \( x = 4 \) and \( y = 7 \) by analyzing the symmetry and the points plotted on the graph).

Step3: Compare maximum values

The maximum value of a quadratic function \( f(x)=ax^{2}+bx + c \) (when \( a<0 \)) is the y - coordinate of the vertex. For Function 1, the maximum value is 9. For Function 2, the maximum value is 7. So Function 1 has a larger maximum value.

Answer:

(a) The vertex of Function 1 is \((-3, 9)\)
(b) The vertex of Function 2 is \((4, 7)\)
(c) The function with the larger maximum value is Function 1, and the larger maximum value is 9.