Questions: Quadratic Functions Use the vertex and intercepts to sketch the given equation of the parabola's axis of symmetry, and identify the function's domain and range. f(x) = x^2 - 6x - 16 Use the graphing tool to graph the equation. Include the intercepts when drawing the graph. The axis of symmetry is (Type an equation.)

The given quadratic function is \( f(x) = x^2 - 6x - 16 \). The vertex form of a quadratic function is \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex. To find the vertex, we use the formula \( h = -\frac{b}{2a} \).

For the function \( f(x) = x^2 - 6x - 16 \), we have \( a = 1 \) and \( b = -6 \).

\[ h = -\frac{-6}{2 \times 1} = 3 \]

Substitute \( x = 3 \) into the function to find \( k \):

\[ k = f(3) = 3^2 - 6 \times 3 - 16 = 9 - 18 - 16 = -25 \]

Thus, the vertex is \((3, -25)\).

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