Answer:
To determine the vertex of the quadratic function \( H(x) = (x - 8)(x + 2) \), let's first expand the expression:
\[ H(x) = x^2 - 6x - 16 \]
The vertex of a quadratic function in the form \( f(x) = ax^2 + bx + c \) is given by the point \( \left( -\frac{b}{2a}, f\left( -\frac{b}{2a} \right) \right) \).
In this case, \( a = 1 \) (coefficient of \( x^2 \)) and \( b = -6 \) (coefficient of \( x \)). Substituting these values into the formula:
\[ x_{\text{vertex}} = -\frac{-6}{2(1)} = 3 \]
Now, substitute \( x = 3 \) into the function to find the y-coordinate:
\[ H(3) = (3)^2 - 6(3) - 16 = 9 - 18 - 16 = -25 \]
So, the vertex of the function is at \( (3, -25) \).
