Given the following function, (a) find the vertex; (b) determine whether there is a maximum or a minimum value, and find the value; (c) find the range; and (d) find the intervals on which the function is increasing and the intervals on which the function is decreasing. \( f(x)=-8 x^{2}-40 x+3 \) (a) The vertex is (Type an ordered pair, using integers or fractions.)

To find the vertex of the function \( f(x) = -8x^2 - 40x + 3 \), we can use the formula for the vertex of a quadratic function, which is given by \( x = -\frac{b}{2a} \). Here, \( a = -8 \) and \( b = -40 \). Calculating the x-coordinate of the vertex: \[ x = -\frac{-40}{2 \cdot -8} = -\frac{40}{-16} = 2.5 \] Plugging \( x = -2.5 \) back into the function to find the y-coordinate: \[ f(-2.5) = -8(-2.5)^2 - 40(-2.5) + 3 = -8(6.25) + 100 + 3 = -50 + 100 + 3 = 53 \] Thus, the vertex is \( (-2.5, 53) \). (b) Since the coefficient of \( x^2 \) is negative \((-8)\), this indicates that the function opens downwards, meaning there is a maximum value at the vertex. The maximum value is \( 53 \). (c) The range of the function is from the maximum value down to negative infinity, written as \( (-\infty, 53] \). (d) To find the intervals of increase and decrease, we look at the vertex. Since the function is a downward-opening parabola, it increases on the interval \( (-\infty, -2.5) \) and decreases on the interval \( (-2.5, \infty) \).