Answer
The vertex is (25, 625).
The range is $0\leq A \leq 625$
Work Step by Step
$A = -x^{2} + 50x$
The standard form for a quadratic equation is
$y = ax^{2} + bx + c$ So a= -1, b= 50, and c= 0
Axis of symmetry: In this question the Axis of symmetry gives us the width to get the maximum Area.
The formula for axis of symmetry is
$x= \frac{-b}{2a}$
$x= \frac{-(50)}{2(-1)}$
$x= \frac{-50}{-2}$
x= 25
Vertex: In this question the vertex gives us the maximum width.
Plug in the x value of the axis of symmetry to find the y value of the vertex.
$A = -x^{2} + 50x$
$A = -(25)^{2} + 50(25)$
y= 625 ft
The vertex is (25, 625)
Since the width starts st zero and goes up to 625 the range is $0\leq A \leq 625$
