## Algebra 1

The vertex is (25, 625). The range is $0\leq A \leq 625$
$A = -x^{2} + 50x$ 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 $0\leq A \leq 625$