## Algebra 1

Published by Prentice Hall

# Chapter 9 - Quadratic Functions and Equations - Concept Byte - Collecting Quadratic Data - Page 547: 2

#### Answer

a) 40 cm b) $20-w = l$ c) $A = -w^2+20w$ d) Please see the graph. e) (10,100) This means that the maximum area of the rectangle is 100 $cm^2$ #### Work Step by Step

a) I started with a 40 cm string. b) $perimeter = 2*w+2*l$ $P = 2w+2l$ $40 = 2w+2l$ $40/2 = (2w+2l)/2$ $20 = w+l$ $20-w = w+l-w$ $20-w = l$ c) $area = l*w$ $area = (20-w)*w$ $A = 20w-w^2$ $A = -w^2+20w$ d) The graph was made using graphing software. Vertex: $x=-b/2a$ $x = -20/2*-1$ $x = -20/-2 = 10$ $A = -(10)^2+20*10$ $A = -100+200 = 100$ Axis of symmetry is $x=10$ (per the work above for finding the vertex) One other point on the curve: $x=0$ $A = -(0)^2+20*0$ $A = -0 +0 = 0$ e) Per part d), the vertex is (10,100).

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