Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.1 Graph Quadratic Functions in Standard Form - 4.1 Exercises - Problem Solving - Page 242: 58

Answer

See below

Work Step by Step

To find the maximum height the mouse can jump, we will find the maximum value of the given function $y=-0.2x^2+1.3x$ Coefficients: $a=-0.2\\b=1.3\\c=0$ The x-coordinate of the vertex is $x=\frac{-b}{2a}=\frac{-1.3}{2.(-0.2)}=3.25$ Substitute this into the given equation to find y: $$y=-0.2(3.25)^2+1.3(3.25)\\ \rightarrow y=-0.2(10.5625)+4.225\\ \rightarrow y=2.1125$$ Hence, the vertex is (3.25,2.1125). The maximum value of the function is $y=2.1125$ The maximum height the mouse can jump is $2.1125$ft. Since the fence is 3ft high, the mouse can not jump over it.
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