Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.2 Mathematical Models: A Catalog of Essential Functions - 1.2 Exercises: 10


Function on Left: $y=2(x-3)^2$ Function on Right: $y=\frac{-1}{2}x^2-2x+1$

Work Step by Step

Left Function: Using the basic factored form of the parabola $y=a(x-b)^2+c$, we can determine the equation of the graph from the two points given (vertex and other point). The vertex is (3,0), which tells us that the graph has shifted 3 units to the right of the origin and 0 units up - hence b = 3 and c = 0. So we get $y=a(x-3)^2$ . To find a, just sub (4,2) into the equation to get $2=a(4-3)^2 $ -> $a=2$. Therefore $y=2(x-3)^2$ Right Function: Use the other form of the parabola: $y=ax^2+bx+c$. From the graph, we can tell the $y-int = 1$, hence $c=1$, so the equation becomes $y=ax^2+bx+1$. Now sub both the points (-2,2) and (1,-2.5) into the equation to get a pair of simultaneous equations: (1) $2 = 4a -2b$ (2) $-2.5=a+b$ --> $a=-b-2.5$ sub (2) into (1) $2=4(-b-2.5)-2b$ $2=-4b-10-2b$ $-6b = 12$ $b=-2$ ... sub into (1) $4a = -2$ $a = \frac{-1}2$ And finally the equation is: $y=\frac{-1}{2}x^2-2x+1$
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