Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 1 - Section 1.2 - Mathematical Models: A Catalog of Essential Functions. - 1.2 Exercises - Page 33: 12

Answer

$g(x)=-x^2-2.5x+1$

Work Step by Step

The $y$-intercept of the parabola on the right is $(0,1)$, so an equation is $g(x)=ax^2+bx+1$……….$(1)$ Since the points $(-2,2)$ and $(1,-2.5)$ are on the parabola, we will substitute in equation $(1)$ the values $-2$ for $x$ and $2$ for $y$ as well as $1$ for $x$ and $-2.5$ for $y$ to obtain two equations with two unknowns $a$ and $b$. $$\begin{cases} 4a-2b+1=2\\ a+b+1=-2.5 \end{cases}$$ Rewrite the equations: $$\begin{cases} 4a-2b=1\\ a+b=-3.5 \end{cases}$$ AMultiply the second equation by $2$ and add it to the first to determine $a$: $4a-2b+2a+2b=1-7$ $6a=-6$ $a=-1$ Calculate $b$: $b=-3.5-a=-3.5-(-1)=-2.5$ The function is: $g(x)=-x^2-2.5x+1$
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