Unformatted text preview: = 0. For 2x2 + x − 3 < 0, we
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need the intervals which we have a (−). Checking the sign diagram, we see this is − 2 , 1 .
We know 2x2 + x − 3 = 0 when x = − 3 and x = 1, so or ﬁnal answer is − 3 , 1 . To check our
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solution graphically, we refer to the original inequality, 2x2 ≤ 3 − x. We let g (x) = 2x2 and
h(x) = 3 − x. We are looking for the x values where the graph of g is below that of h (the
solution to g (x) < h(x)) as well as the two graphs intersect (the solutions to g (x) = h(x).)
The graphs of g and h are given on the right with the sign chart on the left.
2 We have to choose something in each interval. If you don’t like our choices, please feel free to choose diﬀerent
numbers. You’ll get the same sign chart. 162 Linear and Quadratic Functions
y
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6
5 (+) 0 (−) 0 (+)
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−2
−2 4 1
0 3 2 2 −2 −1 1 2 x 2. Once again, we rewrite x2 > 2x + 1 as x2 − 2x − 1 > 0 and we identify f (x) = x2 − 2x − 1.
When we go to ﬁnd the zeros of f , we ﬁnd, to...
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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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