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Unformatted text preview: + b = 220. Solving, we get b = 250. Hence, we get p(x) = −1.5x + 250. We can
check our formula by computing p(20) and p(40) to see if we get 220 and 190, respectively.
Incidentally, this equation is sometimes called the price-demand6 equation for this venture.
6 Or simply the demand equation 120 Linear and Quadratic Functions 2. To determine the applied domain, we look at the physical constraints of the problem. Certainly, we can’t sell a negative number of PortaBoys, so x ≥ 0. However, we also note that the
slope of this linear function is negative, and as such, the price is decreasing as more units are
sold. Another constraint, then, is that the price, p(x) ≥ 0. Solving −1.5x + 250 ≥ 0 results
in −1.5x ≥ −250 or x ≤
= 166.6. Since x represents the number of PortaBoys sold in a
week, we round down to 166. As a result, a reasonable applied domain for p is [0, 166]. 3. The slope m = −1.5, once again, represents the rate of change of the price of a system with
respect to weekly sales of PortaBoys. Since the slope is negative, we ha...
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