Stitz-Zeager_College_Algebra_e-book

410 hooked on conics y 1 x 1 1 1 2 3 the vertex lies

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Unformatted text preview: , we use the fact that after 2 hours, the roast is 125◦ F, which means T (2) = 125. This gives rise to the equation 350 − 310e−2k = 125 which yields k = − 1 ln 45 ≈ 0.1602. The temperature function is 2 62 T (t) = 350 − 310e 2 ln( 62 ) ≈ 350 − 310e−0.1602t . t 45 384 Exponential and Logarithmic Functions 2. To determine when the roast is done, we set T (t) = 165. This gives 350 − 310e−0.1602t = 165 1 whose solution is t = − 0.1602 ln 37 ≈ 3.22. It takes roughly 3 hours and 15 minutes to cook 62 the roast completely. If we had taken the time to graph y = T (t) in Example 6.5.4, we would have found the horizontal asymptote to be y = 350, which corresponds to the temperature of the oven. We can also arrive at this conclusion by applying a bit of ‘number sense’. As t → ∞, −0.1602t ≈ very big (−) so that e−0.1602t ≈ very small (+). The larger the value of t, the smaller e−0.1602t becomes so that T (t) ≈ 350 − very small (+), which indicates the...
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