## Precalculus (6th Edition) Blitzer

a. $261\ ft$, overestimates by 1 ft. b. $x\gt40\ mph$
a. Given $g(x) = 0.125x^2 + 2.3x + 27$ for $x=35$, we have $g(35) = 0.125(35)^2 + 2.3(35) + 27\approx261\ ft$ The bar graph gives a value of $260\ ft$ for the web pavement and 35 mph; thus the value from the model overestimates by 1 ft. b. For $f (x) = 0.125x^2 − 0.8x + 99$, let $f(x)\gt267$ We have $0.125x^2 − 0.8x + 99\gt267$ or $0.125x^2 − 0.8x -168\gt0$ This inequality can be solved graphically as shown in the figure; we have the solution as $x\gt40\ mph$