## Precalculus (6th Edition) Blitzer

a. $f(35)\approx160\ ft$. $g(35) \approx185\ ft$. b. the upper blue curve is for the wet pavement; the lower red curve is for the dry pavement. c. very well. d. $x\gt76\ mph$; see explanations.
a. With $x=35\ mph$, for the dry pavement, we have $f(35)=0.0875(35)^2-0.4(35)+66.6=159.7875\approx160\ ft$ Similarly, for the wet pavement, we have $g(35)=0.0875(35)^2+1.9(35)+11.6=185.2875\approx185\ ft$. b. Clearly, the stopping distance is larger for web the pavement than that for the dry pavement. Based on the graph, we can identify that the upper blue curve is for the wet pavement, while the lower red curve is for the dry pavement. c. The actual data shows that the stopping distances for dry pavement is 160 feet and for the wet pavement is 185 feet when the speed is 35 mph. Thus, the model in part-a fits the data very well. d. Let $f(x)\gt540$; we have $0.0875x^2-0.4x+66.6\gt540$, or $0.0875x^2-0.4x-473.4\gt0$ Solving the inequality graphically as shown in the figure, we have $x\gt76\ mph$. The corresponding point can be found on the graph by locating $x=76\ mph$ and then by finding the vertical coordinate on the red curve, which should give a value of about 540 feet.