## Precalculus (6th Edition) Blitzer

a. $f(55) \approx309\ ft$. $g(55) \approx381\ ft$. b. upper blue curve-(a) is for the wet pavement; lower red curve-(b) is for the dry pavement. c. very well. d. $x\gt68\ mph$; see explanations.
a. With $x=55\ mph$, for the dry pavement, we have $f(55)=0.0875(55)^2-0.4(55)+66.6\approx309\ ft$. Similarly, for the wet pavement, we have $g(55)=0.0875(55)^2+1.9(55)+11.6\approx381\ ft$. b. Clearly, the stopping distance is larger for the web pavement than that for the dry pavement. Based on the graph, we can identify that the upper blue curve-(a) is for the wet pavement, while the lower red curve-(b) is for the dry pavement. c. The actual data shows that the stopping distances for the dry pavement is 310 feet and for the wet pavement is 380 feet when the speed is 55 mph. Thus the model results in part-a fit the data very well. d. Letting $g(x)\gt540$, we have $0.0875x^2+1.9x+11.6\gt540$ or $0.0875x^2+1.9x-528.4\gt0$ Solving the inequality graphically as shown in the figure, we have $x\gt68\ mph$. The corresponding point can be found on the graph by locating $x=68\ mph$ and then by finding the vertical coordinate on the blue curve-(a), which should give a value about 540 feet.