Answer:
See below
Step-by-step explanation:
A) The quadratic function that models this data is [tex]f(x)=-0.02x^2+1.6x+2[/tex] which you can view in the graph attached (done by regression).
B) [tex]f(65)=-0.02(65)^2+1.6(65)+2=21.5[/tex], or 21.5 miles
C) Set [tex]f(x)=16[/tex] and solve for x:
[tex]f(x)=-0.02x^2+1.6x+2[/tex]
[tex]16=-0.02x^2+1.6x+2[/tex]
[tex]0=-0.02x^2+1.6x-14[/tex]
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-1.6\pm\sqrt{(1.6)^2-4(-0.02)(-14)}}{2(-0.02)}[/tex]
[tex]x=\frac{-1.6\pm\sqrt{2.56-1.12}}{-0.04}[/tex]
[tex]x=\frac{-1.6\pm\sqrt{1.44}}{-0.04}[/tex]
[tex]x=\frac{-1.6\pm1.2}{-0.04}[/tex]
[tex]x_1=\frac{-1.6+1.2}{-0.04}[/tex]
[tex]x_1=\frac{-0.4}{-0.04}[/tex]
[tex]x_1=10<55[/tex]
[tex]x_2=\frac{-1.6-1.2}{-0.04}[/tex]
[tex]x_2=\frac{-2.8}{-0.04}[/tex]
[tex]x_2=70>55[/tex]
Since [tex]70>55[/tex], then 10mph is the maximum speed you could drive and still reach the gas station that is 16 miles away when the speed limit is 55.
