Practice Test 3

Based on the passage, rate of change is the average change in speed per second over a chosen interval. A car’s speed is modeled by the polynomial $p(t)=-t^2+6t+8$ (m/s) for $0\le t\le 6$. The model increases at first, since $p(0)=8$ and $p(2)=16$ show acceleration. It then decreases after the peak because the negative quadratic term reduces the speed. For example, $p(4)=16$ and $p(6)=8$ show deceleration later. A separate drag-limited model is $r(t)=\frac{12}{t-7}+6$, approaching $6$ as $t$ increases. The rational model is undefined at $t=7$ due to a vertical asymptote. Referring to the context, how does the rate of change of $r(t)$ differ near its vertical asymptote compared with large $t$?​

Referring to the context, rate of change means average change in speed per second over a stated interval. A car’s speed is modeled by the polynomial $p(t)=-0.5t^2+5t+4$ (m/s) for $0\le t\le 10$. The speed increases early, since values like $p(0)=4$ and $p(4)=16$ show acceleration. Later, the speed decreases because the negative quadratic term reduces speed growth. A second sensor follows the rational model $r(t)=\frac{18}{t-6}+9$, which approaches $9$ as $t$ increases. The rational model has a vertical asymptote at $t=6$, where the expression is undefined. Based on the passage, how does the rate of change of $r(t)$ behave as $t$ gets large?