Answer
a. between 2000 and 2005.
b. 2002, 480 and 480
Work Step by Step
a. Let $f(x)$ represent the difference (imprisonments-violent crimes) between the two crimes per 100,000 where $x$ is the year after 1990. Based on the graph, for year 2000, we have $f(10)\lt0$, and for year 2005, we have $f(15)\gt0$. As $f(x)$ is a continuous function, according to the Intermediate Value Theorem, there is a $x$ value between $10$ and $15$ such that $f(x)=0$. Thus, the event occurred between the years 2000 and 2005.
b. Given the two equations $y = 0.6x^2 − 28x + 730$ and $–15x + y = 300$, use $y=15x+300$ from the second equation; the first one becomes $15x+300 = 0.6x^2 − 28x + 730$ or $ 0.6x^2 − 43x + 430=0$. This quadratic can be solved using a formula or graphically to get $x\approx12$ and $x\approx60$. It can can seen that $x=12$, or year 2002, fits the data and agrees with the predication from part-(a). The number per 100,000 for both was about 480.
