Answer
A) $954,409$
B) $2012$
Work Step by Step
A) For this problem, you must solve for $A(18)$ because $2018$ is $18$ years after $2000$. This means that you must solve $.638(18)^{2}+6.671(18)+627.619$ which results in $954.409$ thousand people. Because the unit is thousands of people, multiply your result by $1,000$ to get $954,409$ people.
B) This time, you are given the expected population and asked to find the year the population is expected. So you have the value for $A(x)$ but need to find the value of $x$. Write an equation of $800=.638x^{2}+6.671x+627.619$, where $A(x)=800$ because $A(x)$ has the unit of $1,000$ people. Solve the equation,
$.638x^{2}+6.671x+627.619=800$
$.638x^{2}+6.671x-172.381=0$ (Subtract $800$ from both sides)
$x=\frac{-6.671+\sqrt {6.671^{2}-4(-172.381)(.638)}}{2(.638)}$ (Use the quadratic formula, but only the positive root as you want the year after 2000)
$x\approx\frac{-6.671+22.01}{1.276}$ (Start to simplify, round to nearest thousandth)
$x\approx\frac{15,339}{1.276}$ (Simplification)
$x\approx12.021$ (Answer rounded to nearest thousandth)
This answer means that shortly after $2012$ starts, the population will reach $800,000$, so $2012$ is the answer.
