SOLUTION: Between 1989 and 1997 the percentage of households with incomes of $100,000 or more (in 1997 dollars) is modeled by f(x) = .071x2 - .426x + 8.05, where x = 0 represents the year 1
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: Between 1989 and 1997 the percentage of households with incomes of $100,000 or more (in 1997 dollars) is modeled by f(x) = .071x2 - .426x + 8.05, where x = 0 represents the year 1
Log On
Question 627543: Between 1989 and 1997 the percentage of households with incomes of $100,000 or more (in 1997 dollars) is modeled by f(x) = .071x2 - .426x + 8.05, where x = 0 represents the year 1989. Based on this model when did the percent of people in this income level reach its minimum? Answer by ewatrrr(24785) (Show Source):
Hi,
f(x) = .071x2 - .426x + 8.05 ||y = ax^2 + bx + c ⇒ y = a(x -(-b/2a)^2) - a(-b/2a)^2 + c
f(x) = .071(x-3)^2 - .071·9 + 8.05
f(x) = .071(x-3)^2 + 7.411 Vertex(minimum point) is (3,7.411)
Based on this model when did the percent of people in this income level reach its minimum? 1989 + 3 = 1992