Given: f(x)=0.04x²-8.5x+500 ··· Eq.1 Eq.1 shows that the average cost per a head per a year, if x heads of horses are breeded together.
i). Plug x=50 into Eq.1 We have: f(50)=175 (What is the currency?). The average cost is 175 per a head per a year, if 50 heads of horses are breeded together.
ii). Draw a graph of Eq.1 to examine what the graph suggests. The coefficient of x² is positive(0.04), so the graph is smooth but rather slow curve because of the coefficient(0.04), and concave downwards(holds water). To examine the minimam cost, change Eq.1 into a vertex- form.
We have: f(x)=0.04(x-106.25)²+48.4375 ··· Eq.2 Thus, At x=106.25, Eq.2 takes its minimum value: 48.4375 So, if 106 heads of horses are breeded together, the average cost, the lowest, is 44.44 per ahead per a year.
The graph suggests that breeding around 106 heads of horses is the most cost effective.
Plug x=150, 100, 50, 30 and 0 into Eq.1, we have f(150)=125, f(100)=50, f(50)=175, f(30)=281 and f(0)=500. This also suggests that there must be some limitations for the ranges x takes.
