# Question # The following quadratic function in general form, displaystyle{S}{left({t}right)}={5.8}{t}^{2}—{81.2}{t}+{1200} models the number of luxury home sales

Modeling data distributions
ANSWERED The following quadratic function in general form, $$\displaystyle{S}{\left({t}\right)}={5.8}{t}^{2}—{81.2}{t}+{1200}$$ models the number of luxury home sales, S(t), in a major Canadian urban area, according to statistical data gathered over a 12 year period. Luxury home sales are defined in this market as sales of properties worth over \$3 Million (inflation adjusted). In this case, $$\displaystyle{\left\lbrace{t}\right\rbrace}={\left\lbrace{0}\right\rbrace}\ \text{represents}\ {2000}{\quad\text{and}\quad}{\left\lbrace{t}\right\rbrace}={\left\lbrace{11}\right\rbrace}$$represents 2011. Use a calculator to find the year when the smallest number of luxury home sales occurred. Without sketching the function, interpret the meaning of this function, on the given practical domain, in one well-expressed sentence. 2020-11-08
Given:
The number of luxury home sales S(t) in a major Canadian urban area over a period of 12 year is given by:
$$\displaystyle\Rightarrow{S}{\left({t}\right)}={5.8}{t}^{2}-{81.2}{t}+{1200}$$
For minimum number of sales:
$$\displaystyle\Rightarrow\frac{{{d}{S}{\left({t}\right)}}}{{\left.{d}{t}\right.}}={0}$$
$$\displaystyle\Rightarrow\frac{{{d}{\left({5.8}{t}^{2}-{81.2}{t}+{1200}\right)}}}{{\left.{d}{t}\right.}}={0}$$
$$\displaystyle\Rightarrow{11.6}{t}-{81.2}={0}$$
$$\displaystyle\Rightarrow{11.6}{t}={81.2}$$
$$\displaystyle\Rightarrow{t}=\frac{{{81.2}}}{{{11.6}}}$$
$$\displaystyle\Rightarrow{t}={7}$$
So, minimum number of sales is given by
$$\displaystyle\Rightarrow{S}{\left({7}\right)}={5.8}{\left({7}\right)}^{2}-{81.2}{\left({7}\right)}+{1200}$$
$$\displaystyle\Rightarrow{S}{\left({7}\right)}={284.2}-{568.2}+{1200}$$
$$\displaystyle\Rightarrow{S}{\left({7}\right)}{915.8}$$
$$\displaystyle\Rightarrow{S}{\left({7}\right)}\stackrel{\sim}{=}{916}$$