[tex]\bf f(x)=0.35x^2-2.1x+15.8\qquad
\begin{cases}
x=\textit{year since 1995}\\
f(x)=\textit{population amount}
\end{cases}[/tex]
the equation is a quadratic one, and it has a positive coefficient on the leading term, meaning, is opening upwards, so it has a "burrow" for the vertex.
the minimum or lowest point for a quadratic opening upwards is, well, the vertex point :), the "x" value is the year, the "y" or f(x) value is the population, we're asked for the year, or the x-coordinate of the vertex
well [tex]\bf \begin{array}{llll}
f(x)=&0.35x^2&-2.1x&+15.8\\
&\quad \uparrow &\quad \uparrow&\uparrow \\
&\quad a&\quad b &c
\end{array}
\\\\
\\\\
\qquad \textit{vertex of a parabola}\\ \quad \\
\qquad
\left(\boxed{-\cfrac{{{ b}}}{2{{ a}}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
