Question

The function of (x) varies directly with x^2, and f(x)=96 when x=4 what is the value of f(2)

Answer 1

$\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array}\\\\ -------------------------------$

$\bf \textit{f(x) varies directly with }x^2\qquad f(x)=kx^2 \\\\\\ \textit{we also know that } \begin{cases} f(x)=96\\ x=4 \end{cases}\implies 96=k(4)^2\implies 96=16k \\\\\\ \cfrac{96}{16}=k\implies 6=k\qquad therefore\qquad \boxed{f(x)=6x^2} \\\\\\ \textit{now, when }\stackrel{f(2)}{x=2}\textit{ what is \underline{f(x)}?}\qquad f(2)=6(2)^2$