## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\color{blue}{y=15x^2}$
RECALL: (1) When $y$ varies directly as $x$, the variation is direct and is represented by the equation $y=kx$ where $k$ is the constant of variation. (2) When $y$ varies inversely as $x$, the variation is inverse and is represented by the equation $y=\frac{k}{x}$ where $k$ is the constant of variation. $y$ varies directly as he square of $x$ so the equation is $y=kx^2$ with $k$=constant of variation. To find the value of $k$, substitute the given values into $y=kx^2$ to obtain: $$y=kx^2 \\0.15=k(0.1^2) \\0.15 =k(0.01) \\\frac{0.15}{0.01}=\frac{k(0.01)}{0.01} \\15=k$$ Thus, the equation of the variation is: $\color{blue}{y=15x^2}$.