## University Calculus: Early Transcendentals (3rd Edition)

$L(x)=\frac{\sqrt{20x^2-20x+25}}{4}$
$\mathrm{See\:the\:figure\:below.}$ When you have any point in a plane, you can use the Pythagorean theorem to determine the distance to the origin. $L=\sqrt{x^2+y^2}$ From the given function $\ 2y+4y=5,\$ we can express $\ y\$ as a function of $\ x.$ $2x+4y=5$ $\Rightarrow\ 4y=5-2x$ $\Rightarrow\ y=\frac{5-2x}{4}$ Plug in the expression for $\ y\$ to get $\ L\$ as a function of $\ x.$ $L=\sqrt{x^2+(\frac{5-2x}{4})^2}$ $\Rightarrow\ L=\frac{\sqrt{20x^2-20x+25}}{4}$