Answer
$y=4x$, $y=2$
Work Step by Step
Step 1. Given the equation for the Newton’s serpentine $y=\frac{4x}{x^2+1}$, we can find the derivative function as $y'=\frac{(x^2+1)4-4x(2x)}{(x^2+1)^2}=\frac{4-4x^2}{(x^2+1)^2}$
Step 2. At the origin $(0,0)$, the slope of the tangent is $m_1=\frac{4-4(0)^2}{((0)^2+1)^2}=4$ and the line equation can be written as $y=4x$
Step 3. At point $(1,2)$, the slope of the tangent is $m_2=\frac{4-4(1)^2}{((1)^2+1)^2}=0$ and the line equation can be written as $y=2$
