Write the equations of the tangents to the curve y =

Write the equations of the tangents to the curve $y = \bigl|4x-x^{2}\bigr|$ at the point $x = 0$ and find the angle between them.

Step 1: Remove the absolute value near the origin

Since $4x-x^{2} = x\bigl(4-x\bigr)$ and $4-x\gt 0$ near $x = 0$, the sign is that of $x$: \[y = \begin{cases} 4x-x^{2}, & x\ge0,\\ x^{2}-4x, & x\le0, \end{cases} \qquad y(0) = 0.\]

Step 2: The one-sided derivatives

\[ y'_{+}(0) = \lim_{h\to0^{+}}\frac{4h-h^{2}}{h} = 4, \qquad y'_{-}(0) = \lim_{h\to0^{-}}\frac{h^{2}-4h}{h} = -4. \]

They differ, so the graph has a corner at the origin and two distinct one-sided tangents.

Step 3: The two tangents

Both pass through $\bigl(0,0\bigr)$: $y = 4x \quad\text{(from the right)}, \qquad y = -4x \quad\text{(from the left)}$.

Step 4: The angle between them

\[ \tan\varphi = \left|\frac{k_{1}-k_{2}}{1+k_{1}k_{2}}\right| = \left|\frac{4-(-4)}{1+4(-4)}\right| = \left|\frac{8}{-15}\right| = \frac{8}{15}, \]

\[ \varphi = \arctan\frac{8}{15} = 28^{\circ}04' = 28.0725^{\circ}. \]

Step 5: Verification

\[\text{Direction vectors of the two branches at } O:\ \bigl(1,4\bigr) \text{ and } \bigl(-1,4\bigr)\]

\[\cos\varphi = \frac{-1+16}{\sqrt{17}\sqrt{17}} = \frac{15}{17} \;\Longrightarrow\; \tan\varphi = \frac{8}{15} \;\;\checkmark\]

\[\text{Alternatively } \varphi = \pi-2\arctan4,\ \text{and } \tan\bigl(2\arctan4\bigr) = \frac{8}{1-16} = -\frac{8}{15} \;\;\checkmark\]