Use multiplicity to distinguish crossings and contacts

Set each factor to zero: $x=-1$ has multiplicity 1 (crossing), while $x=1$ has multiplicity 4 (contact without a sign change). At $x=0$, $y=1$. This is a degree-5 polynomial with positive leading coefficient, so its left end goes down and its right end goes up. The Teacher completes the Student’s blank axes with these intercepts and the turning-point label $(-0.600,2.62)$.

Independent calculus verifies the source sketch:

$\begin{aligned}&y'\\ & =(x-1)^3\big((x-1)+4(x+1)\big)\\ & =(x-1)^3(5x+3).\end{aligned}$

The derivative changes $+\to-$ at $x=-3/5$, and $-\to+$ at $x=1$. Thus the maximum is $(-3/5,8192/3125)=(-0.6,2.62144)$ and the minimum is $(1,0)$. The Teacher’s $2.62$ is rounded; retain the exact point for calculation.

Both copies already print this completed sketch. The even multiplicity 2 gives contact at $(-1,0)$; the odd multiplicity 3 gives a crossing at $(1,0)$. The $y$-intercept is $(0,-1)$. Again the degree is 5 with positive leading coefficient. The printed lower turning point is $(-0.200,-1.11)$.

Independent calculus gives

$\begin{aligned}&y'\\ & =(x+1)(x-1)^2\big(2(x-1)+3(x+1)\big)\\ & =(x+1)(x-1)^2(5x+1).\end{aligned}$

The derivative changes $+\to-$ at $x=-1$ (local maximum), then $-\to+$ at $x=-1/5$ (local minimum). Its exact lower point is $(-1/5,-3456/3125)=(-0.2,-1.10592)$, agreeing with the source’s rounded ordinate. At $x=1$ the squared derivative factor leaves the sign positive on both sides, so the curve crosses with a horizontal tangent and does not turn. Locally $y=(x+1)^2(x-1)^3$ has a nonzero cubic leading term about $x=1$, giving a stationary inflection.