Sketch each curve from its domain, intercepts and asymptotes

The source gives intercepts (0,−2) and approximately (2.46,0); local maximum (0.107,−1.95) and local minimum (1.56,−5.02). A polynomial has no vertical, horizontal or oblique line asymptotes here. For verification, $y'=6x^2-10x+1$ has zeros $(5\pm\sqrt{19})/6$; the smaller gives the maximum and the larger the minimum.

As $x\to-\infty$, $4x^2e^x\to0^+$, so y=0 is a horizontal asymptote on the left. Since $e^x>0$, $4x^2e^x=0$ only when x=0; hence (0,0) is the only axial intercept. Turning points are a minimum (0,0) and local maximum (−2.00,2.17), the latter exactly $(-2,16/e^2)$. The product limit uses exponential decay dominating polynomial growth; eˣ→0 alone is not a general product-limit argument.

For x>0, asymptotes are x=0 and y=0; the only axial intercept is (1,0), with no y-intercept. The maximum is approximately (2.72,0.184), exactly $(e,1/(2e))$. As x→0⁺ the curve tends to−∞; as x→∞ it tends to 0 from above.

$y=1-5/[(x+3)(x-1)]$. Thus asymptotes are x=−3,x=1,y=1. Intercepts are (−4,0),(2,0),(0,8/3); the middle branch has a minimum (−1,2.25). Outside the two poles the curve approaches y=1 from below; between the poles it lies above y=1.

The absolute-value bars are present in the source image. On 0≤x≤2, $y=2x-x^2$; outside this interval, $y=x^2-2x$. Intercepts are (0,0),(2,0), with a smooth local maximum (1,1). The source lists this maximum; the two intercepts are also non-differentiable cusp minima. They must be retained when “turning points” includes changes of direction. There are no asymptotes.

$ax+1=a(x-b)+(1+ab)$, so $y=a+(1+ab)/(x-b)$, with x≠b. Provided 1+ab≠0, the asymptotes are x=b,y=a, and intercepts are $(0,-1/b)$ and $(-1/a,0)$. This retains the source long-division quotient a and remainder 1+ab. For either nondegenerate regime, $y'=-(1+ab)/(x-b)^2$ is never zero, so there are no turning points.

For −1<a<0 and 0<b<1, $-1/a>1$, hence $b<-1/a$. Also 1+ab>0: the branch right of x=b runs from+∞down toward y=a, while the left branch lies below y=a.

For a<−1 and b>1, $0<-1/a<1$, so $-1/a<b$. Here 1+ab<0: the branch left of x=b rises toward+∞near the pole, and the right branch lies below y=a.

If ab+1=0, then $y=a+0/(x-b)=a$ only for x≠b. The graph is the horizontal line y=a with a hole at (b,a). The source shows both a<0,b>0 and a>0,b<0 sign cases. This cancellation is a general extension: neither of the two strict parameter regimes above allows ab=−1. There is no vertical asymptote at the removable hole.