The following four points are known to lie
on a closed curve in the (x, y)-plane: (−1,0), (−1/2,3), (1/2,−3), (1,0)
and the goal of this question is to fit a piecewise-polynomial approximation of the curve that passes through all four points. 
One approach could be to consider two halves separately, that is, define two curves that are both given by the graph of a (polynomial) function of the x-coordinates x0 = −1, x1 = −1/2, x2 = 1/2, and x3 = 1. More precisely, the interpolating polynomial pt(x) for the top half is constructed based on nodes x0, x1 and x3, while the interpolating polynomial pb(x) for the bottom half is based on x0, x2 and x3.
- Show that it is not possible to use quadratic (degree-two) splines for pt(x) and pb(x) such that the combined four-component piecewise-polynomial approximation is a quadratic spline.
I am getting stuck in this question particularly why I cannot use quadratic spline in this term, can someone help me with this question
Best Answer
The easiest way to show that quadratic splines don't work, is to actually construct them !
Note that both splines go through $(-1,0)$ and $(1,0)$. Hence the splines are of the form $C (1-x^2)$, where $C$ is a constant. Substituting the third point gives $C = 4$ for $pt(x)$ and $C = -4$ for $pb(x)$.
As already pointed out by Eric Towers in the comments, the derivatives of these splines in $(-1,0)$ and $(1,0)$ are unequal and hence the tangents intersect.