I'm stuck on coordinate graph involving tangents of parabolas...

Question

Consider the function $f(x) = \max \{-11x - 37, x - 1, 9x + 3\}$ defined for all real $x.$ Let $p(x)$ be a quadratic polynomial tangent to the graph of $f$ at three distinct points with $x$-coordinates $x_1,$ $x_2,$ $x_3.$ Find $x_1 + x_2 + x_3.$

I have found a piecewise function representing $f(x)$, and I do not know what else to do. Any suggestions?

Answer 1

This can be solved without explicitly finding the equation of the parabola. Its axis is parallel to the $y$-axis, so it is uniquely determined by the three tangent lines. Taking the first two tangent lines, we use the two-points-two-tangents property of parabolas to equate the difference in $y$-coordinates of the points of tangency with the difference in $y$-coordinates of the other intersections of the two lines with the vertical lines through those points: $$(-11x_1-37)-(x_2-1)=(x_1-1)-(-11x_2-37).$$ Applying the same construction to the other two pairs of tangent lines gives you a system of three linear equations in the $x_i$, which I hope you can solve on your own.

Answer 2

The following image displays 3 line portions :

Let the quadratic function have equation $y=f(x)=ax^2+bx+c$ with derivative $f'(x)=2ax+b$.

Assuming that the points $(x_k;f(x_k))$ of contact of the parabola are one on each line portion, one has two families of contraints : the fact that the parabola passes through the points, and the fact that, at these points, the slopes are the given slopes :

$$\begin{cases}ax_1^2+bx_1+c&=&-11x_1-37\\ax_2^2+bx_2+c&=&x_2-1\\ax_3^2+bx_3+c&=&9x_3+3\end{cases}$$

$$\begin{cases}2ax_1+b=-11\\2ax_2+b=1\\2ax_3+b=9\end{cases}$$

which makes 6 equations with 6 unknowns.

Submitting them to a CAS (Computer Algebra system), I have directly obtained a unique solution :

$$x_1=-9/2, \ x_2=-3/2, \ x_3=1/2, \ a=2, \ b=7, \ c=7/2$$

therefore

$$x_1+x_2+x_3=-11/2$$

Remark : There must be a way to obtain this sum in a simpler way, but I don't see how.

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Edit : in fact, @amd has found this way. His solution that you can see below is a jewel and IMHO should have been selected instead of mine :

For instructive purposes, here is a way to program it (with Matlab facilities for handling symbolic variables)

syms x1 x2 x3
f=@(x)(-11*x-37);
g=@(x)(x-1);
h=@(x)(9*x+3);
S=solve(f(x1)-g(x2)==g(x1)-f(x2),...
        f(x1)-h(x3)==h(x1)-f(x3),...
        g(x2)-h(x3)==h(x2)-g(x3),...
        x1,x2,x3);
[S.x1,S.x2,S.x3], % gives x_1=-9/2, x_2=-3/2, x_3=1/2