The graphs of \(y=|x|\) and \(y=-x^2-3x-2\) are drawn. For every \(x\), a vertical segment connecting these two graphs can be drawn as well. Find the smallest possible length of one of these vertical segments.

Guest Dec 27, 2019

#1**+2 **

It is (maybe) easiest to start with a sketch.

\(y=-x^2-3x-2\\ y=-(x^2+3x+2)\\ y=-(x+2)(x+1)\)

This is a concave down parabola, roots are -2 and -1.

Vertex is at x= -3/2

When x=-3/2 y=-9/4 +9/2 -2 = 0.25

Vertex is (-1.5, 0.25)

A quick look at the graph reveals that y=|x| is always above the parabola.

Also it is easy to see that the shortest distance betweem them will definitely be somewhere in the domain \( -1.5 \le x\le 0\)

so

the distance betweem them is

\(Q=|x|-(-x^2-3x-2)\\ Q=|x|+x^2+3x+2\\ \)

But x will be negative so

\(Q=-x+x^2+3x+2\\ Q=x^2+2x+2\\ \)

Q is a concave up parabola so any stat point will be a minimum.

\(\frac{dQ}{dx}=2x+2\)

This equals 0 when x=-1

When x=-1 Q=1-2+2 = 1

So the smallest vertical distance between the graphs is 1 and it occurs when x=-1

Here is the graph:

Coding:

Q=|x|-(-x^2-3x-2)\\

Q=|x|+x^2+3x+2\\

Q=-x+x^2+3x+2\\

Q=x^2+2x+2\\

Melody Dec 28, 2019