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# Help!

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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.

Dec 27, 2019

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

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\\

Dec 28, 2019