Let $f(x)= x+1$ and $g(x)= x-3$. Let $h(x)= f(x)g(x)= (x+1)(x-3)$.
My question is : geometrically, is there any way to predict that the product $f(x)g(x)$ will be at a minimum at $x=1$?
I know one can show that algebraically by rewriting the product as the equation of a parabola $a(x-k)²+h$ , or using calculus, by solving $h'(x)=0$, but geometrically, the couple of points $(1, f(1))$ and $(1, g(1))$ do not seem to have anything special explaining why the line passing through them contains the point where the product is at a minimum.
A geometrical/graphical way of finding the minimum product of factors $ \ (mx + a)·(mx + b) \ $ is to plot the parallel lines and note the location of the points on those lines that are equidistant from the $ \ x-$axis. The value of $ \ x \ $ at which this occurs is the value which gives the minimum product, and said product is found by multiplying together the $ \ y-$coordinates of the two points.
We can first use your example function $ \ (x + 1)·(x - 3 ) \ \ , \ $ plotting the lines $ \ y \ = \ x + 1 \ $ and $ \ y \ = \ x - 3 \ \ . $ The points $ \ (1 \ , \ 2) \ $ and $ \ (1 \ , \ -2) \ $ are the only ones on the lines that are equidistant from the $ \ x-$axis and the product of the $ \ y-$coordinates is $ \ -4 \ \ . \ $ We see that this agrees with the minimun of the product function represented by the parabola.
The $ \ y-$intercepts of the lines need not have opposite signs. For the function $ \ (x + 1)·(x + 3) \ \ , \ $ the points on the lines equidistant from the $ \ x-$axis are $ \ (-2 \ , \ 1) \ $ and $ \ (-2 \ , \ -1) \ \ , \ $ giving the minimum of this product function as $ \ -1 \ \ . \ $
This technique is not limited to factors/lines with unit slopes. For the product $ \ (3x + 5)·(3x - 7) \ \ , \ $ the "vertically equidistant" points are $ \ \left( \frac13 \ , \ 6 \right) \ $ and $ \ \left( \frac13 \ , \ -6 \right) \ \ . \ $ So the minimum of this product is $ \ -36 \ $ (the vertex of the parabola in violet is beyond the bottom of the graph shown).
Why does this "trick" work? If we "complete the square" for the product function, we obtain $$ (mx \ + \ a)·(mx \ + \ b) \ \ = \ \ m^2·x^2 \ + \ m·(a + b)·x \ + \ ab $$ $$ = \ \ m^2 · \left[ \ x \ + \ \frac{(a + b)}{2m} \ \right]^2 \ + \ \left[ \ ab \ - \ m^2·\frac{(a + b)^2}{4m^2} \ \right] $$ $$ = \ \ m^2 · \left[ \ x \ + \ \frac{(a + b)}{2m} \ \right]^2 \ - \ \frac{(a - b)^2}{4} \ \ = \ \ m^2 · \left[ \ x \ + \ \frac{(a + b)}{2m} \ \right]^2 \ - \ \left(\frac{ a - b }{2} \right)^2 \ \ . $$
This shows that the minimal value of the product $ \ (mx + a)·(mx + b) \ $ occurs at the value of $ \ x \ $ half-way between the $ \ x-$intercepts of the two lines, $ \ -\frac{a}{m} \ $ and $ \ -\frac{b}{m} \ \ , \ $ which are also the $ \ x-$intercepts of the parabola corresponding to the product function (zeroes of the quadratic polynomial) and thus also the value of $ \ x \ $ at which the symmetry axis of the parabola is located (as remarked by MathWork). This minimal value is the negative of the square of half the vertical distance between the "vertically equidistant" points, or $$ -\left(\frac{ a - b }{2} \right)^2 \ \ = \ \ -\frac{ (a - b) }{2} \ · \ \frac{ (a - b) }{2} \ \ = \ \ \left( \ -\frac{a + b}{2} + b \ \right) \ · \ \left( \ -\frac{a + b}{2} + a \ \right) $$ $$ = \ \ \left( \ m· \left[-\frac{a + b}{2m} \right] \ + \ b \ \right) \ · \ \left( \ m· \left[-\frac{a + b}{2m} \right] \ + \ a \ \right) \ \ , $$ and thus also the product of the $ \ y-$coordinates of the "vertically equidistant" points.
Geometrically, these "vertically equidistant" points and the $ \ x-$intercepts of the two parallel lines constitute the vertices of a parallelogram, the diagonals of which bisect one another, granting us the requisite symmetry for this technique. Plainly, there is no comparably simple approach for the product $ \ (mx + a)·(nx + b) \ \ , \ $ which would involve lines with distinct slopes.
To apply this trick to the last example, the product $ \ (3x + 5)·(3x - 7) \ $ is found to have zeroes $ \ -\frac53 \ $ and $ \ +\frac73 \ \ , \ $ for which the "midpoint" is $ \ +\frac13 \ \ ; \ $ we see from the graph (or by calculation with the product factors) that the minimal value is $ \ 6·(-6) \ = \ -36 \ \ . \ $ (We confirm this by completing the square for $$ (3x + 5)·(3x - 7) \ \ = \ \ 9x^2 \ - \ 6x \ - \ 35 \ \ = \ \ 9·\left(x \ - \ \frac13 \right)^2 \ - \ 35 \ - \ 9·\left(\frac13 \right)^2 $$ $$ = \ \ 9·\left(x \ - \ \frac13 \right)^2 \ - \ 36 \ \ . $$
[I had never learned of a geometrical method either, but it isn't difficult to figure one out. I'd hardly claim this to be original: I would imagine someone within the century after Descartes introduced analytic geometry (in 1637) noticed something along these lines.]