Recover f from a transformed input graph before reciprocating

Zeros 6,10,18 become reciprocal vertical asymptotes at those x-values. The negative minimum (8,−2) becomes a maximum (8,−1/2); the positive maximum (14,2) becomes a minimum (14,1/2). The left horizontal asymptote y=−4 becomes y=−1/4; the unbounded right tail gives reciprocal horizontal asymptote y=0.

G is undefined at its original vertical asymptote x=0. Its reciprocal tends to 0 there, but(0,0) is an open limiting point, not the filled intercept in the printed sketch. There are no actual axial intercepts of 1/G.

The source shows how f produces G: $f(x)\to f(3+x)$ by a translation 3 left; then $f(3+x/2)$ by horizontal scale 2; then $f(3-x/2)$ by reflection in the y-axis.

Starting from G, reflect in the y-axis to get $f(3+x/2)$; scale horizontally by 1/2 to get $f(3+x)$; translate 3 right to get f. Equivalently a source point (X, Y) of G corresponds to $(3-X/2, Y)$ on f.

The stationary points become (−1,−2), (−4,2). Zeros 6,10,18 become 0,−2,−6. The pole x=0 maps to x=3; the left G horizontal asymptote becomes f’s right horizontal asymptote y=−4. Thus f has maximum (−4,2), minimum (−1,−2), x-intercepts −6,−2,0, and y-intercept 0.