Porisma

Worked solution

Map labelled points and preserve reciprocal domains

The given graph is y=f(x)y=f(x), with maximum A=(2,2), B=(4,0), C=(0,1) and vertical asymptote x=5; its left tail approaches y=0. Sketch on separate diagrams:

(i) y=f(2x−1)y=f(2x-1); marks

(ii) y=f(x/2)−2y=f(x/2)-2; marks

(iii) y=1/f(x)y=1/f(x); marks

Show the corresponding A,B,C coordinates and all asymptotes where applicable.

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Source schematic y=f(x), A(2,2) maximum,C(0,1),B(4,0), vertical asymptote x=5 and left tail tending to y=0. Only the supplied branch x<5 is reconstructed. Interpolation is qualitative.
Source schematic y=f(x), A(2,2) maximum,C(0,1),B(4,0), vertical asymptote x=5 and left tail tending to y=0. Only the supplied branch x<5 is reconstructed. Interpolation is qualitative.
Source schematic y=f(x), A(2,2) maximum,C(0,1),B(4,0), vertical asymptote x=5 and left tail tending to y=0. Only the supplied branch x<5 is reconstructed. Interpolation is qualitative.

CJC · Review problems · Question 2

Answer

(i) A(3/2,2), B(5/2,0), C(1/2,1), x=3.

(ii) A(4,0), B(8,−2), C(0,−1), x=10, y=−2.

(iii) A(2,1/2), C(0,1), vertical asymptote x=4; (5,0) excluded.

(i) Follow the source two-step transformation

Translate the original one unit in the positive x-direction: y=f(x−1)y=f(x-1). Scale parallel to the x-axis by 1/2: y=f(2x−1)y=f(2x-1). The point map is (x,y)↦((x+1)/2,y)(x,y)\mapsto((x+1)/2, y), giving A=(3/2,2), B=(5/2,0), C=(1/2,1). The vertical asymptote moves from x=5 to x=3; the left limiting horizontal asymptote remains y=0.

How did this part go?

(ii) Stretch horizontally, then translate down

Scale parallel to x by 2 to obtain y=f(x/2)y=f(x/2), then translate 2 units down to y=f(x/2)−2y=f(x/2)-2. The point map (x,y)↦(2x,y−2)(x,y)\mapsto(2x,y-2) gives A=(4,0), B=(8,−2), C=(0,−1). The asymptotes are x=10 and y=−2.

How did this part go?

(iii) Reciprocate ordinates where defined

A becomes (2,1/2), a minimum, and C remains (0,1). B has f(4)=0, so x=4 is a reciprocal vertical asymptote, not a finite image point. On the shown branch, 1/f(x)→0−1/f(x)\to0^- as x→5−, but f is undefined at x=5. Therefore (5,0) is an open limiting point, not an x-intercept. The reciprocal has no x-intercepts; its y-intercept is (0,1). The shown reciprocal left branch grows without bound as x→−∞, so y=0 is not a horizontal asymptote of that branch.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Three source-derived panels. First A(3/2,2),B(5/2,0),C(1/2,1), pole x=3. Second A(4,0),B(8,-2),C(0,-1), asymptotes x=10,y=-2. Reciprocal A(2,1/2),C(0,1), pole x=4; open limiting point (5,0) replaces the source filled endpoint because f is undefined at its vertical asymptote. There are no actual x intercepts of the reciprocal.
Three source-derived panels. First A(3/2,2),B(5/2,0),C(1/2,1), pole x=3. Second A(4,0),B(8,-2),C(0,-1), asymptotes x=10,y=-2. Reciprocal A(2,1/2),C(0,1), pole x=4; open limiting point (5,0) replaces the source filled endpoint because f is undefined at its vertical asymptote. There are no actual x intercepts of the reciprocal.
Three source-derived panels. First A(3/2,2),B(5/2,0),C(1/2,1), pole x=3. Second A(4,0),B(8,-2),C(0,-1), asymptotes x=10,y=-2. Reciprocal A(2,1/2),C(0,1), pole x=4; open limiting point (5,0) replaces the source filled endpoint because f is undefined at its vertical asymptote. There are no actual x intercepts of the reciprocal.

Common mistakes

The printed reciprocal sketch fills(5,0), although f is undefined at its vertical asymptote x=5. A limiting value does not add a domain point.

Solution notes

All printed asks and both affine transformation chains retained. The source reciprocal endpoint is explicitly corrected to an open point. Only the branch supplied in the PDF is reconstructed; no unsupported branch beyond x=5 is invented. Original native text and original closed-dot diagram remain the source evidence.

Read the hint