Porisma

Worked solution

Transform a piecewise curve and crop to the requested interval

(i) State transformations taking y=x2y=x^2 to y=14(x−3)2y=\frac14(x-3)^2.

Define

f(x)={1,0≤x≤1,14(x−3)2,1<x≤3,0,otherwise.f(x)=\begin{cases}1,&0\le x\le1,\\ \frac14(x-3)^2,&1<x\le3,\\0,&\text{otherwise}.\end{cases}

(ii) Sketch f for −1≤x≤4.

(iii) Separately sketch y=1+f(x/2)y=1+f(x/2) for −1≤x≤4.

[2015/I/5]

CJC · Discussion · Question 9

Answer

(i) Translate 3 units right, then apply a vertical scale factor of 1/4.

(iii) 1 on [−1,0), 2 on [0,2], and 1+(x−6)2/161+(x-6)^2/16 on (2,4], ending at (4,5/4), with the jump at 0.

(i) Translate, then scale vertically

Replace x by x−3: y=(x−3)2y=(x-3)^2, a translation 3 units right. Replace y by 4y: 4y=(x−3)24y=(x-3)^2, hence y=14(x−3)2y=\frac14(x-3)^2, a vertical scale factor of 1/4.

How did this part go?

(ii) Retain every piece and endpoint

On −1≤x<0, f=0. At x=0, f jumps to 1, so (0,0) is open and (0,1) is filled. On 0≤x≤1, f=1; on 1<x≤3, draw the quadratic arc (x−3)2/4(x-3)^2/4 from height 1 to 0; on 3<x≤4, f=0. The requested endpoints (−1,0) and (4,0) are included.

How did this part go?

(iii) First source stage: horizontal scale 2

The graph f(x/2)f(x/2) has the same ordinates at twice the original x-values. In the source’s illustrated expanded interval [−2,8], it is 0 for x<0, 1 on [0,2], then (x−6)2/16(x-6)^2/16 on (2,6], and 0 after 6. The source retains an open point (0,0) and a filled point (0,1).

How did this part go?

(iii) Second source stage: translate up 1

Adding 1 raises the expanded graph: the baseline is 1, the plateau is 2, and the curved part is 1+(x−6)2/161+(x-6)^2/16. The jump becomes an open point (0,1) and a filled point (0,2). The expanded sketch is an intermediate window, not the requested final domain.

How did this part go?

(iii) Crop last and calculate the endpoint

Now restrict to −1≤x≤4. The final graph is 1 on [−1,0), 2 on [0,2], and 1+(x−6)2/161+(x-6)^2/16 on (2,4]. At x=4, 1+f(2)=1+14(2−3)2=5/41+f(2)=1+\frac14(2-3)^2=5/4. Include (−1,1), (0,2) and (4,5/4); exclude (0,1). The source note stresses that the last segment is a curve, not a straight line.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Exact source piecewise f: 1 on [0,1],(x-3)^2/4 on (1,3],0 otherwise. Four panels retain the original [-1,4] window, stretched [-2,8] window, raised [-2,8] window, and final requested [-1,4] crop. Open and filled points distinguish each jump at 0. Final curve ends at (4,5/4), not a straight segment; all requested endpoint values retained.
Exact source piecewise f: 1 on [0,1],(x-3)^2/4 on (1,3],0 otherwise. Four panels retain the original [-1,4] window, stretched [-2,8] window, raised [-2,8] window, and final requested [-1,4] crop. Open and filled points distinguish each jump at 0. Final curve ends at (4,5/4), not a straight segment; all requested endpoint values retained.
Exact source piecewise f: 1 on [0,1],(x-3)^2/4 on (1,3],0 otherwise. Four panels retain the original [-1,4] window, stretched [-2,8] window, raised [-2,8] window, and final requested [-1,4] crop. Open and filled points distinguish each jump at 0. Final curve ends at (4,5/4), not a straight segment; all requested endpoint values retained.

Common mistakes

Do not leave the final graph on the expanded source window[−2,8], or replace its quadratic arc with a straight line.

Solution notes

All 3 asks, equation substitutions, all 4 source graph panels, open/filled endpoints and final curved-segment annotation retained. Exact piecewise plots use the supplied formula. No marks are printed on this sheet.

Read the hint