Porisma

Worked solution

Distinguish an absolute input from the order of affine transformations

(a) The supplied graph y=f(x)y=f(x) has asymptotes y=0, y=2, x=−1, x=1, maximum (3,3), minimum at O and the right zero (2,0) shown. Sketch:

(i) y=f(∣1+x∣)y=f(|1+x|); marks

(ii) y=1/f(x)y=1/f(x). marks

(b) Alfred and Betty are to transform y=f(x)y=f(x) to y=5f(2x+1)−3y=5f(2x+1)-3, with original f(x)=x2f(x)=x^2 and target y=5(2x+1)2−3y=5(2x+1)^2-3.

Alfred’s steps, in order: 1.Scale parallel to y by 5; 2.Translate down 3; 3. Scale parallel to x by 1/2; 4. Translate along x by −1.

Betty’s steps, in order: 1.Translate along x by −1; 2. Scale parallel to x by 1/2; 3. Scale parallel to y by 5; 4. Translate down 3.

One person made an error in one step.

(i) Identify the person and incorrect step. marks

(ii) Give the final equation resulting from that incorrect sequence. mark

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Source graph with asymptotes x=-1,x=1, left horizontal y=0 and right horizontal y=2, minimum O(0,0), maximum (3,3) and right zero (2,0). Independent qualitative interpolation, not an exact supplied formula.
Source graph with asymptotes x=-1,x=1, left horizontal y=0 and right horizontal y=2, minimum O(0,0), maximum (3,3) and right zero (2,0). Independent qualitative interpolation, not an exact supplied formula.
Source graph with asymptotes x=-1,x=1, left horizontal y=0 and right horizontal y=2, minimum O(0,0), maximum (3,3) and right zero (2,0). Independent qualitative interpolation, not an exact supplied formula.

[2012/SAJC/Promo/10]

CJC · Review problems · Question 8

Answer

(a)(i) Symmetry x=−1, poles x=−2,0, maxima (−4,3), (2,3), minimum (−1,0), other zeros −3,1, y=2.

(ii) Poles 0,2, right asymptote y=1/2, minimum (3,1/3), open points (±1,0). (b) Alfred Step 4; incorrect result y=5(2x+2)2−3y=5(2x+2)^2-3.

(a)(i) Start with the shifted input, then take its absolute value

Let u=|1+x|≥0; retain the original f branch for nonnegative inputs and reflect it about x=−1. The new graph is symmetric about x=−1. The source pole u=1 gives x=−2,0; the horizontal asymptote is y=2 at both ends. The source maximum u=3 gives (−4,3), (2,3). The original minimum u=0 gives (−1,0). The source zero u=2 gives (−3,0), (1,0).

How did this part go?

(a)(ii) Reciprocate without filling excluded inputs

Zeros of f at 0,2 give vertical asymptotes x=0,2. The maximum (3,3) becomes the minimum (3,1/3). The right horizontal asymptote is y=1/2; as x→−∞, f→0+, so 1/f grows without bound. Original pole inputs x=−1,1 remain excluded, with open limiting points (−1,0), (1,0). They are not actual x-intercepts. There are no axis intercepts, since f(0)=0.

How did this part go?

(b)(i) Identify the incorrect ordered step

Alfred’s Step 4 is incorrect: after horizontal scaling by 1/2, the required horizontal translation is −1/2, not −1. Betty’s sequence correctly gives f(x)→f(x+1)→f(2x+1)→5f(2x+1)→5f(2x+1)−3f(x)\to f(x+1)\to f(2x+1)\to5f(2x+1)\to5f(2x+1)-3.

How did this part go?

(b)(ii) Follow Alfred’s actual operations

For f(x)=x2f(x)=x^2, Alfred obtains 5x25x^2, then 5x2−35x^2-3, then 5(2x)2−35(2x)^2-3, then 5[2(x+1)]2−3=5(2x+2)2−35[2(x+1)]^2-3=5(2x+2)^2-3. This is the printed incorrect final graph.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

First panel f(|1+x|) is symmetric about x=-1, with poles x=-2,0, minimum (-1,0), maxima (-4,3),(2,3), other zeros (-3,0),(1,0), and horizontal y=2. Second reciprocal has vertical asymptotes x=0,2 and horizontal y=1/2 on the right, minimum (3,1/3), and open zero-limit points (-1,0),(1,0) rather than true x intercepts.
First panel f(|1+x|) is symmetric about x=-1, with poles x=-2,0, minimum (-1,0), maxima (-4,3),(2,3), other zeros (-3,0),(1,0), and horizontal y=2. Second reciprocal has vertical asymptotes x=0,2 and horizontal y=1/2 on the right, minimum (3,1/3), and open zero-limit points (-1,0),(1,0) rather than true x intercepts.
First panel f(|1+x|) is symmetric about x=-1, with poles x=-2,0, minimum (-1,0), maxima (-4,3),(2,3), other zeros (-3,0),(1,0), and horizontal y=2. Second reciprocal has vertical asymptotes x=0,2 and horizontal y=1/2 on the right, minimum (3,1/3), and open zero-limit points (-1,0),(1,0) rather than true x intercepts.

Common mistakes

f(|1+x|) is not f(1+|x|). The source reciprocal sketch’s unmarked zero-limit locations at ±1 must remain excluded.

Solution notes

All (a)/(b) asks, both students’ complete four-step sequences, source graph features and source conclusion retained. Input absolute-value bars restored by comparing the PDF image; extraction omits them. Reciprocal domain holes at ±1 are explicit. Both source solution sketches reconstructed.

Read the hint