Porisma
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Graph transformations

Practice section 9 of 9

Questions 49–53 in this collection

Practice questions

Review practice · 1 of 5

Compare modulus, negative scaling and reciprocal graphs

The given graph y=f(x)y=f(x) has asymptotes x=−1x=-1, y=−2y=-2, turning point A=(1,3)A=(1,3), and zeros O=(0,0)O=(0,0), B=(−2,0)B=(-2,0), C=(4,0)C=(4,0).

(a) State the turning point of y=1+13f(x)y=1+\frac13f(x). [1]

(b) Sketch y=∣f(x)∣y=|f(x)|, labelling asymptotes and the points corresponding to A,B,C. [2]

(c) Sketch y=−f(2x)y=-f(2x) with those labels. [3]

(d) Sketch y=1/f(x)y=1/f(x), labelling asymptotes, turning points and axis crossings. [3]

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Qualitative reconstruction checked against the actual given graph in lecture Student PDF 38 / printed 110 and Teacher PDF 39 / printed 111: A(1,3), B(−2,0), O(0,0), C(4,0), asymptotes x=−1,y=−2, with the source branch directions. The standalone solution PDF omits that original figure; no unique formula is inferred.
Qualitative reconstruction checked against the actual given graph in lecture Student PDF 38 / printed 110 and Teacher PDF 39 / printed 111: A(1,3), B(−2,0), O(0,0), C(4,0), asymptotes x=−1,y=−2, with the source branch directions. The standalone solution PDF omits that original figure; no unique formula is inferred.
Qualitative reconstruction checked against the actual given graph in lecture Student PDF 38 / printed 110 and Teacher PDF 39 / printed 111: A(1,3), B(−2,0), O(0,0), C(4,0), asymptotes x=−1,y=−2, with the source branch directions. The standalone solution PDF omits that original figure; no unique formula is inferred.

[2022/YIJC/Promo/6]

CJC · Review problems · Question 5

Solution

Review practice · 3 of 5

Reverse a translation, reflection and vertical scale

A translates 2 units in the negative x-direction; B reflects in the x-axis; C scales parallel to y by 2 with the x-axis invariant. Applied in order A,B,C, the resulting equation is y=−2(x+3)/(x+2)y=-2(x+3)/(x+2). Find the equation before the transformations. [3]

[2012/RI/Promo/2]

CJC · Review problems · Question 7

Solution

Review practice · 4 of 5

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|); [3]

(ii) y=1/f(x)y=1/f(x). [3]

(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. [2]

(ii) Give the final equation resulting from that incorrect sequence. [1]

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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

Solution

Review practice · 5 of 5

Recover a signed graph from its two modulus graphs

The supplied graphs are y=f(∣x∣)y=f(|x|) and y=∣f(x)∣y=|f(x)|. Both have maximum (4, a) and vertical tangents x=0, x=6. The first is even, with maxima (−4, a), (4, a), zeros (−6,0), (0,0), (6,0), and slant asymptotes y=x+2 on the left and y=−x+2 on the right. The second is nonnegative, with zeros (0,0), (6,0), and slant asymptotes y=−x+2 on the left and y=x−2 on the right.

Given f itself has only one oblique asymptote, sketch y=f(x)y=f(x), stating asymptotes, turning points and axial intercepts wherever possible. [2]

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Two independent reconstructions of the supplied f(|x|) and |f(x)| graphs. Maximum height a remains symbolic and is not inferred numerically from the drawing. The first is even, with maxima (±4,a), zeros 0,±6, slant asymptotes y=x+2 left,y=-x+2 right. The second is nonnegative, with maximum (4,a), zeros 0,6, slant asymptotes y=-x+2 left,y=x-2 right. Vertical tangents are not asymptotes.
Two independent reconstructions of the supplied f(|x|) and |f(x)| graphs. Maximum height a remains symbolic and is not inferred numerically from the drawing. The first is even, with maxima (±4,a), zeros 0,±6, slant asymptotes y=x+2 left,y=-x+2 right. The second is nonnegative, with maximum (4,a), zeros 0,6, slant asymptotes y=-x+2 left,y=x-2 right. Vertical tangents are not asymptotes.
Two independent reconstructions of the supplied f(|x|) and |f(x)| graphs. Maximum height a remains symbolic and is not inferred numerically from the drawing. The first is even, with maxima (±4,a), zeros 0,±6, slant asymptotes y=x+2 left,y=-x+2 right. The second is nonnegative, with maximum (4,a), zeros 0,6, slant asymptotes y=-x+2 left,y=x-2 right. Vertical tangents are not asymptotes.

[2020/RI/Promo/8(ii)]

CJC · Review problems · Question 9

Solution

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