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Transformations of Functions · Paper 1 — Free Maths AI HL Practice Questions

1FoundationSAQ-STranslation, reflection, stretching and compression5 marksPaper 1~8 min
The graph of f(x)=x2f(x) = x^2 is transformed to the graph of g(x)=−(x+3)2+5g(x) = -(x+3)^2 + 5.
(a)
State the single transformation that the negative sign in g(x)g(x) represents. [1 mark]
(b)
Determine the coordinates of the vertex of g(x)g(x). [2 marks]
(c)
The graph of g(x)g(x) is translated by (4−3)\begin{pmatrix} 4 \\ -3 \end{pmatrix}. Write down the equation of the resulting function h(x)h(x) in the form y=a(x−h)2+ky = a(x - h)^2 + k, and hence state the range of h(x)h(x). [2 marks]
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Solutions

2MasterySAQ-STranslation, reflection, stretching and compression5 marksPaper 1~8 min
The function ff is defined by f(x)=x2−4x+7f(x) = x^2 - 4x + 7, for x∈Rx \in \mathbb{R}. The graph of ff is translated by the vector (3−2)\begin{pmatrix} 3 \\ -2 \end{pmatrix} to give the graph of a function gg.
(a)
Show that g(x)=x2−10x+26g(x) = x^2 - 10x + 26. The graph of gg is then reflected in the xx-axis to give the graph of a function hh. [3 marks]
(b)
Write down the coordinates of the vertex of the graph of hh. [1 mark]
(c)
The graph of hh is translated by the vector (0k)\begin{pmatrix} 0 \\ k \end{pmatrix} so that the resulting graph has exactly one xx-intercept. Determine the value of kk. [1 mark]
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Solutions

3ChallengeSAQ-LTranslation, reflection, stretching and compression8 marksPaper 1~12 min
A function ff is defined by f(x)=ln⁡(x2−4x+5)f(x) = \ln(x^2 - 4x + 5), for x∈Rx \in \mathbb{R}.
(a)
Show that the minimum point of ff is at (2, 0)(2,\, 0). [2 marks]
(b)
The function gg is defined by g(x)=−f(x+2)+3g(x) = -f(x+2) + 3. Determine the coordinates of the minimum point of gg. [2 marks]
(c)
The graph of ff is transformed as follows: - reflected in the yy-axis, - translated by the vector (1−2)\begin{pmatrix} 1 \\ -2 \end{pmatrix}, - stretched vertically by scale factor 44. Determine the equation of the resulting function h(x)h(x), writing your answer in the form h(x)=aln⁡(x2+bx+c)+dh(x) = a\ln(x^2 + bx + c) + d, where a, b, c, d∈Ra,\, b,\, c,\, d \in \mathbb{R}. [4 marks]
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Solutions

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