Unit 12 Review Exercises

Graph Transformations

These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. AHL marks questions for HL students only.

1 Paper 1 The point \((-2, 6)\) lies on the graph of \(y = f(x)\). Write down its image on the graph of

(a) \(y = f(x - 3) + 1\)

(b) \(y = 2f(x)\)

(c) \(y = f(-x)\)

(a) \((-2 + 3, 6 + 1) = (1, 7)\).

(b) \((-2, 12)\).

(c) \((2, 6)\).

See 12.1.

2 Paper 1 Describe a sequence of two transformations that maps the graph of \(y = x^2\) onto the graph of \(y = 4 - (x - 1)^2\). Write down the vertex of the final graph.

\(4 - (x - 1)^2 = -(x - 1)^2 + 4\). A reflection in the \(x\)-axis (\(y = -x^2\)), followed by a translation by the vector \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\).

The vertex \((0, 0)\) moves to \((1, 4)\), and it is now a maximum point.

See 12.1 and 12.2.

3 Paper 1 The graph of \(y = x^3 + 2x - 1\) is translated by the vector \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\). Find the equation of the new graph in the form \(y = x^3 + px^2 + qx + r\).

Replace \(x\) by \(x - 1\) and subtract \(3\):

\[ y = (x - 1)^3 + 2(x - 1) - 1 - 3 = x^3 - 3x^2 + 3x - 1 + 2x - 2 - 4 = x^3 - 3x^2 + 5x - 7 . \]

See 12.1.

4 Paper 1 Let \(f(x) = x^2 - 6x + 5\) and \(g(x) = f(2x)\).

(a) Describe the transformation that maps the graph of \(f\) onto the graph of \(g\).

(b) Find the \(x\)-intercepts, the vertex and the \(y\)-intercept of the graph of \(g\).

(a) A stretch parallel to the \(x\)-axis with scale factor \(\frac{1}{2}\).

(b) \(f(x) = (x - 1)(x - 5) = (x - 3)^2 - 4\): roots \(1\), \(5\); vertex \((3, -4)\); \(y\)-intercept \((0, 5)\).

Halve the \(x\)-coordinates: roots \(\frac{1}{2}\) and \(\frac{5}{2}\); vertex \(\left(\frac{3}{2}, -4\right)\); \(y\)-intercept \((0, 5)\) (invariant).

See 12.1.

5 Paper 1 The graph of \(y = f(x)\) has a maximum point \((-2, 6)\) and the horizontal asymptote \(y = 4\). Find the maximum point and the horizontal asymptote of the graph of \(y = \frac{1}{2}f(x) - 3\).

\((x, y) \longmapsto \left(x, \frac{1}{2}y - 3\right)\).

Maximum: \(\left(-2, \frac{1}{2} \times 6 - 3\right) = (-2, 0)\). Asymptote: \(y = \frac{1}{2} \times 4 - 3 = -1\).

See 12.2.

6 Paper 1 AHL Describe how the graph of \(y = \ln(3x + 6)\) can be obtained from the graph of \(y = \ln x\). State the vertical asymptote and the axis intercepts of \(y = \ln(3x + 6)\).

\(\ln(3x + 6) = \ln\big(3(x + 2)\big)\): a stretch parallel to the \(x\)-axis with scale factor \(\frac{1}{3}\), then a translation of \(2\) units in the negative \(x\)-direction.

Asymptote: \(x = \frac{0 - 6}{3} = -2\). \(\;x\)-intercept: \(3x + 6 = 1 \Rightarrow x = -\frac{5}{3}\). \(\;y\)-intercept: \((0, \ln 6)\).

See 12.3.

7 Paper 1 AHL Sketch the graph of \(y = |x^2 - 4x + 3|\). Hence solve \(|x^2 - 4x + 3| = 1\).

\(x^2 - 4x + 3 = (x - 1)(x - 3)\) has vertex \((2, -1)\). Reflect the part between \(1\) and \(3\): the graph touches the axis at \((1, 0)\) and \((3, 0)\) with sharp points, has a local maximum \((2, 1)\), and \(y\)-intercept \((0, 3)\).

The line \(y = 1\) passes through the maximum \((2, 1)\) and cuts each outer branch once: three solutions.

\(x^2 - 4x + 3 = 1 \Rightarrow x^2 - 4x + 2 = 0 \Rightarrow x = 2 \pm \sqrt{2}\). \(\quad x^2 - 4x + 3 = -1 \Rightarrow (x - 2)^2 = 0 \Rightarrow x = 2\).

Solutions: \(x = 2 - \sqrt{2},\; 2,\; 2 + \sqrt{2}\).

See 12.4.

8 Paper 1 AHL Let \(f(x) = x^2 - 2x - 3\). Sketch the graph of \(y = f(|x|)\), giving the axis intercepts and the minimum points.

For \(x \ge 0\): \(f(x) = (x + 1)(x - 3)\) has the root \(3\), the minimum \((1, -4)\) and the \(y\)-intercept \((0, -3)\). Reflect this part in the \(y\)-axis.

Intercepts \((-3, 0)\), \((3, 0)\) and \((0, -3)\). Minimum points \((-1, -4)\) and \((1, -4)\). (The point \((0, -3)\) is a local maximum.)

See 12.4.

9 Paper 1 AHL Let \(f(x) = (x - 1)(x - 5)\). Sketch the graph of \(y = \dfrac{1}{f(x)}\). State the asymptotes, the turning point and the \(y\)-intercept.

Zeros \(1\) and \(5\) \(\Rightarrow\) vertical asymptotes \(x = 1\) and \(x = 5\). \(f(x) \to \infty\) as \(x \to \pm\infty\) \(\Rightarrow\) asymptote \(y = 0\).

The minimum of \(f\) is \((3, -4)\), so \(\frac{1}{f}\) has a maximum \(\left(3, -\frac{1}{4}\right)\). \(y\)-intercept: \(\frac{1}{f(0)} = \frac{1}{5}\), so \(\left(0, \frac{1}{5}\right)\).

The middle branch is below the axis; the outer branches are above.

See 12.5.

10 Paper 2 AHL Let \(f(x) = x^3 - 3x^2 + 4\). The graph of \(y = f(x)\) has a maximum point \((0, 4)\) and a minimum point \((2, 0)\).

(a) Show that \(f(x) = (x + 1)(x - 2)^2\).

(b) Sketch \(y = [f(x)]^2\), giving the turning points.

(c) Write down the equations of the vertical asymptotes of \(y = \dfrac{1}{f(x)}\) and the coordinates of its local minimum point.

(d) Use your GDC to solve \(\dfrac{1}{f(x)} = 1\).

(a) \((x + 1)(x^2 - 4x + 4) = x^3 - 4x^2 + 4x + x^2 - 4x + 4 = x^3 - 3x^2 + 4\) ✓.

(b) The graph touches the \(x\)-axis at \((-1, 0)\) and \((2, 0)\) (minimum points). The maximum \((0, 4)\) becomes \((0, 16)\).

(c) \(f(x) = 0\) at \(x = -1\) and \(x = 2\), so the asymptotes are \(x = -1\) and \(x = 2\). The maximum \((0, 4)\) of \(f\) gives the local minimum \(\left(0, \frac{1}{4}\right)\).

(d) \(\frac{1}{f(x)} = 1 \iff f(x) = 1 \iff x^3 - 3x^2 + 3 = 0\). GDC (Menu → Analyze Graph → Intersection, or nSolve): \(x = -0.879\), \(x = 1.35\), \(x = 2.53\) (3 s.f.).

See 12.5.