This note covers how the graph of a function changes when you translate, reflect or stretch it, including combinations of these, and how to use the logistic model for growth that levels off. Transformations let you build new models from familiar ones, and the logistic function is the standard model for populations and spreading effects that have a natural limit.
Key ideas
Transformations of \(y = f(x)\). Changes made outside the function act on the \(y\)-values. Changes made inside the brackets act on the \(x\)-values, and they work in the opposite way to what you might expect.
- Vertical translation, \(y = f(x) + k\): moves the graph up by \(k\) (down if \(k\) is negative).
- Horizontal translation, \(y = f(x - h)\): moves the graph right by \(h\) (left if \(h\) is negative). So \(f(x + 3)\) moves the graph 3 units left.
- Vertical stretch, \(y = af(x)\): multiplies every \(y\)-value by \(a\), with scale factor \(a\) and the \(x\)-axis fixed. A negative \(a\) also reflects the graph in the \(x\)-axis.
- Horizontal stretch, \(y = f(bx)\): divides every \(x\)-value by \(b\), so the scale factor is \(\dfrac{1}{b}\) and the \(y\)-axis is fixed. For example, \(f(2x)\) squashes the graph to half its width, and \(f\left(\dfrac{x}{2}\right)\) stretches it to double the width.
- Reflection in the \(x\)-axis, \(y = -f(x)\): changes the sign of every \(y\)-value.
- Reflection in the \(y\)-axis, \(y = f(-x)\): changes the sign of every \(x\)-value.
Image of a point. If \((p, q)\) is on \(y = f(x)\), then \((p + h, q + k)\) is on \(y = f(x - h) + k\). For a general transformation \(y = af(b(x - h)) + k\), the point \((p, q)\) goes to \(\left(\dfrac{p}{b} + h,\; aq + k\right)\).
Order of composite transformations. In \(y = af(b(x - h)) + k\), the \(x\)-changes happen in this order: stretch by factor \(\dfrac{1}{b}\), then translate \(h\) right. For the \(y\)-values: stretch by factor \(a\), then translate \(k\) up. The order matters. Stretching and then translating gives a different graph from translating and then stretching. To find an equation, apply each step to the equation one at a time, replacing \(x\) or the whole function as required.
Logistic models. Many populations grow quickly at first, then slow down as they approach a limit set by space or resources. The logistic function describes this:
\[ f(t) = \dfrac{L}{1 + Ce^{-kt}}, \quad L, C, k \gt 0. \]
- \(L\) is the carrying capacity. It is the limiting value that \(f(t)\) approaches as \(t \to \infty\), so \(y = L\) is a horizontal asymptote. The graph also has the asymptote \(y = 0\) as \(t \to -\infty\).
- The starting value is \(f(0) = \dfrac{L}{1 + C}\).
- \(k\) controls how fast the curve rises. Larger \(k\) means a steeper S-shape.
- The graph is S-shaped. It rises most steeply when \(f(t) = \dfrac{L}{2}\), which happens at \(t = \dfrac{\ln C}{k}\). This is the point of fastest growth, and after it the growth rate decreases.
To solve problems, set \(f(t)\) equal to a given value and solve with the GDC, or rearrange using \(\ln\). Use the GDC graph to find the maximum rate of growth (the steepest point). Logistic models are also fitted to data by logistic regression on the GDC.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(f(x) + k\) | Translation \(k\) units up. |
| \(f(x - h)\) | Translation \(h\) units to the right. |
| \(af(x)\) | Vertical stretch, scale factor \(a\). If \(a \lt 0\), also reflect in the \(x\)-axis. |
| \(f(bx)\) | Horizontal stretch, scale factor \(\dfrac{1}{b}\). |
| \(-f(x)\), \(f(-x)\) | Reflection in the \(x\)-axis, and in the \(y\)-axis. |
| \((p, q) \mapsto \left(\dfrac{p}{b} + h, aq + k\right)\) | Image of a point under \(y = af(b(x - h)) + k\). |
| \(f(t) = \dfrac{L}{1 + Ce^{-kt}}\) | Logistic model. \(L\) is the carrying capacity. |
| \(f(0) = \dfrac{L}{1 + C}\) | Initial value of a logistic model. |
| \(t = \dfrac{\ln C}{k}\) | Time of fastest growth, when \(f = \dfrac{L}{2}\). |
Worked examples
Example 1 The point \((2, 3)\) lies on the graph of \(y = f(x)\). Find the image of this point on each graph: (a) \(y = f(x + 3) - 4\); (b) \(y = -2f(x)\); (c) \(y = f\left(\dfrac{x}{2}\right)\); (d) \(y = 3f(2(x - 1)) + 1\).
(a) \(f(x + 3)\) moves the graph 3 left, and \(-4\) moves it 4 down. The point goes to \((2 - 3, 3 - 4) = (-1, -1)\).
(b) The \(y\)-value is multiplied by \(-2\), and the \(x\)-value is unchanged. The point goes to \((2, -6)\).
(c) The horizontal scale factor is \(\dfrac{1}{1/2} = 2\), so the \(x\)-value doubles. The point goes to \((4, 3)\).
(d) Here \(a = 3\), \(b = 2\), \(h = 1\), \(k = 1\). The new \(x\)-value is \(\dfrac{2}{2} + 1 = 2\). The new \(y\)-value is \(3 \times 3 + 1 = 10\).
Check: at \(x = 2\), \(y = 3f(2(2 - 1)) + 1 = 3f(2) + 1 = 3(3) + 1 = 10\).
Answer: (a) \((-1, -1)\); (b) \((2, -6)\); (c) \((4, 3)\); (d) \((2, 10)\).
Example 2 The graph of \(y = \ln x\) is reflected in the \(y\)-axis, and then translated 4 units to the right. Find the equation of the new graph, its domain, and its vertical asymptote.
Reflection in the \(y\)-axis replaces \(x\) by \(-x\): \(y = \ln(-x)\).
A translation 4 units right replaces \(x\) by \(x - 4\) in this new equation: \(y = \ln(-(x - 4)) = \ln(4 - x)\).
The log needs a positive input, so \(4 - x \gt 0\) and \(x \lt 4\).
The original graph has the vertical asymptote \(x = 0\). After reflection it is still \(x = 0\), and after moving 4 right it is \(x = 4\). This agrees with the domain.
Answer: \(y = \ln(4 - x)\), domain \(x \lt 4\), asymptote \(x = 4\).
Example 3 The number of fish in a lake after \(t\) years is modelled by \(P(t) = \dfrac{2400}{1 + 11e^{-0.5t}}\). Find (a) the initial number, (b) the number after 6 years, (c) the limiting number, (d) when the population grows fastest.
(a) \(P(0) = \dfrac{2400}{1 + 11} = 200\).
(b) \(P(6) = \dfrac{2400}{1 + 11e^{-3}} = \dfrac{2400}{1 + 11(0.049787)} = \dfrac{2400}{1.54766} = 1550.7\ldots\) fish, so about 1550.
(c) As \(t \to \infty\), \(e^{-0.5t} \to 0\), so \(P \to 2400\). The carrying capacity is 2400.
(d) The fastest growth occurs when \(P = \dfrac{2400}{2} = 1200\). Solve \(1 + 11e^{-0.5t} = 2\), so \(e^{-0.5t} = \dfrac{1}{11}\) and \(t = \dfrac{\ln 11}{0.5} = 4.796\ldots\) years.
Answer: (a) 200; (b) about 1550; (c) 2400; (d) at \(t = 4.80\) years.
Common errors
- Moving \(f(x + 3)\) three units to the right. It moves three units to the left.
- Using the scale factor \(b\) for a horizontal stretch of \(f(bx)\) instead of \(\dfrac{1}{b}\).
- Applying the steps of a composite transformation in the wrong order, for example translating before stretching when the equation needs the reverse.
- Giving \(L\) as the initial value of a logistic model. The initial value is \(\dfrac{L}{1 + C}\), and \(L\) is the limit.
- Saying the logistic model grows fastest at the start. Its fastest growth is when the value is half of \(L\).
Exam tip: To check a transformation, track one or two key points of the original graph, such as an intercept or a turning point, and make sure their images agree with your equation.
Check yourself
- The point \((-1, 4)\) is on \(y = f(x)\). Find its image on \(y = f(x - 2) + 1\).
Answer
Move 2 right and 1 up: \((-1 + 2, 4 + 1) = (1, 5)\). - The point \((3, 5)\) is on \(y = f(x)\). Find its images on \(y = -f(x)\) and on \(y = f(-x)\).
Answer
For \(y = -f(x)\), the \(y\)-value changes sign: \((3, -5)\). For \(y = f(-x)\), the \(x\)-value changes sign: \((-3, 5)\). - Find the equation of the graph of \(y = x^2\) after a vertical stretch with scale factor 2 and then a translation of 3 units up. How does it differ if the translation is done first?
Answer
Stretch: \(y = 2x^2\). Then translate: \(y = 2x^2 + 3\). If the translation comes first, \(y = x^2 + 3\), and then the stretch gives \(y = 2(x^2 + 3) = 2x^2 + 6\), which is a different graph. - A logistic model is \(P(t) = \dfrac{500}{1 + 4e^{-0.3t}}\). Find the initial value, the limiting value, and the time of fastest growth.
Answer
\(P(0) = \dfrac{500}{5} = 100\). The limit is 500. The fastest growth is at \(P = 250\), which gives \(1 + 4e^{-0.3t} = 2\), so \(e^{-0.3t} = \dfrac{1}{4}\) and \(t = \dfrac{\ln 4}{0.3} = 4.62\).