MathematicsGrade 9–11

Power functions y = a · xⁿ

Even or odd, positive or negative: y = a · xⁿ changes with every exponent, and only a few points always stay the same.

Loading simulation …

What is it about?

A power function has the form

f(x)=a⋅xn(a≠0,  n an integer)f(x) = a \cdot x^n \qquad (a \neq 0,\; n \text{ an integer})

The exponent nn decides the shape of the graph. The stretch factor aa stretches (∣a∣>1|a| > 1) or compresses (∣a∣<1|a| < 1) it vertically; for a<0a < 0 it is also reflected in the xx-axis. The point Q(1∣a)Q(1 \mid a) always lies on the graph because 1n=11^n = 1.

Depending on nn there are four classes (shown in different colours in the simulation):

Exponent Graph (for a=1a = 1) Symmetry Domain DD Range WW
nn even, n>0n > 0 parabola of order nn, both branches up about the yy-axis R\mathbb{R} R0+\mathbb{R}_0^+
nn odd, n>0n > 0 parabola of order nn (for n=1n = 1 a line through the origin), from bottom left to top right about the origin R\mathbb{R} R\mathbb{R}
nn even, n<0n < 0 hyperbola, both branches up about the yy-axis R∖{0}\mathbb{R} \setminus \{0\} R+\mathbb{R}^+
nn odd, n<0n < 0 hyperbola, branches in quadrants I and III about the origin R∖{0}\mathbb{R} \setminus \{0\} R∖{0}\mathbb{R} \setminus \{0\}
  • Symmetry: For even nn, f(−x)=f(x)f(-x) = f(x): the graph is symmetric about the yy-axis. For odd nn, f(−x)=−f(x)f(-x) = -f(x): the graph is symmetric about the origin.
  • Negative exponents: x−n=1xnx^{-n} = \frac{1}{x^n}. The function is not defined at x=0x = 0 (pole); the axes x=0x = 0 and y=0y = 0 are asymptotes.
  • Special cases: n=1n = 1 gives the line y=a⋅xy = a \cdot x through the origin. For n=0n = 0, x0=1x^0 = 1 (for x≠0x \neq 0), so f(x)=af(x) = a: a line parallel to the xx-axis. The expression 000^0 is not considered here – the simulation leaves 00 out of the domain (hollow point).
  • Common points: All graphs of y=xny = x^n pass through (1∣1)(1 \mid 1). For even nn they pass through (−1∣1)(-1 \mid 1), for odd nn through (−1∣−1)(-1 \mid -1), and for n>0n > 0 through the origin.
  • Between −1-1 and 11: the larger the (positive) exponent, the flatter the graph, e.g. 0.54=0.0625<0.52=0.250.5^4 = 0.0625 < 0.5^2 = 0.25. Outside this interval it is the other way round.

Try it

  1. Move the slider for nn slowly from −4-4 to 66, or click Run through all exponents. The graph morphs smoothly. For which nn does the class change? Watch the colour and the label at the top right of the key features.
  2. Click Show symmetry for n=3n = 3: the right part is rotated through 180°180° about the origin. Try the same for n=4n = 4. Load example
  3. Drag the point Q(1∣a)Q(1 \mid a) up and down. What happens to the range and the monotonicity for a<0a < 0? Load example
  4. Drag PP along the graph and compare the calculation in the picture with the table of values. At the positions in the table, the matching row is highlighted.
  5. Choose the family view and tap the individual exponents. The other graphs of the family are faint and labelled with their exponent. Which points always stay the same? Even exponents · Odd exponents
  6. Compare xx and x3x^3 and let the vertical line run with Compare from left to right. Where does x3x^3 lie above xx? Load example
  7. Set n=−1n = -1 and n=−2n = -2. How do the function values behave near 00 on the left and on the right? Load example

Tasks

Task 1: Checking points

Do the points A(−2∣−4)A(-2 \mid -4) and B(3∣13)B(3 \mid 13) lie on the graph of f(x)=0.5⋅x3f(x) = 0.5 \cdot x^3?

Show solution

f(−2)=0.5⋅(−2)3=0.5⋅(−8)=−4f(-2) = 0.5 \cdot (-2)^3 = 0.5 \cdot (-8) = -4, so AA lies on the graph. f(3)=0.5⋅27=13.5≠13f(3) = 0.5 \cdot 27 = 13.5 \neq 13, so BB does not. Show in the simulation

Task 2: Finding the stretch factor

The graph of f(x)=a⋅x4f(x) = a \cdot x^4 passes through the point (2∣8)(2 \mid 8). Find aa and state the range. Load task

Show solution

a⋅24=8⇒16a=8⇒a=0.5a \cdot 2^4 = 8 \Rightarrow 16a = 8 \Rightarrow a = 0.5. Because nn is even and a>0a > 0, the range is W=R0+W = \mathbb{R}_0^+. Show in the simulation

Task 3: Key features

Describe f(x)=−2⋅x−3f(x) = -2 \cdot x^{-3}: domain, range, symmetry, monotonicity, behaviour as x→±∞x \to \pm\infty and near 00. Compute f(1)f(1) and f(−1)f(-1).

Show solution

D=W=R∖{0}D = W = \mathbb{R} \setminus \{0\}, symmetric about the origin (nn odd). Because a<0a < 0, the hyperbola 1x3\frac{1}{x^3} is reflected in the xx-axis: the branches lie in quadrants II and IV, and ff is increasing for x<0x < 0 and for x>0x > 0. As x→±∞x \to \pm\infty, f(x)→0f(x) \to 0; to the left of 00, f(x)→+∞f(x) \to +\infty, to the right f(x)→−∞f(x) \to -\infty. Asymptotes x=0x = 0 and y=0y = 0. f(1)=−2f(1) = -2, f(−1)=−2⋅1−1=2f(-1) = -2 \cdot \frac{1}{-1} = 2. Show in the simulation

Task 4: Which is bigger?

For which xx is x4<x2x^4 < x^2? Give a numerical example.

Show solution

The graphs intersect at x=−1x = -1, 00 and 11. For 0<∣x∣<10 < |x| < 1, x4<x2x^4 < x^2, e.g. 0.54=0.0625<0.25=0.520.5^4 = 0.0625 < 0.25 = 0.5^2. For ∣x∣>1|x| > 1 it is the other way round, e.g. 24=16>42^4 = 16 > 4. Show in the simulation

Task 5: Common points

Which points do all graphs of y=xny = x^n with odd n>0n > 0 have in common? Which one is lost if n=−1n = -1 and n=−3n = -3 are also allowed?

Show solution

(−1∣−1)(-1 \mid -1), (0∣0)(0 \mid 0) and (1∣1)(1 \mid 1), because (−1)n=−1(-1)^n = -1, 0n=00^n = 0 and 1n=11^n = 1. For negative nn, 00 is not in the domain, so the origin is lost. Show in the simulation

Task 6: Recognising a graph

A graph of f(x)=xnf(x) = x^n lies entirely above the xx-axis, is symmetric about the yy-axis and has a pole at x=0x = 0. What do you know about nn? Give an example.

Show solution

nn is even and negative, e.g. n=−2n = -2: f(x)=1x2f(x) = \frac{1}{x^2} with W=R+W = \mathbb{R}^+. Show in the simulation

Ideas for the classroom

  • Use: About grades 9/10, power functions with natural exponents and extending powers, then polynomial functions. Negative exponents build on the hyperbola y=1xy = \frac{1}{x} from Year 8; please check the curriculum for whether and how far they are covered.
  • Discovery: The family view is a good way to collect conjectures (“What do they all have in common?”), which can then be justified with 1n1^n, (−1)n(-1)^n and 0n0^n.
  • Justifying symmetry: The animation shows what symmetry about an axis and about a point mean; the reasoning (−x)n=xn(-x)^n = x^n or (−x)n=−xn(-x)^n = -x^n can follow.
  • Common stumbling blocks:
    • “x4x^4 is always bigger than x2x^2.” – Only for ∣x∣>1|x| > 1; between −1-1 and 11 it is the other way round.
    • x−2x^{-2} is read as −x2-x^2. – In fact x−2=1x2>0x^{-2} = \frac{1}{x^2} > 0.
    • For n=0n = 0, 000^0 is simply set to 11. – The simulation deliberately leaves out 00; a short discussion is worthwhile here.
    • Monotonicity for odd negative nn: 1x\frac{1}{x} is decreasing on each part, but not on the whole domain (e.g. f(−1)=−1<f(1)=1f(-1) = -1 < f(1) = 1).
  • Task mode: With info=0 and Share → Hide results, the key features, the table of values and the calculation are replaced by “?”; the graph stays visible.