2.01 Power functions

Functions with a form like $y=x^2$y=x2, $y=2x^2$y=2x2 and $y=3x^3$y=3x3, etc. are known as power functions.

Exploration

The following applet lets us see the effect of increasing the powers of $x$x  for the function $y=x^n$y=xn. 

• Note how for odd powers the graph moves in opposite directions at the extremities, and for even powers the graph moves off in the same direction at the extremities. 
• Also look carefully at the shape of the curves as they come close to the origin.
• What else do you notice about the curves?

For odd powers greater than $1$1, the curves rise toward the origin with a decreasing positive slope until, when the origin is reached, they become momentarily horizontal and then continue upward with an ever increasing slope. That "terracing" shape around the origin is known as a horizontal inflection.

For even powers, the curve changes direction at the origin. This is called a minimum turning point.

 

Concepts of power functions

Power functions have the general form $y=ax^n$y=axn where $n$n is any number. When $n$n is a positive integer, we can learn to sketch these functions by considering a few simple principles of powers.  

Whenever a non-zero number is raised to an even integer power, the result is always positive. So for example $\left(3\right)^4=+81$(3)4=+81 and also $\left(-3\right)^4=+81$(−3)4=+81.  This results in the graphs of all power functions that have even integral powers to have some similar properties and general shape.

Whenever a non-zero number is raised to an odd integer power, the result is positive when that number is also positive, and negative when that number is also negative. So $\left(3\right)^3=+27$(3)3=+27 and $\left(-3\right)^3=-27$(−3)3=−27.  This results in the graphs of all power functions that have odd integer powers to have some similar properties and general shape.

Whenever a number lies in the interval $-1−1<x$<$<$1$1, squaring it or cubing it (or indeed raising it to any integer power) reduces the absolute value of its size. So $\left(-0.5\right)^3=-0.125$(−0.5)3=−0.125 and $\left(0.9\right)^2=0.81$(0.9)2=0.81 . 

To demonstrate the effect of these principles on power graphs, we have graphed the functions given by $y=x$y=x, $y=x^2$y=x2, $y=x^3$y=x3 and $y=x^4$y=x4 between $x=-2$x=−2 and $x=2$x=2. 

 

Effect of the coefficient $a$a

The coefficient $a$a in the power function form $y=ax^n$y=axn scales the function values of $y=x^n$y=xn by a factor of $a$a. So for example, if $a=\frac{1}{2}$a=12​, then each value of the function $y=x^n$y=xn is halved, so the graph looks compressed. If $a=2$a=2, the function values are doubled and the graph looks vertically stretched instead. 

If $a$a happens to be negative, the effect on the graph is a reflection across the $x$x-axis. That is, function values that are negative become positive and function values that are positive become negative.

As an example compare the graphs of $y=x^2,y=2x^2,y=-\frac{1}{2}x^2$y=x2,y=2x2,y=−12​x2 and $y=-3x^2$y=−3x2 as shown here:

 

Practice questions

Question 1

Question 2

Question 3

Transformations of power functions 

A transformation of a curve might involve a distortion in the shape of the curve. It might involve a reflection of the curve in the $x$x-axis or $y$y-axis.  It might simply involve shifting the curve either horizontally or vertically. The important point, however, is that the essential character of the curve doesn't change.

 

 

Watch how the graph of the function $y=x^3$y=x3 progressively changes to the graph of $y=\frac{1}{2}\left(x-3\right)^3+5$y=12​(x−3)3+5.

From the graph of $y=x^3$y=x3 we:

• Halve each $y$y value to compress the curve.
• Shift the graph to the right by $3$3 units.
• Lift the curve up by $5$5 units.

 

Explor

Now its your turn. The following applet shows $y=a\left(x-h\right)^n+k$y=a(x−h)n+k and you can change all four constants $a$a, $h$h, $n$n and $k$k to see the effects. It's important that you experiment with different combinations of constants to really understand the way transformations work.

 

Practice questions

QUESTION 4

QUESTION 5

QUESTION 6

Applications of power functions

We encounter many laws of science expressible as power functions, and one famous one was discovered by Johannes Kepler (1571-1630).

Kepler's third law

Kepler's third law of planetary motion states that the time taken for a planet to orbit the Sun (known as the planets period) is related to its distance away from the Sun.

Specifically, if $p$p is the period of the planet and $d$d is the average distance away from the Sun, then Kepler's 3rd law states that $d^3\propto p^2$d3∝p2.

By taking cube roots, and introducing a constant of variation $k$k,  the law can be mathematically written as $d=k\times p^{\frac{2}{3}}$d=k×p23​ .

Finding k

We can find the constant of variation by considering the data for the planet we live on.

Assuming that the period of Earth is given by $p=365.25$p=365.25 Earth days, and that its distance from the Sun is given by  $d=1.496\times10^8$d=1.496×108 km, then we can write:

$d$d	$=$=	$k\times p^{\frac{2}{3}}$k×p23​
$1.496\times10^8$1.496×108	$=$=	$k\times365.25^{\frac{2}{3}}$k×365.2523​
$\therefore k$∴k	$=$=	$\frac{1.496\times10^8}{365.25^{\frac{2}{3}}}$1.496×108365.2523​​
	$=$=	$2.928\times10^7$2.928×107

So, based on the Earth's period and radial distance, we have developed a power model for other planets.

Kepler's 3rd law becomes:

 $d=\left(2.928\times10^7\right)\times p^{\frac{2}{3}}$d=(2.928×107)×p23​

where $d$d is the mean distance from the Sun in km and $p$p is the period of the orbit.

 

The Mars-Sun distance

We now can predict the distance Mars' is away from our Sun.  

Mars has a period of $687$687 Earth days and so using this number, our function predicts  $d=\left(2.928\times10^7\right)\times687^{\frac{2}{3}}=2.2797\times10^8$d=(2.928×107)×68723​=2.2797×108 km. 

From the internet, Mars has a mean distance from the Sun of $2.279\times10^8$2.279×108 km, so the prediction is quite close.

 

Alternative approach to the 3rd law

Since $d^3\propto p^2$d3∝p2, we could just as easily made the period $p$p the subject, so that for a different constant of variation, say $k_2$k2​,  we have $p=k_2\times d^{\frac{3}{2}}=k_2\times d\sqrt{d}$p=k2​×d32​=k2​×d√d. 

Using the same strategy above, this new constant would have the approximate value of $k_2=1.996\times10^{-10}$k2​=1.996×10−10, so that the rearranged law would given as:

$p=\left(1.996\times10^{-10}\right)\times d\sqrt{d}$p=(1.996×10−10)×d√d.

 

Period of Venus

Considering Venus this time, with a average distance from the Sun of $1.082\times10^8$1.082×108 km, we would estimate the orbital period as:

$p=\left(1.996\times10^{-10}\right)\times\left(1.082\times10^8\right)\sqrt{\left(1.082\times10^8\right)}$p=(1.996×10−10)×(1.082×108)√(1.082×108)

When simplified, we find that $p=224.6$p=224.6 days, which is close to the accepted value of $225$225 days. 

 

 

Practice questions

QUESTION 7

QUESTION 8