The Graphs of y = axⁿ

One family of graphs

Several familiar graphs share one form, $y=a{x}^n$y = axⁿ: a number a multiplied by a power of x. The parabola $y=x^2$y = x² has a = 1 and n = 2. The cubic $y=x^3$y = x³ has n = 3. The reciprocal graph $y=\frac{1}{x}$y = 1/x is $y=x^{-1}$y = x⁻¹, because a negative index means one divided by the power, so it has n = −1.

The power n sets the shape of the graph, and the number a stretches it or flips it. When two power terms are added, the graph mixes their two shapes.

The graph of $y=\frac{1}{x^2}$y = 1/x²

For n = −2 the equation is $y=x^{-2}$y = x⁻², which is $y=\frac{1}{x^2}$y = 1/x². Work out some values. At x = 1, $y=\frac{1}{1}=1$y = 1/1 = 1. At x = 2, $y=\frac{1}{4}$y = 1/4, and at x = 3, $y=\frac{1}{9}$y = 1/9. As x grows, $x^2$x² grows faster still, so $\frac{1}{x^2}$1/x² shrinks toward zero. It never reaches zero, because one divided by a positive number is never zero.

Near zero the opposite happens. At $x=\frac{1}{2}$x = 1/2, $x^2=\frac{1}{4}$x² = 1/4 and $y=1÷\frac{1}{4}=4$y = 1 ÷ 1/4 = 4. At x = 0.1, $x^2=0.01$x² = 0.01 and y = 100. The curve climbs steeply as x closes in on zero, and at x = 0 there is no value at all, because $\frac{1}{0}$1/0 is not a number. The curve runs up alongside the y-axis without touching it and along the x-axis without touching it. A line that a curve gets closer and closer to, without ever reaching it, is called an asymptote, so both axes are asymptotes of $y=\frac{1}{x^2}$y = 1/x².

Now try negative values of x. Squaring a negative number gives a positive number: ${\left(-2\right)}^2=4$(−2)² = 4, so at x = −2, $y=\frac{1}{4}$y = 1/4, exactly as at x = 2. Every point on the right has a partner at the same height on the left, and neither arm ever goes below the x-axis. That is the difference from $y=\frac{1}{x}$y = 1/x, whose left arm is below the axis because 1 divided by a negative number is negative.

$y=\frac{1}{x^2}$y = 1/x². The points (−1, 1) and (1, 1) are at the same height, and both arms stay above the x-axis.

Even powers and odd powers

Whether the power is even or odd decides the symmetry of the graph. For an even power, a negative x gives the same value as the positive one: ${\left(-2\right)}^2=2^2=4$(−2)² = 2² = 4. So the graphs of $y=x^2$y = x² and $y=\frac{1}{x^2}$y = 1/x² are symmetric in the y-axis: the left half is the mirror image of the right half.

For an odd power, a negative x gives the negative of the value: ${\left(-2\right)}^3=-8$(−2)³ = −8, while $2^3=8$2³ = 8. So the point (2, 8) on $y=x^3$y = x³ has a partner at (−2, −8). Turn the graph half a turn about the origin and it lands on itself. The graphs of y = x, $y=x^3$y = x³ and $y=\frac{1}{x}$y = 1/x all have this half-turn symmetry.

A negative a flips the graph

The number a multiplies every value of $x^n$xⁿ. In $y=2x^2$y = 2x², each point is twice as high as on $y=x^2$y = x², so the parabola is stretched upward. In $y=-x^2$y = −x², each value is multiplied by −1: at x = 2, $y=-\left(2^2\right)=-4$y = −(2²) = −4, where $y=x^2$y = x² has 4.

Every point of $y=x^2$y = x² moves to the same distance on the other side of the x-axis, so $y=-x^2$y = −x² is the reflection of $y=x^2$y = x² in the x-axis. It opens downward, with its highest point at the origin.

Be careful with the order of operations here. In $-x^2$−x² the power acts first and the minus sign after it, so at x = 3, $-x^2=-\left(3^2\right)=-9$−x² = −(3²) = −9. That is not the same as ${\left(-3\right)}^2$(−3)², which is 9.

The gold curve is $y=-x^2$y = −x², and the white curve is $y=x^2$y = x². The point (2, 4) reflects in the x-axis to (2, −4).

Adding two powers

A graph can be a sum of power terms, such as $y=x^3-2x$y = x³ − 2x. Which term matters more depends on the size of x. Compare the two terms at a few values.

Near zero, $x^3$x³ is tiny and −2x makes almost all of y. Far from zero, $x^3$x³ is so large that −2x hardly changes it.

Close to the origin, the cube of a small number is very much smaller than the number itself, so $y=x^3-2x$y = x³ − 2x behaves like the straight line y = −2x and runs downhill through the origin. Far from the origin, $x^3$x³ grows much faster than 2x, so the graph behaves like $y=x^3$y = x³: it climbs steeply on the right and falls steeply on the left.

Put the two together and the curve rises from the bottom left, turns, runs downhill through the origin, turns again, and climbs away to the top right. It crosses the x-axis where $x^3-2x=x\left(x^2-2\right)=0$x³ − 2x = x(x² − 2) = 0, which is at x = 0 and at $x=\sqrt{2}$x = √2 and $x=-\sqrt{2}$x = −√2, about 1.41 and −1.41. Both terms are odd powers, so the whole graph has half-turn symmetry: at x = 1, y = 1 − 2 = −1, and at x = −1, y = −1 + 2 = 1.

The gold curve is $y=x^3-2x$y = x³ − 2x. The white straight line is y = −2x, which it follows near the origin, and the white curve is $y=x^3$y = x³, which it follows further out.

The usual mistakes

Giving $\frac{1}{x^2}a$1/x² a negative value for a negative x. At x = −3, $x^2=9$x² = 9, so $y=\frac{1}{9}$y = 1/9, not $-\frac{1}{9}$−1/9. A negative value would need an odd power.

Turning the fraction over. $\frac{1}{x^2}$1/x² is one divided by the square, so at x = 3 it is $\frac{1}{9}$1/9, not 9.

Reading $-x^2$−x² as ${\left(-x\right)}^2$(−x)². At x = 3, $-x^2$−x² is −9, because the square is worked out before the minus sign is applied.

Dropping the x from a term. At x = 2, $y=x^3-2x$y = x³ − 2x is 8 − 2 × 2 = 4. Subtracting a bare 2 gives 6, and adding the 2x gives 12.