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Revision:OCR Core 1 - Graphs of nth power functions

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 1 - Graphs of nth power functions

1 8. Graphs of nth power functions
2 Also See
3 Comments

8. Graphs of nth power functions

Graphs of negative integer powers

It is now known that the following is true:

a > 0

$x^{-a} = \frac{1}{x^{a}}$

Hence, the graph of a negative power:

$y = x^{-a}$

Is equivalent to the following:

$y = \frac{1}{x^{a}}$

Consider the graph below.

Figure 005

This shows the graphs of $y = \frac{1}{x}$ , and $y = \frac{1}{x^{2}}$ .

One can see that both of these graphs have a common point at (1, 1), and also there is a pattern of flattening (it seems) as with the graphs of $y = x^{a}$ .

These graphs do not touch either of the axes, they become infinitesimally close, but never cross them, this is due to the fractional powers of zero.

Consider that the interception with the x-axis occurs only when one has a value of "x" for which "y" is zero. So consider the following:

$y = \frac{1}{x^{a}} = 0$

If one is to multiply by $x^{a}$ it produces 1 = 0 which is contradictory, hence there are no values of "x" for which "y" is zero, hence there are no x-axis intercepts.

Consider that for a y-axis intercept, "x" must be zero, hence:

$y = \frac{1}{x^{a}} = \frac{1}{0^{a}} = \frac{1}{0}$

This is undefined, and hence it shows that there is not y-axis intercept.

Not one can consider the flattening pattern. It is rather self evident that as one has a value of above 1, and the power of the function decreases, there will be a flattening; putting this more formally makes it clearer.

$a \in \mathbb{R}$

a > 0

$y = x^{-a} = \frac{1}{x^{a}}$

Hence, for:

x > 1

One can predict that as $a \to \infty$ , the value of "y" will decrease.

One might also notice that there is a difference in the quadrants which the graphs occupy. It is evident that for:

$a \equiv 0 \pmod{2}$

The value of "y" is always positive (as one is conducting a division of 1 by a positive value, see the previous section for a more detailed explanation).

Conversely, the values of "y" for a graph where:

$a \equiv 0 \pmod{2}$

Can be negative. This results in the graph occupying the same quadrants as the graph of $y = x^{a}$ , for the same constraints on "a".

Differentiation with negative integer indices

There is no special rule for the negative indices (in the context of differentiation). It is merely a case where one can use the general rule for the differentiation of a function of x, of the form:

$f(x) = x^{n}$

One must, however, be careful to ensure that one takes from the power, hence, in the cases that one will discuss in this section of the notes, this will result in no limitation (in contrast to the differentiation of a function whose power is a positive integer; as decrementing a positive integer with a decrementation of 1 will result in a power of 0, and hence a constant value, which differentiates to zero).

Example

1. Calculate the equation to the tangent of the curve $y = \frac{1}{x}$ , at the point where x = 1 .

One will follow the same type of process as one did with the previous examples (previously in these notes) of calculating the equation of the tangent of a curve at a given point.

First one can calculate the differential:

$y = \frac{1}{x} = x^{-1}$

$\frac{dy}{dx} = -x^{-2} = \frac{-1}{x^{2}}$

(It is important that one makes sure one is able to convert between the negative powers, and the reciprocal notation, as this will aid the calculation of the gradient, numerically speaking).

Now calculate the specific gradient, one can calculate the y-coordinate of the point at this stage also.

$y = \frac{1}{x} = \frac{1}{1} = 1$

Hence one is dealing with the point of contact (1, 1).

Now, find the gradient of the tangent:

$m = \frac{-1}{x^{2}} = \frac{-1}{1^{2}} = -1$

Hence, the equation is of the form:

y = -x + c

Hence, one can now substitute:

1 = -1 + c

c = 2

Hence, the equation is:

x + y = 2

It is a simple process as previously (one could demonstrate the general rule, and how it works through the use of a similar method to that used in the notes upon differentiation, however this is not necessary as it has been shown, using binomial expansion, and other such methods, that the general rule is true for all positive "n"; and one can express these examples in terms of a positive power, in reciprocal form).

Graphs of $y = x^{n}$ for fractional n

Suppose one has the graph:

$y = x^{\frac{1}{n}}$

n > 0

One can simplify, and rearrange this:

$y = x^{\frac{1}{n}} = \sqrt[n]{x}$

Hence:

$y^{n} = x$

This means that if one is to graph the original graph, one can merely "swap" the x-axis with the y-axis (if they are the same scale), and draw a graph of $y = x^{n}$ .

For example, see the graph below (and reference it to Figure 003).

Figure 006

(Compare the similarity between a graph of $y = x^{2}$ ).

The above graph (Figure 006) depicts $y = x^{\frac{1}{2}} = \sqrt{x}$ .

These graphs will all pass through the point (1, 1). Also, as one might expect (from the previous transformatory method of depiction, or prediction thereof) those fractions whose denominator and numerator are even will produce positive "y" values, and the odd values will produce both positive, and negative values of "y".

Other than this information, one must be aware that these powers will use the same rules for differentiation as usual.

Example

1. Calculate the equation of the tangent to the curve $y = \sqrt{x}$ at the point (1, 1).