• Revision:OCR Core 1 - Investigating the shapes of graphs

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11. Investigating the shapes of graphs

It is a useful skill to be able to draw an accurate (representative) sketch graph of a function of a variable. It can aid in the understanding of a topic, and moreover, it can aid those who might find the mental envisaging of some of the more complex functions very difficult.

Stationary points

Often one refers to the "vertex" of a quadratic function, but what is this?

The vertex is the point where the graph changes the direction, but with the new skills of differentiation this can be generalised (rather helpfully):

A stationary point is a point on a graph of:

 y = f(x) ,

such that:

 f^{'}(x) = 0 .

This is simple to explain in words. One is basically finding all values of "x" (and hence the coordinates of the corresponding points, if it is required) of the places on the graph where the gradient is equal to zero.


1. Calculate the coordinates of the vertex of the graph:  y = x^{2} + 2x + 3 .

First one must calculate the derivative, such that one is able to calculate the value of "x" for which this is zero, and hence the gradient is zero.


 \frac{dy}{dx} = 2x + 2


 2x + 2 = 0

 2x = -2

 x = -1

Hence, one now uses the original function to obtain the "y" value, and thence the coordinate of the point:

 y = (-1)^{2} + (2 \times (-1)) + 3 = 2

Hence there is a stationary point, or vertex at (-1, 2).

One can check this using the rules about the transformation of graphs, along with the completion of the square technique.

Maximum and minimum points

It is evident that there are different types of these stationary points. One can envisage simply that there are those point whose gradient is positive, and then becomes zero, after which they are negative (maxima), and those points whose gradient change is from negative to zero to positive (minima)

One could perform an analysis upon these points to check whether they are maxima, or minima.


1. For the stationary point calculated in the previous example, deduce whether it is a point of local maximum, or local minimum.

One obtained the point (-1, 2) on the graph of:

 y = x^{2} + 2x + 3

One can therefore take an "x" value either side of this stationary point, and calculate the gradient.

 \frac{dy}{dx} = 2x + 2

Hence, one can now take an "x" value of  -1 - \epsilon , giving a gradient:

 \epsilon > 0

 m = 2x + 2 = 2(-1 - \epsilon) + 2 = -2 - 2 \epsilon + 2 = -2 \epsilon

This is evidently negative.

Now take the value of "x" as  -1 + \epsilon , giving a gradient:

 m = 2x + 2 = 2(-1 + \epsilon) + 2 = -2 + 2 \epsilon + 2 = 2 \epsilon

This is evidently positive.

Hence the gradient has gone from negative, to zero, to positive; and therefore the stationary point is a local minimum.

It is important that one understands that these "minima" and "maxima" are with reference to the local domain. This means that one can have several points of local minimum, or several points of local maximum on the same graph (the maximum is not the single point whose value of "y" is greatest, and the minimum is not the single point whose value of "y" is least).

An application to roots of equations

It is evident, and has been shown previously, that one can obtain the roots of an equation through the analysis, and calculation of the points of intersection of two functions (when graphed). It is evident why this is true; for example:

 y = f(x)

 f(x) = 4x + 2

It is therefore simple to deduce that:

 y = 4x + 2

This is only true while the value of  f(x) is equivalent to "y", and hence one might conclude that the intersections of the lines of:

 y = f(x) ,


 y = 4x + 2

Are the real roots to the equation.

This is correct, and is useful knowledge when conjoined with the knowledge of stationary points, and basic sketching skills.

Consider that one wishes to calculate the roots of the equation:

 f(x) = g(x)

These roots (if they are real) are graphically described as the intersections of the lines:

 y = f(x) ,


 y = g(x) .

Hence one would plot both graphs, and calculate the points of intersection.

However, it is often the case that one will merely want to know how many real roots there are to an equation, and hence the work on sketch graphs is relevant.

One does not need to know the accurate roots, merely the number of them, and hence it is useful to learn how to plot a good sketch graph.

First one would calculate the stationary points of one of the functions, and then one could deduce their type. This could then be sketched onto a pair of axes. Repetition of this with the second function would lead to a clear idea of where the intersections may (or may not be), and therefore one can not only give the number of real roots to the equation, but also approximations as to the answer (these are usually given as inequalities relating to the positioning of the x-coordinate of the intersection with those of the stationary points).

Second derivatives

There is a much better, and usually more powerful technique for calculating the type of stationary point one is dealing with than the method described earlier.

If one is to think of the derivative of a function to be the "rate of change" of the function, the second derivative is the "rate of change, of the rate of change" of a function.

This is a difficult sounding phrase, but it is a rather easy concept. In some functions, one will have noticed that the derivative involves a value of "x", and hence there is a change to the gradient. A straight line has a constant value as the derivative, and hence it has a constant gradient. A constant gradient is not changing, there is no "rate of change of the gradient, other than zero".

A derivative that has a term of a variable within it will have a second derivative other than zero. This is because at any given value of "x" on the curve, there is a different gradient, and hence one can calculate how this changes.


1. What is the second derivative of the function of "x":

 f(x) = x^{3} ?

This is simple; being by calculating the first derivative:

 f^{'} (x) = 3x^{2}

Now, one would like to know what the rate of change, of this gradient is, hence one can calculate the derivative of the derivative (the second derivative):

 f^{''}(x) = 6x

One should be aware that the second derivative is notated in two ways:

 f^{''}(x) ,



The former is pronounced "f two-dashed x", and the latter, "d two y, d x squared".

The application of this to minima and maxima is useful.

If a graph is bending "upwards" (like a quadratic function whose coefficient of  x^{2} is positive) has a positive second derivative, and will have a local minimum.

If a graph is bending "downwards" (like a quadratic function whose coefficient of  x^{2} is negative) has a negative second derivative, and will have a local maximum.

In many cases, this is a much more powerful tool than the original testing with values above and below the stationary point.


1. Demonstrate why (through the use of second derivative) the stationary point calculated in the example earlier (in this section) produced a local minimum.

First one can assert the function:

 y = x^{2} + 2x + 3

Now, the derivative:

 \frac{dy}{dx} = 2x + 2

Finally, the second derivative:

 \frac{d^{2}y}{dx^{2}} = 2

 2 > 0

Hence the point is a local minimum, and moreover, the graph bends upwards, and does not have a local maximum.

Graphs of other functions

Functions other than the simple polynomial one that has been considered can use the same method.

One is already aware that these graphs of fractional, or negative indices of "x" are differentiable using the same rule for differentiating powers of "x" as the positive, integer powers use.

One does have to be slightly more careful however, as there are some points on these graphs that are undefined (the square root of any negative value, for instance, is not defined in the set of real numbers).

One should simply apply the same principles.


1. Calculate the coordinates, and types of any stationary points on the curve of  y =  x^{3} + \frac{1}{x^{3}} .

First one must find the derivative (it might be a good idea to find the second derivative at the same time, so as to do all of the calculus first):

 \frac{dy}{dx} = 3x^{2} - \frac{3}{x^{2}}

(Note, in this example one might wish to write down the original expression of "y" as a positive, and negative power of "x", it will aid, one would imagine, the understanding of the situation).


 \frac{d^{2}y}{dx^2} = 6x + \frac{1}{6x}

Now one can find the stationary points:

 \frac{dy}{dx} = 3x^{2} - \frac{3}{x^{2}} = 0


 3x^{2} = \frac{3}{x^{2}}

 3x^{4} = 3

 x^{4} = 1

 x = \pm 1

Hence, there are stationary points at (-1, -2), and (1, 2).

Now one can identify them, in turn:

 x = -1

 \frac{d^{2}y}{dx^{2}} < 0

 x = 1

 \frac{d^{2}y}{dx^{2}} > 0

Hence there is a point of local maximum at (-1, -2), and a point of local minimum at (1, 2).

One thing that one should be aware of is that sometimes one will encounter a change in the gradient of a curve that is from a positive to a positive, or from a negative to a negative, this is a point of inflexion. Although this is not strictly in the syllabus, it is useful to know, and can help to explain the stationary point that is found in graphs such as  y = x^{3} .

Also See

Read these other OCR Core 1 notes:

  1. Coordinates, points, and lines
  2. Surds
  3. Some important graphs
  4. Quadratics
  5. Differentiation
  6. Inequalities
  7. Index Notation
  8. Graphs of nth power functions
  9. Polynomials
  10. Transforming graphs
  11. Investigating the shapes of graphs
  12. Applications of differentiation
  13. Circles


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