9.03 Properties of graphs

Connections between a function and its derivative

We have seen we can find the gradient of a function by using its derivative. Can we glean any more information from the derivative function? The applet below shows the graphs of several different functions (linear, quadratic, cubic and the hyperbola $y=\frac{1}{x}$y=1x​ ) together with the graph of their derivative. Explore each function and look for connections between the features of the function and the graph of the derivative such as:

• The steepness of the curve
• Where the function is increasing or decreasing
• Key points such as maximums or minimums
• The shape of the derivative function

Created with Geogebra

What connections did you find? Let's look further at the connections for polynomials and their gradient functions. The shape of a polynomial determines the shape of its derivative, since when we differentiate a polynomial of degree $n$n we obtain a derivative of degree $n-1$n−1. Hence, a linear function will produce a constant derivative, a quadratic function will produce a linear derivative, a cubic function will produce a quadratic derivative and so forth.

Another observation which becomes apparent after looking at a few functions together with their derivatives is that turning points(maximum and minimum) in the function correspond to $x$x-intercepts in the gradient function. Check the graphs below which display this connection. This occurs since the tangent to a turning point will be horizontal and thus have a gradient of zero. (for further discussion see stationary points)

Quadratic	Cubic	Quartic

 

Increasing and decreasing

When a function is increasing, that is as the $x$x values increase the $y$y values increase, this relates to a positive gradient. Thus, when the function is increasing the derivative will be above the $x$x-axis.

When a function is decreasing, that is as the $x$x values increase the $y$y values decrease, this relates to a negative gradient. Thus, when the function is decreasing the derivative will be below the $x$x-axis.

For the graphs below observe where the function is increasing (indicated in green sections), where it is decreasing decreasing (indicated in blue) and where the derivative function is located.

 

Stationary points

A stationary point is where the derivative is zero, that is a point $\left(a,f(a)\right)$(a,f(a)) is said to be a stationary point if $f'(a)=0$f′(a)=0. At this point the tangent is horizontal and thus, the instantaneous rate of change is zero - so the function is momentarily stationary. We have three types of stationary points:

• Local minimum: at this point $f(x)$f(x) has the lowest value in the region of this point. The graph changes from decreasing to increasing at this point. Hence, the derivative will change from negative to positive.
• Local maximum: at this point $f(x)$f(x) has the highest value in the region of this point. The graph changes from increasing to decreasing at this point. Hence, the derivative will change from positive to negative.

• Stationary point of inflection: at this point the gradient does not change either side of the point, it can be either positive either side of the stationary point or negative either side. At points of inflection the rate of change switches from increasing to decreasing or vice versa. So unlike the maximum or minimum it is not the graph itself changing between increasing and decreasing but the gradient of the graph. So the gradient changes from becoming less steep to becoming steeper or vice versa. This will be explored more in further studies of calculus.

Local maximum and minimum are also referred to as turning points, since the function changes between increasing and decreasing at these points.

The reason for the emphasis on "local" minimum or maximum is that they may give the minimum or maximum value within a region of the point but not over the entire domain of the graph. If a point gives the maximum or minimum value over the entire domain of a graph we refer to the point as a global maximum or global minimum. In the first graph below we have both a global and local minimum, as well as a local maximum, however, as the graph is unbounded there is no global maximum. The second graph below shows a restricted domain where the global maximum is in fact at an end-point and not a turning point.

For many functions a local maximum or minimum is also a global maximum or minimum, such as turning points for an unrestricted quadratic graph.

 

Practice questions

Question 1

Question 2

 

Sketching polynomials

We can sketch the general shape of a given polynomial using facts about:

• The degree of the polynomial
• The coefficient of the leading term and behaviour of the function at the extremities
• The $y$y-intercept
• The $x$x-intercepts from factored forms or using technology
• The multiplicity of roots and the shape of the function about the root

Previously we had to use technology to find the location and nature of stationary points and where a function is increasing or decreasing. We can now use calculus to achieve this.

To find stationary points for a function $y=f(x)$y=f(x):

1. Find the derivative $f'(x)$f′(x)
2. Solve $f'(x)=0$f′(x)=0 for $x$x.
3. Substitute the $x$x-coordinate of any stationary points found back into the original function $f(x)$f(x) to find the $y$y-coordinate of the point.

To determine the nature of a stationary point, we can evaluate the derivative just either side of the stationary point. Thus, if the derivative changes from negative to positive - we have identified a local minimum. If the derivative changes from positive to negative - we have identified a local maximum. And if the sign of the derivative does not change either side of the stationary point - we have identified a stationary point of inflection. A table or diagram indicating the sign and/or shape (increasing, decreasing or constant) is helpful to keep track and visualise the nature of turning points for graphing purposes.

Steps for sketching polynomials:

1. Determine $x$x-intercepts, solve $f(x)=0$f(x)=0 for $x$x. This can be done through factorisation or use of technology. If the function does not factorise easily and you do not have access to technology you can create a rough sketch using the other key features found and approximating the location of  the $x$x-intercepts.
2. Determine the $y$y-intercept by letting $x=0$x=0.
3. Find the derivative and use it to determine the location and nature of the turning points. 
4. Determine the behaviour at the extremities by looking at the leading term. Recall the degree of the polynomial, $n$n, together with the leading coefficient, $a$a, dictate the overall shape and behaviour of the function at the extremities:

$n$n	$a>0$a>0	$a<0$a<0
Even
Odd

Worked example 

Example 1

For the function $f(x)=\left(x-5\right)\left(x-2\right)\left(x+3\right)$f(x)=(x−5)(x−2)(x+3), find:

a) The $x$x-intercepts.

b) The $y$y-intercept.

c) The location and nature of any turning points.

d) Use the information to sketch the function.

Think: This is cubic and if we expanding the brackets we would have a positive leading coefficient. This tells us the general shape with the tails of the function going from bottom left to top right.

Do:

a) Find the $x$x-intercepts: Solving $f(x)=0$f(x)=0:

$\left(x-5\right)\left(x-2\right)\left(x+3\right)=0$(x−5)(x−2)(x+3)=0

As the function is in factored form we can see the solutions are $x=5$x=5, $2$2 and $-3$−3. So the $x$x-intercepts are $\left(5,0\right)$(5,0), $\left(2,0\right)$(2,0) and $\left(-3,0\right)$(−3,0).

b) Find the $y$y-intercept: When $x=0$x=0:

$f\left(0\right)$f(0)	$=$=	$\left(0-5\right)\left(0-2\right)\left(0+3\right)$(0−5)(0−2)(0+3)
	$=$=	$30$30

Thus, the $y$y-intercept is $\left(0,30\right)$(0,30).

c) Find the location and nature of turning points: Find the derivative and solve $f'(x)=0$f′(x)=0.

Expanding $f(x)$f(x) we obtain:

$f(x)$f(x)	$=$=	$\left(x-5\right)\left(x-2\right)\left(x+3\right)$(x−5)(x−2)(x+3)
	$=$=	$\left(x-5\right)\left(x^2+x-6\right)$(x−5)(x2+x−6)
	$=$=	$x^3-4x^2-11x+30$x3−4x2−11x+30
Hence, $f'(x)$f′(x)	$=$=	$3x^2-8x-11$3x2−8x−11

Solving $f'(x)=0$f′(x)=0:

$3x^2-8x-11$3x2−8x−11	$=$=	$0$0
$\left(3x-11\right)\left(x+1\right)$(3x−11)(x+1)	$=$=	$0$0
$\therefore x$∴x	$=$=	$\frac{11}{3}$113​ or $-1$−1

Substituting these values into $f(x)$f(x) we obtain $f\left(\frac{11}{3}\right)=-\frac{400}{27}\approx-14.8$f(113​)=−40027​≈−14.8 and $f\left(-1\right)=36$f(−1)=36. So we have two stationary points at $\left(\frac{11}{3},-\frac{400}{27}\right)$(113​,−40027​) and $\left(-1,36\right)$(−1,36). We can use information about the shape of the graph (positive cubic) to ascertain that the point to the left, $\left(-1,36\right)$(−1,36), will be the maximum and the second point, $\left(\frac{11}{3},-\frac{400}{27}\right)$(113​,−40027​), will be the minimum.

Alternatively, we can use calculus to test the behaviour of the derivative either side and between these two points. Choose convenient $x$x-values before $x=-1$x=−1, (such as $x=-2$x=−2), between $x=-1$x=−1 and $\frac{11}{3}$113​ (such as$x=0$x=0) and after $x=\frac{11}{3}$x=113​ (such as $x=4$x=4). Substitute the values into the gradient function - we are concerned with the sign and whether the function is increasing or decreasing, so you can choose to simply record the sign in the table and not include the value.

$x$x	$-2$−2	$-1$−1	$0$0	$\frac{11}{3}$113​	$4$4
$f'(x)$f′(x)	$17$17	$0$0	$-11$−11	$0$0	$5$5
sign	$+$+	$0$0	$-$−	$0$0	$+$+
shape

The table shows that at $x=-1$x=−1 the graph changes from increasing to decreasing and hence, $\left(-1,36\right)$(−1,36) is a maximum. And at $x=\frac{11}{3}$x=113​ the graph changes from decreasing to increasing and hence, $\left(\frac{11}{3},-\frac{400}{27}\right)$(113​,−40027​) is a minimum.

d) Plotting the points we have so far and sketching a smooth curve connecting them we obtain the graph:

 

Practice questions

Question 3

Question 4

Question 5

 

Sketching functions from given information

In the examples above we had to find the key features from an equation but we could be given pieces of information and be required to identify an appropriate graph or sketch a possible graph.

 

Worked example

Example 2

Sketch a possible graph that has the following properties:

• $f'(2)=0$f′(2)=0 and $f'(1)=0$f′(1)=0
• $f(2)=0$f(2)=0
• $f(1)=1$f(1)=1
• $f(0)=-4$f(0)=−4
• $f'(x)<0$f′(x)<0 for $11<x<2 and $f'(x)>0$f′(x)>0 elsewhere

Think: Carefully consider if the information given is about the function or its derivative. 

The first clue tells us that the graph has stationary points at $x=2$x=2 and $x=1$x=1.

The second piece of information tells us the graph passes through $\left(2,0\right)$(2,0). This is also our stationary point from the first clue and an $x$x-intercept.

Third piece of information tells us that the graph also passes through $\left(1,1\right)$(1,1), which is our other stationary point.

Dot point $4$4, tells us that when $x=0$x=0, $y=-4$y=−4. That is the $y$y-intercept is $\left(0,-4\right)$(0,−4).

Dot point $5$5 tells us that between the two stationary points we have a decreasing function and elsewhere we have an increasing function.

Do: Plotting the points and indicating the known areas of increasing and decreasing, we obtain:

Joining the dots with a smooth curve and following the arrow trends we have a graph of a function with the given properties - this is not the only possibility, can you draw a few more?

 

Practice questions

Question 6

Question 7

Question 8