6.03 Apply calculus to graphs

We have seen that the first derivatives of a function have been shown to give us a wealth of information about the shape of the curve. Combining this with the previous understanding of functions, such as domain and range, properties of polynomials, features of exponential and trigonometric functions, we are able to create a checklist of skills that allow us to sketch many functions that are familiar and unfamiliar. 

 

Sketching functions

Strategy	Description
Determine any special characteristics	Investigate values of $x$x for which $f\left(x\right)$f(x) is not defined, such as within fractions or square root functions. Check if there are any asymptotes or discontinuities.
Domain and range	Consider the $x$x and $y$y values that the function can take. If a domain is specified, clearly indicate the coordinates of the end points on the function sketch.
Find axis intercepts	Let $x=0$x=0 to find the $y$y-intercept, and let $y=0$y=0 and solve to find any $x$x-intercepts, where possible.
Determine stationary points	Consider $\frac{dy}{dx}=0$dydx​=0.
Determine the nature of stationary points	Consider $\frac{dy}{dx}$dydx​ left and right of each stationary point.
Limiting behaviours	Consider the graphs behaviour as $x$x approaches $\pm\infty$±∞.  Also consider the graphs behaviour about any asymptotes, if the graph has any.  (See below for further detail)

Create a sketch with clearly labelled and appropriately scaled axes, as well as labelling all key features. When available make effective use of technology to help sketch a function.

 

Limiting behaviours of functions 

When sketching curves it is useful to consider what happens to the function as $x\rightarrow\pm\infty$x→±∞. This gives us an indication of what the graph is doing as $x$x gets very large ($x\rightarrow\infty$x→∞) and very small ($x\rightarrow-\infty$x→−∞). Also, if we have identified any asymptotes in the graph, it is important to consider the behaviour of the graph as it approaches critical values.

• The limiting behaviour of polynomials as $x\rightarrow\pm\infty$x→±∞ is determined by the leading term, that is, the one with the highest index. For example, for the polynomial $P\left(x\right)=2x^3+3x^2-10x$P(x)=2x3+3x2−10x, as $x\rightarrow\pm\infty$x→±∞ the term $2x^3$2x3 dictates that the function will be very large and positive as $x\rightarrow\infty$x→∞ and the function will be very large and negative as $x\rightarrow-\infty$x→−∞ . The table below summarises the behaviour of polynomials with leading term $ax^n$axn.

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

• It is useful to understand the limiting behaviour of $\frac{1}{x}$1x​ as $x\rightarrow\pm\infty$x→±∞ for functions with a variable in the denominator, such as hyperbolic functions. If we divide $1$1 by a large positive number, we get a positive number very close to $0$0. And we can see in the graph below that as $x\rightarrow+\infty$x→+∞, the function approaches zero from above. Similarly, if we divide $1$1 by a large negative number, we get a negative number very close to $0$0. Looking at the graph we can see that as$x\rightarrow-\infty$x→−∞, the function approaches zero from below. 

• In the graph shown above we also have a vertical asymptote. For functions that contain a fraction with a variable in the denominator, identify any values that cause the denominator to be zero. These are likely candidates for vertical asymptotes and can be further identified by observing the behaviour of the graph as $x$x approaches any critical values. In the case above, a value of $x=0$x=0 would cause the function $y=\frac{1}{x}$y=1x​ to be undefined and we can look at the behaviour either side to understand if this is an asymptote or single point discontinuity. If we divide $1$1 by a very small positive number we get a very large positive number, that is, as $x$x approaches $0$0 from above $y$y approaches $\infty$∞. Similarly, as $x$x approaches $0$0 from below $y$y approaches $-\infty$−∞.
• Exponential functions, such as $y=e^x$y=ex also display interesting behaviour as $x$x approaches $\pm\infty$±∞. As the $x$x increases, the $y$y-values increase at an increasing rate. Thus, as $x\rightarrow\infty$x→∞, $y\rightarrow\infty$y→∞. We also have a horizontal asymptote of $y=0$y=0, since as $x\rightarrow-\infty$x→−∞, $y$y approaches $0$0 from above.

 

Worked examples

Example 1

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

a) Find the axes intercepts.

Think: Find the intercepts by letting the opposite variable equal to zero and solve. The cubic is already factorised so we can use the null factor law to solve for the $x$x-intercepts.

Do:

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).

 

Find the $x$x-intercepts: Solving $f\left(x\right)=0$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 location and nature of any turning points.

Think: We need to find the derivative, solve for $f'\left(x\right)=0$f′(x)=0 and then determine the nature of the turning points. This can be achieved using calculus and testing the gradient either side of the critical points. Alternatively, we can use our knowledge of the shape of the type of polynomial we have. 

Do: Expanding $f\left(x\right)$f(x) we obtain:

$f\left(x\right)$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, $f'\left(x\right)$f′(x)	$=$=	$3x^2-8x-11$3x2−8x−11

Solving $f'\left(x\right)=0$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 $f\left(x\right)$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). 

Using 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
$f'\left(x\right)$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.

As mentioned, we could alternatively use information about the type of graph we have to ascertain which stationary point is maximum and which is minimum. This method can be more efficient for familiar polynomial functions. Here we have a cubic and with the brackets expanded we see that we have a positive leading coefficient. This tells us the general shape of the graph with the tails of the function going from bottom left to top right. Thus, 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.

 

c) Use the information to sketch the function.

Think: Given the points found, think of an appropriate scale to show the key features. Then plot each of the points we have and sketch in a smooth curve connecting them to obtain the graph.

Do:

Practice questions

Question 1

Question 2

Question 3