Non-Linear Graphs and Tables of Values

By now you should be familiar with a number of common functions, including lines and quadratics. You may even be able to construct their graphs fairly quickly by immediately identifying features of the graph such as slopes, turning points, intercepts, or transformations.

However, for difficult functions, or functions that are new to you, it can be useful to just select some values of $x$x, and construct a table of values for $f\left(x\right)$f(x). From this table, we can gain useful information about the graph.

Consider the cubic function $f\left(x\right)=2x^3+8x^2$f(x)=2x3+8x2. The shape and location of the graph might not be immediately obvious. Let's complete a table of values to help us.

$x$x	$-8$−8	$-4$−4	$-2$−2	$0$0	$1$1	$4$4
$f\left(x\right)$f(x)

We substitute the given values of $x$x from the table into the function $f\left(x\right)=2x^3+8x^2$f(x)=2x3+8x2.

So, for $x=-8$x=−8 we have $f\left(-8\right)=2\left(-8\right)^3+8\left(-8\right)^2=-512$f(−8)=2(−8)3+8(−8)2=−512 and for $x=-4$x=−4 we have $f\left(-4\right)=2\left(-4\right)^3+8\left(-4\right)^2=0$f(−4)=2(−4)3+8(−4)2=0.

In this way, we can fill out the table as follows.

$x$x	$-8$−8	$-4$−4	$-2$−2	$0$0	$1$1	$4$4
$f\left(x\right)$f(x)	$-512$−512	$0$0	$16$16	$0$0	$10$10	$256$256

Hence, from the table, we can see that we have the points $\left(-8,-512\right)$(−8,−512), $\left(-4,0\right)$(−4,0), $\left(-2,16\right)$(−2,16), $\left(0,0\right)$(0,0), $\left(1,10\right)$(1,10), $\left(4,256\right)$(4,256) on our graph.

And if we fit a cubic to these points, we get the following graph.

Without having to do any tricky work involving slopes, turning points, or algebraic manipulations, we have come up with the required graph.

We can see the $x$x-intercepts, the $y$y-intercept, where the function is positive and negative, and when the curve is increasing and decreasing. As you can see, tables of values can be a quick and efficient way to come up with a graph.

Examples

QUESTION 1

Consider the function $f\left(x\right)=\frac{1}{x}$f(x)=1x​.

a) Complete the following table of values. Give the answers as improper fractions.

$x$x	$-4$−4	$-1$−1	$-\frac{1}{10}$−110​	$\frac{1}{10}$110​	$1$1	$4$4
$f\left(x\right)$f(x)

b) For the function $g\left(x\right)=\frac{1}{x}+3$g(x)=1x​+3, complete the following table of values. Give the answers as improper fractions.

$x$x	$-4$−4	$-1$−1	$-\frac{1}{10}$−110​	$\frac{1}{10}$110​	$1$1	$4$4
$g\left(x\right)$g(x)

c) Judging from the tables, what did the $+3$+3 change about the function? Choose all the correct statements.

(A) Vertical Translation

(B) Horizontal Translation

(C) Reflection

(D) Dilation

d) Plot $f\left(x\right)=\frac{1}{x}$f(x)=1x​.

e) Hence, plot $g\left(x\right)=\frac{1}{x}+3$g(x)=1x​+3.

SOLUTION

a) As before, we substitute the given values of $x$x from the table into the function $f\left(x\right)=\frac{1}{x}$f(x)=1x​.

For $x=-4$x=−4 we have $f\left(-4\right)=\frac{1}{-4}=-\frac{1}{4}$f(−4)=1−4​=−14​ and for $x=-1$x=−1 we have $f\left(-1\right)=\frac{1}{-1}=-1$f(−1)=1−1​=−1.

Notice that for $x=-\frac{1}{10}$x=−110​ we will get:

$f\left(-\frac{1}{10}\right)$f(−110​)	$=$=	$\frac{1}{-\frac{1}{10}}$1−110​​
	$=$=	$-\frac{1}{\frac{1}{10}}$−1110​​
	$=$=	$-10$−10

We can fill out the table as follows.

$x$x	$-4$−4	$-1$−1	$-\frac{1}{10}$−110​	$\frac{1}{10}$110​	$1$1	$4$4
$f\left(x\right)$f(x)	$-\frac{1}{4}$−14​	$-1$−1	$-10$−10	$10$10	$1$1	$\frac{1}{4}$14​

b) Similarly, we can fill out the table for $g\left(x\right)=\frac{1}{x}+3$g(x)=1x​+3.

$x$x	$-4$−4	$-1$−1	$-\frac{1}{10}$−110​	$\frac{1}{10}$110​	$1$1	$4$4
$f\left(x\right)$f(x)	$\frac{11}{4}$114​	$2$2	$-7$−7	$13$13	$4$4	$\frac{13}{4}$134​

c) Notice that for each value of $x$x in the above tables, $g\left(x\right)$g(x) is $3$3 greater than $f\left(x\right)$f(x) because of the $+3$+3. In other words, $g\left(x\right)$g(x) is $3$3 units higher than $f\left(x\right)$f(x) at any value of $x$x. The $+3$+3 represents vertical translation.

Hence, (A) is the correct answer.

d) From the table, we can see that we have the points $\left(-4,-\frac{1}{4}\right)$(−4,−14​), $\left(-1,-1\right)$(−1,−1), $\left(-\frac{1}{10},-10\right)$(−110​,−10), $\left(\frac{1}{10},10\right)$(110​,10), $\left(1,1\right)$(1,1), $\left(4,\frac{1}{4}\right)$(4,14​) on our graph for $f\left(x\right)=\frac{1}{x}$f(x)=1x​.

We fit our function to the dots as follows.

e) Given what we know about $g\left(x\right)$g(x) being a vertical translation of $f\left(x\right)$f(x), we don't need to plot points for $g\left(x\right)$g(x). We can just move our previous graph up by $3$3 units.

Question 2

Question 3