From the past lessons, we have already established a clear understanding of what a function is. But aside from equations and tables, functions can also be represented in the form of graphs. For example, a function of $f(x)=x^2$, can be tabulated by assigning values to $x$.
\[\begin{array}{c|rrrrrrr} x & -3 & -2 & -1 & 0 & 1 &2 &3 \\ \hline f(x) & 9 & 4 & 1 & 0 & 1 & 4 & 9 \end{array}\]Moreover, this can be plotted on a cartesian plane, with your input (independent variable) lying on the horizontal axis and the output (dependent variable) on the vertical axis.
Figure 1: Graphical representation of $f(x)=x^2$
This topic may have been covered in previous prerequisite courses. However, to refresh your memory, consider the following examples.
$\example{1}$ Graph the equation of the line $y=-x+3$
$\solution$
There are a lot of ways of plotting a graph in a cartesian plane, but one of the simplest method is by assigning values to the independent variable, which in this case is $x$,
\[\begin{array}{c|cccc} x & 0 & 1 & 2 & 3 \\ \hline y & 3 & 2 & 1 & 0 \end{array}\]Trace the points on the cartesian plane after plotting them.
$\example{2}$ Express the volume of a cube as a function of the length of its edge. Graph the function.
$\solution$
$\text{Let: } V = \text{volume of the cube} $
$\phantom{\text{Let: }} x = \text{length of an edge of the cube}$
Hence, the equation for the volume of a cube is,
\[\begin{align*} V &= x^3 \tagans \end{align*}\]Assign values to $x$, then tabulate the result,
\[\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4\\ \hline V & 0 & 1 & 8 & 27 & 64 \end{array}\]Plot the points on the cartesian plane,
$\example{3}$ Express the length of the edge of a cube as a function of its volume. Graph the function.
$\solution$
$\text{Let: } V = \text{volume of the cube} $
$\phantom{\text{Let: }} x = \text{length of an edge of the cube}$
Hence, the equation for the volume of a cube is,
\[V = x^3\]Then express the length of the edge as a function of its volume, \(x = V^{1/3} \tagans\)
Assign values to $V$, then tabulate the result,
\[\begin{array}{c|ccccc} V & 0 & 1 & 8 & 27 & 64\\ \hline x & 0 & 1 & 2 & 3 & 4 \end{array}\]Plot the points on the cartesian plane,
$\example{4}$ A right triangle has a hypotenuse of 5 cm. Express its area as a function of its base. Graph the function.
$\solution$
$\text{Let: } A = \text{area} $
$\phantom{\text{Let: }} a = \text{altitude}$
$\phantom{\text{Let: }} b = \text{base}$
$\phantom{\text{Let: }} c = \text{hypotenuse}$
We can solve for the altitude $a$ of the triangle using the Pythagorean theorem,
\[\begin{align*} c^2 &= a^2 + b^2 \\ a^2 &= c^2 -b^2 \\ a &= \sqrt{5^2-b^2} \\ &= \sqrt{25-b^2} \end{align*}\]Solving for the area will give us,
\[\begin{align*} A &= \frac{1}{2} ab \\ A &= \frac{1}{2} b\sqrt{25-b^2} \tagans \end{align*}\]Assign values to $b$, then tabulate the result,
\[\begin{array}{c|cccccc} b & 0 & 1 & 2 & 3 & 4 & 5\\ \hline A & 0 & 2.45 & 4.58 & 6 & 6 & 0 \end{array}\]Plot the points on the cartesian plane,
