# Graphs of Functions and The Vertical Line Test

**The Vertical Line Test: **

The The Vertical Line: A curve in the \(xy\) - plane is the graph of a function \(x\) if and only if no vertical line intersects the curve more than once. |

The reason for the truth of the Vertical Line Test can be seen in Figure 13. If each vertical line \(x = a\) intersects a curve only once, at \(\left( {a,b} \right)\), then exactly one functional value is defined by \(f\left( a \right) = b\). But if a line \(x = a\) intersects the curve twice, at \(\left( {a,b} \right)\) and \(\left( {a,c} \right)\), then the curve can’t represent a function because a function can’t assign two different values to *\(a\).*

For example, the parabola \(x = {y^2} - 2\) shown in Figure 14(a) is not the graph of a function of *\(x\)* because, as you can see, there are vertical lines that intersect the parabola twice. The parabola, however, does contain the graphs of two functions of *\(x\)*. Notice that the equation \(x = {y^2} - 2\) implies \({y^2} = x + 2\), so \(y = \pm \sqrt {x + 2} \). Thus, the upper and lower halves of the parabola are the graphs of the functions \(f\left( x \right) = \sqrt {x + 2} \) and \(g\left( x \right) = - \sqrt {x + 2} \).

[See figures 14(b) and (c)]. We observe that if we reverse the roles of *x *and *y*, then the equation \(x = h\left( y \right) = {y^2} - 2\) does define *x* as a function of *y *(with *y *as the independent variable and *x* as the dependent variable) and the parabola now appears as the graph of the function \(h\)*.*

The functions in the following examples are defined by different formulas in different parts of their domains. Such functions are called piecewise defined functions.

**Example- 5**

A function *\(f\) * is defined by

\[f\left( x \right) = \left\{ \begin{align}&1 - x & {\rm{if \;\;}}x \le 1\\&{x^2} & {\rm{if \;\;}}x > 1\end{align} \right.\]

Evaluate \(f\left( 0 \right),\;f\left( 1 \right),\;{\rm{and }}f\left( 2 \right)\) and sketch the graph.

**Solution: ** Remember that a function is a rule. For this particular function the rule is the following. First look at the value of the input *x*. If it happens that \(x \le 1\), then the value of \(f\left( x \right)\) is \(1 - x.\) On the other hand, if \(x > 1\), then the value of \(f\left( x \right)\) is \({x^2}\).

Since \(0 \le 1,\) we have \(f\left( 0 \right) = 1 - 0 = 1\)

Since \(1 \le 1,\) we have \(f\left( 1 \right) = 1 - 1 = 0\)

Since \(2 > 1,\) we have \(f\left( 2 \right) = {2^2} = 4\)

How do we draw the graph of \(f\,?\) We observe that if \(x \le 1\), then \(f\left( x \right) = 1 - x,\) so the part of the graph of *f* that lies to the left of the vertical line \(x = 1\) must coincide with the line \(y = 1 - x,\) which has slope –1 and *y – *intercept 1. If \(x > 1, \) then \(f\left( x \right) = {x^2}\), so the part of the graph of *f* that lies to the right of the line \(x = 1\) must coincide with the graph of \(y = {x^2}\), which is a parabola. This enables us to sketch the graph in the figure. The solid dot indicates that the point (1, 0) is included on the graph; the open dot indicates that the point (1, 1) is excluded from the graph.

The next example of a piecewise defined function is the absolute value function. Recall that the **absolute value **of a number *a*, denoted by |*a*|, is the distance from *a* to 0 on the real number line. Distances are positive or 0, so we have

\(\left| a \right| \ge 0\) for every number *a*

For example,

\[\begin{align}&\left| 3 \right| = 3\;\;\;\;\;\left| { - 3} \right| = 3\;\;\;\;\;\left| 0 \right| = 0\;\;\;\;\;\left| {\sqrt 2 - 1} \right| = \sqrt 2 - 1 \;\;\;\;\;\left| {3 - \pi } \right| = \pi - 3\end{align}\]

In general, we have

\(\fbox{${\left| a \right| = a\quad {\rm{\;if}}\;a \ge 0\\\left| a \right| = - a\quad {\rm{\;if \;}}a < 0}$} \)

(Remember that if \(a\) is negative, then \(–a\) is positive).

**Example- 6**

Sketch the graph of the absolute value function \(f\left( x \right) = \left| x \right|\)

**Solution: ** From the preceding discussion we know that

\[\left| x \right| = \left\{ \begin{align}&\quad x & {\rm{if\;}}\;x \ge 0\\&- x & {\rm{if \;\;}}x < 0\end{align} \right.\]

Using the same method as in the previous example, we see that the graph of *f* coincides with the line \(y = x\) to the right of the *y* -axis and coincides with the line \(y = - x\) to the left of the *y*-axis

**Example- 7**

Find a formula for the function *f *graphed in Figure - 17.

**Solution: ** The line through (0, 0) and (1, 1) has slope \(m = 1\) and *y-*intercept \(b = 0\), so its equation is \(y = x\). Thus for the part of the graph *f* that joins (0, 0) to (1, 1), we have

\[f\left( x \right) = x\quad {\rm{if\;\; 0}} \le x \le {\rm{1}}\]

The line through (1, 1) and (2, 0) has slope \(m = - 1\), so its point - slope form is

\[y - 0 = \left( { - 1} \right)\left( {x - 2} \right)\quad {\rm{or \;\;\;}} y = 2 - x\]

So we have

\[\,f\left( x \right) = 2 - x\quad {\rm{if \;\;}}1 < x \le 2\]

We also see that the graph of *f* coincides with the *x*-axis for \(x > 2\). Putting this information together, we have the following three - piece formula for \(f\) :

\[f\left( x \right) = \left\{ \begin{align}&x & & {\rm{if\quad }}0 \le x \le 1\\&2 - x & & {\rm{if \quad}}1 < x \le 2\\&0 & & {\rm{if \quad }}x > 2\end{align} \right.\]

**TRY YOURSELF - I **

**Q.1: **An equilateral triangle is inscribed in circle of radius *r*. Find the area *A* of the triangle as a function of *r*.

**Q.2: **Let \(y = f\left( x \right)\)be such that \(\begin{align}&\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.\end{align}\)Is \(f\left( x \right)\) a function?

**Q.3: **Find a formula for the function graphed below:

**Q.4: **A projectile is thrown with an initial velocity *u * at an angle \(\theta \) to the horizontal. Let \((x,y)\) be its position at some given time *t*.

Find

(a) * x* as a function of *t*

(b) * y* as a function of *t*

(c) * y* as a function of *x*

(d) the distance *r* of the projectile from the origin, as a function of *t*.