>Fundamentals of Functions>How to Calculate Zeros or Roots of a Function

How to Calculate Zeros or Roots of a Function

The zeros or roots of a function tell us where a graph intersects the $x$ -axis. Since we’re talking about intersection with the $x$ -axis, you know that $y = 0$ . That means that you can find the zeros by solving the equation $f (x) = 0$ .

Rule

Zeros

You find the zeros of a function by solving the equation

f (x) = 0 .

The sign chart of $f (x)$ tells you when the graph of $f$ is above or below the $x$ -axis, and where $f (x)$ intersects the $x$ -axis.

Graph showing f(x) and its zeros

Example 1

Find the zeros of $f (x) = x^{2} + 5 x + 6$

You find the zeros by solving the equation $f (x) = x^{2} + 5 x + 6 = 0$ : $\begin{array}{l} x & = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot 6}}{2} \\ = \frac{- 5 \pm \sqrt{25 - 24}}{2} \\ = \frac{- 5 \pm 1}{2} \end{array}$

This gives

\begin{array}{l} x_{1} & = \frac{- 5 - 1}{2} = - 3, \\ x_{2} & = \frac{- 5 + 1}{2} = - 2 . \end{array}

So the graph meets the $x$ -axis at the points i $(- 3, 0)$ and $(- 2, 0)$ .

Example 2

Find the zeros of $g (x) = 2 \sin (2 x) + 1$ in the interval $x \in [0, π)$

Again, you can find the zeros by solving the equation $g (x) = 0$ : $\begin{array}{l} 2 \sin (2 x) + 1 & = 0 \\ \sin (2 x) & = - \frac{1}{2} \end{array}$

The basic equation has the solutions

\begin{array}{l} 2 x & = - \frac{π}{6} + 2 π n, \\ 2 x & = π - (- \frac{π}{6}) + 2 π n . \end{array}

Solving these for $x$ , you get

\begin{array}{l} x & = - \frac{π}{12} + π n, \\ x & = \frac{7 π}{12} + π n . \end{array}

Since the interval is $x \in [0, π)$ , you get that $x \in {\frac{7 π}{12}, \frac{11 π}{12}}$ . The graph meets the $x$ -axis at the points $(\frac{7 π}{12}, 0)$ and $(\frac{11 π}{12}, 0)$ .

How to Calculate and Interpret Intersection Points

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