# How many intercepts does y = x^2 + x − 6  have?

Aug 16, 2015

This function has a total of three intercepts.

#### Explanation:

There are two types of intercepts a function can have, $y$-intercepts, which are calculated for $x = 0$, and $x$-intercepts, which are calculated for $y = 0$.

So, starting with the $y$-intercept, you have $x = 0$, which results in

$y = {\left(0\right)}^{2} + 0 - 6$

$y = - 6$

This means that the $y$-intercept, which is the point where the graph of the function intercepts the $y$-axis, will be $\left(0 , - 6\right)$.

Now for the $x$-intercept. In order to find the $x$-intercept, you need to have $y = 0$, which means that

$y = {x}^{2} + x - 6 = 0$

The discriminant of this quadratic equation will actually tell you how many $x$-intercepts you have.

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

the discriminant is defined as

$\textcolor{b l u e}{\Delta = {b}^{2} - 4 a c}$

For your quadratic, $a = 1$, $b = 1$, and $c = - 6$. The discriminant will be equal to

$\Delta = {1}^{2} - 4 \cdot 1 \cdot \left(- 6\right)$

$\Delta = 1 + 24 = 25$

When $\Delta > 0$, the quadratic equation will have two disctinct real roots, which is another way of saying that the graph of the function will intercept the $x$-axis in two points.

These roots will be equal to

color(blue)(x_(1,2) = (-b +- sqrt(Delta))/(2a)

${x}_{1 , 2} = \frac{- 1 \pm \sqrt{25}}{2 \cdot 1}$
${x}_{1 , 2} = \frac{- 1 \pm 5}{2} = \left\{\begin{matrix}{x}_{1} = \frac{- 1 - 5}{2} = - 3 \\ {x}_{2} = \frac{- 1 + 5}{2} = 2\end{matrix}\right.$
The two $x$-intercepts will thus be $\left(- 3 , 0\right)$ and $\left(2 , 0\right)$.
Therefore, the function will have a total of three intercepts, one $y$-intercept and two $x$-intercepts.