# How many intercepts does y = x^2 − 5x + 6 have?

Aug 13, 2015

Two $x$-intercepts and one $y$-intercept.

#### Explanation:

You can find this function's $x$-intercepts by making equal to zero and its $y$-intercepts by evaluating the function for $x = 0$.

For the $x$-intercepts, you have

${x}^{2} - 5 x + 6 = 0$

You can determine how many solutions this quadratic has by calculating its discriminant, $\Delta$, which, for a general form quadratic equation

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

takes the form

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

In your case, the discriminant will be

$\Delta = {\left(- 5\right)}^{2} - 4 \cdot 1 \cdot 6$

$\Delta = 25 - 24 = \textcolor{g r e e n}{1}$

When $\Delta > 0$, the quadratic equation has two distinct real roots that take the form

$\textcolor{b l u e}{{x}_{1 , 2} = \frac{- b \pm \sqrt{\Delta}}{2 a}}$

In your case, these roots will be

${x}_{1 , 2} = \frac{- \left(- 5\right) \pm \sqrt{1}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{5 \pm 1}{2} = \left\{\begin{matrix}{x}_{1} = \frac{5 + 1}{2} = 3 \\ {x}_{2} = \frac{5 - 1}{2} = 2\end{matrix}\right.$

This means that the function will have two $x$-intercepts at $x = 2$ and $x = 3$.

The $y$-intercept will be

$y = {\left(0\right)}^{2} - 5 \cdot \left(0\right) + 6 = 6$

The function will intercept the $x$-axis in the points $\left(2 , 0\right)$ and $\left(3 , 0\right)$, and the $y$-axis in the point $\left(0 , 6\right)$.

graph{x^2 - 5x + 6 [-10, 10, -5, 5]}