# How many intercepts does y = x^2 − 6x + 9 have?

Aug 12, 2015

One $x$-intercept and one $y$-intercept.

#### Explanation:

You can determine $x$-intercepts by making the function equal to zero and $y$-intercepts by evaluating the function for $x = 0$.

When $y = 0$, you have

${x}^{2} - 6 x + 9 = 0$

In order to determine how many solutions this quadratic equation has, you can calculate the value of its discriminant, $\Delta$.

For a quadratic equation that takes the general form

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

the discriminant is equal to

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

In your case, you have $a = 1$, $b = - 6$, and $c = 9$, which means that the discriminant is equal to

$\Delta = {\left(- 6\right)}^{2} - 4 \cdot 1 \cdot 9$

$\Delta = 36 - 36 = \textcolor{g r e e n}{0}$

When the discriminant is equal to zero, your equation will only have one real solution (a repeated root) that takes the form

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- b \pm 0}{2 a} = - \frac{b}{2 a}$

In your case, the root will be

$x = - \frac{\left(- 6\right)}{2 \cdot 1} = \frac{6}{2} = 3$

This means that the function has one $x$-intercept, $x = 3$.

The $y$-intercept will be

$y = {\left(0\right)}^{2} - 6 \cdot \left(0\right) + 9 = 9$

The function will thus intercept the $y$-axis in the point $\left(0 , 9\right)$ and the $x$-acis in the point $\left(3 , 0\right)$.

graph{x^2 - 6x + 9 [-10, 10, -5, 5]}