# How do you find the vertex and intercepts for y = x^2 - x + 2?

Nov 20, 2015

The x-coordinate of the vertex $= - \frac{b}{2 a}$

#### Explanation:

Remember, the general form of a quadratic is:

$a {x}^{2} + b x + c = 0$

x-coordinate of vertex $= \frac{- b}{2 a} = \frac{- \left(- 1\right)}{2 \times 1} = \frac{1}{2}$

So, when $x = \frac{1}{2}$, then $y = {\left(\frac{1}{2}\right)}^{2} - \left(\frac{1}{2}\right) + 2 = \frac{7}{4}$

VERTEX $= \left(\frac{1}{2} , \frac{7}{4}\right)$

Solve for $y = 0$ to find the y-intercepts :

$\left(x - 2\right) \left(x + 1\right) = 0$

So, $x = 2$ and $x = - 1$

The x-intercept is when $x = 0$

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