# How do you find the vertex and intercepts for y = x^2 - 4x + 1?

Apr 11, 2016

Vertex (2, -3)

#### Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{4}{2} = 2$
y-coordinate of vertex --> when x = 2
y(2) = 4 - 8 + 1 = -3
Vertex (2, -3)
To find y-intercept, make x = 0 --> y = 1
To find x-intercepts, make y = 0, and solve the quadratic equation:
${x}^{2} - 4 x + 1 = 0$ by the improved quadratic formula (Socratic Search)
$D = {d}^{2} = {b}^{2} - 4 a c = 16 - 4 = 12$ --> $d = \pm 2 \sqrt{3}$
There are 2 x-intercepts (real roots):
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{4}{2} \pm \frac{2 \sqrt{3}}{2} = 2 \pm \sqrt{3}$