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

Mar 22, 2016

Vertex (1, -1)

Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{4}{4} = 1$
y-coordinate of vertex:
$\left(1\right) = 2 - 4 + 1 = - 1$
To find y-intercept, make x = 0 --> y = 1.
To find x-intercepts, make y = 0 and solve the quadratic equation:
$2 {x}^{2} - 4 x + 1 = 0.$
$D = {d}^{2} = {b}^{2} - 4 a c = 16 - 8 = 8$ --> $d = \pm 2 \sqrt{2}$
There are 2 x-intercepts (2 real roots):
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{4}{4} \pm \frac{2 \sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}$