Question

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

Answer 1

Vertex (1, -1)

Explanation:

x-coordinate of vertex:
`x = -b/(2a) = 4/4 = 1`
y-coordinate of vertex:
`(1) = 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:
`2x^2 - 4x + 1 = 0.`
`D = d^2 = b^2 - 4ac = 16 - 8 = 8` --> `d = +- 2sqrt2`
There are 2 x-intercepts (2 real roots):
`x = -b/(2a) +- d/(2a) = 4/4 +- (2sqrt2)/4 = 1 +- sqrt2/2`