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

1 Answer
Nghi N.
Apr 11, 2016

Vertex (2, -3)

Explanation:

x-coordinate of vertex:
#x = -b/(2a) = 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 - 4x + 1 = 0 # by the improved quadratic formula (Socratic Search)
#D = d^2 = b^2 - 4ac = 16 - 4 = 12# --> #d = +- 2sqrt3#
There are 2 x-intercepts (real roots):
#x = - b/(2a) +- d/(2a) = 4/2 +- (2sqrt3)/2 = 2 +- sqrt3#

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