How do you write #y=x^2-2x+8# in vertex form and identify the vertex, y intercept and x intercept?

1 Answer
Nghi N.
May 19, 2015

#y = x^2 - 2x + 8#

x of vertex: #x = (-b/2a) = 2/2 #= 1#

y of vertex: #f(1) = 1 - 2 + 8 = 7#

Vertex form: #y = (x - 1)^2 + 7.#

Find x-intercepts by solving quadratic equation:
#y = x^2 - 2x + 8 = 0#.
Use the new AC Method to solve quadratic equation. Compose factor pairs of c = 8: (-2, 4). This sum is -2 + 4 = 2 = -b. Then the 2 real roots (x -intercepts) are: -2 and 4.

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