Find two points on the graph of the parabola other than the vertex and x-intercepts.

Question

k ( x ) = ( x − 1 )^(2) − 6

Answer 1

k(x) = (x−1)^2 − 6
Clearly the vertex is at (1,-6),
The x-intercepts are not integers,
The y-intercept is at (0,-5)

So, pick any integer value except 0 or 1 for x, and plug it in!

k(-1) = -2
k(10) = 75
k(-100) = 10195

You can use decimal or irrational values as well, but they're more trouble to evaluate.