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^2-2x+1
where a=1 b=-2 c=1
but he graph of kx=ax^2+bx+c
the co-ordinate x of th vertex is -b/2a then the y co-ordinate is h(-b/2a)
xco-ordinate=-b/2a=-2/2(1)=-1
y cordinate=-b/2a=h(-1)=(-1)^2-2( -1)+5=0=6
y-intercept now
but y-intercept now is given by y=c
from your graph c=5
hx=x^2-2x+5 has c=5
so the y intercept is given by 5
the x-intercept if they exist i guess are given by setting
ax^2+bx+c=0
it can be determine by
x=-b+_root(b^2-4ac)/2a
opps i see b^2-4ac equivalent
then there may be no x intercept
the parabola may be either above or below
but x^2-2x+1=0
(x^2-x)-(x+1)=0
x(x-1)-1(x-1)=0
(x-1)^2=0
x-1=0
x=1
so the x intercept is 1

Answer 2

Well, you did some calculations (some of them incorrect), but at the end, you didn't even come close to answering the question that was asked.

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.