Find two points on the graph of the parabola other than the vertex and x-intercepts.
k ( x ) = ( x − 1 )^(2) − 6
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.
To find two points on the graph of the parabola \(k(x) = (x - 1)^2 - 6\) other than the vertex and x-intercepts, you can choose any two values of \(x\) and then calculate the corresponding \(y\) values.
Let's choose \(x = 0\) as the first value. Substitute this value into the equation to find the corresponding \(y\) value:
\(k(0) = (0 - 1)^2 - 6\)
\(k(0) = (-1)^2 - 6\)
\(k(0) = 1 - 6\)
\(k(0) = -5\)
So, when \(x = 0\), \(y = -5\). Therefore, the first point is (0, -5).
Now, let's choose \(x = 2\) as the second value. Substitute this value into the equation to find the corresponding \(y\) value:
\(k(2) = (2 - 1)^2 - 6\)
\(k(2) = (1)^2 - 6\)
\(k(2) = 1 - 6\)
\(k(2) = -5\)
So, when \(x = 2\), \(y = -5\). Therefore, the second point is (2, -5).
Therefore, two points on the graph of the parabola \(k(x) = (x - 1)^2 - 6\) other than the vertex and x-intercepts are (0, -5) and (2, -5).