Question

Find two points on the graph of the parabola other than the vertex and x-intercepts. k(x)=(x−4)^2−4

Answer 1

(0,12) and (1,5).

Explanation:

y = (x−4)^2 − 4.

The y-intercept occurs when x = 0 so

(0-4)^2 - 4

= 12

So one point is (0,12).

When x = 1,

y = (1-4)^2 - 4

= (-3)^2 - 4

= 5

So another point is (1, 5).