Answer:
The x-intercepts are: ( 3 + i√5, 0) and (3 - i√5, 0).
The y-intercept is (0,5).
The vertex is at (3, 5).
Explanation:
The given equation y=(x-3)^2+5 has the general form y=(x-h)^2+k, whose vertex is at (h,k).
Thus, the vertex of y=(x-3)^2+5 is (3,5).
Find the x-intercepts by setting y=(x-3)^2+5 = 0 and solving for x. To do this, solve (x-3)^2+5 = 0 for x. Subtracting 5 from both sides, we get:
(x-3)^2 = -5. We want to solve this for x.
Taking the square root of both sides, we get:
x-3 = √-1 * √5. Then x = 3 + i√5 and x = 3 - i√5. The x-intercepts are:
( 3 + i√5, 0) and (3 - i√5, 0).