find x and y intercepts. f(x)=(x-3)^2 + 5.?

f(x) = (x - 3)² + 5
y = (x - 3)² + 5

When y = 0 :
0 = (x - 3)² + 5
(x - 3)² = 5
x - 3 = ±√5
x = 3 ± √5
The x-intercepts are 3 + √5 and 3 - √5

When x = 0 :
y = (0 - 3)² + 5
y = 14
The y-intercept is 14.

y-intercept: (0, 14). Solve when x = 0.

x-intercepts: None. Here's why:

0 = (x - 3)^2 + 5

-5 = (x - 3)^2. STOP HERE.

You can't square a real number and get a negative real number. Not even if you square a negative real number; you'd get a positive one. Also if you notice you got the vertex form; (3, 5) is the vertex and global minimum. So there won't be any x-intercepts.

(x)

x = 2, y = -1