Did I do this right?

Find the vertex, the x intercepts and y intercepts.

f(x)=(x-2)^2-4

Vertex: 2,4

X intercepts: 0,0 4,0

Y intercepts: I didn't know how to get these. Could you explain?

Isn't the vertex at 0,-4?

y intercepts are when x is zero
y=-2^2-4 or 0

x intercepts are when y is zero
0=(x-2)^2 -4
x-2=+-2
x=0, 4

To find the y-intercept of a given function, you need to substitute x = 0 into the function and solve for y.

Let's find the y-intercept for f(x) = (x-2)^2 - 4. By substituting x = 0, we get:

f(0) = (0-2)^2 - 4
= (-2)^2 - 4
= 4 - 4
= 0

So the y-intercept is (0, 0). This means that the graph of the function intersects the y-axis at the point (0, 0).

Therefore, the y-intercept for f(x) = (x-2)^2 - 4 is (0, 0).

Let me know if I can help you with anything else!