How do I find the X-Intercepts of f(x)=(x-2)^2 -9

Please help. Thanks.

(x-2)^2-9
x^2-4x+4-9
=x^2-4x-5
When You solve for the x-intercepts
you factorise the equation.
x^2-4x-5
=(x-5)(x+1)
x=5,-1
Thus,the x-intercepts are (5,0),(-1,0)
Graph this to see if my answer is right.

The function given is in the vertex form. From there, you can get the vertex V:(2,-9).

To get the x-intercepts, equate the function to zero because the x-intercepts are the x values when y=0.

f(x)=(x-2)^2-9
0=x^2-4x+4-9
0=x^2-4x-5
solve by factoring
0=(x-5)(x+1)
Therefore, x=5 or x=-1

The y-intercept is always the constant in standard form, in this case, -5.

for x intercept value of Y (here f(x)) should be zero
=> (x-2)^2 - 9 = 0
=> (x-2)^2 = 9
=> x-2 = 3 or x-2 = -3
=> x = 5 or x = -1
the intercepts are 5 or -1

You can solve it in two ways,
1) let (x-2)^2 = 9
2) expand (x-2)^2 and minus the 9 from it
But no matter what, to find X-intercepts, let (x-2)^2-9=0

Let f(x)=0

0=((X-2)^2)-9
0=X^2 + 4x -5
0=(x-5)(x+1)

therefore, x-intercepts are
(5,0) and (-1,0)

0 = (x-2)^2 -9
9= = (x-2)^2
root(9)= x-2
x= root(9) +2
x= (+/-)3 + 2

when x= 5/-1 f(x)=0

0 = (x-2)^2 - 9

(x-2)^2 = 9

x-2 = +/- 3

x = 2 +/- 3

ie. x = 5 or -1