Answer:
The function has two x-intercepts at (0,0) and (4,0).
Step-by-step explanation:
Consider the provided function.
[tex]y+4=(x-2)^2[/tex]
The x-intercepts are the point, where y is 0.
To find x-intercepts, substitute y = 0 in the provided equation and solve for x.
[tex]0+4=(x-2)^2[/tex]
[tex]4=(x-2)^2[/tex]
[tex]\pm\sqrt{4}=x-2[/tex]
[tex]\pm2=x-2[/tex]
[tex]2=x-2[/tex] or [tex]-2=x-2[/tex]
[tex]2+2=x[/tex] or [tex]-2+2=x[/tex]
[tex]4=x[/tex] or [tex]0=x[/tex]
Therefore, the function has two x-intercepts at (0,0) and (4,0).