Respuesta :
[tex]f(x)=a(x-h)^2+k \Rightarrow \text{vertex}=(h,k)\\\\
f(x)=(x-2)^2+4 \Rightarrow \text{vertex}=(2,4)[/tex]
The range of [tex]f(x)=a(x-h)^2+k[/tex] is
[tex]y\leq k[/tex] for [tex]a<0[/tex]
[tex]y\geq k[/tex] for [tex]a>0[/tex]
The domain of any quadratic function is all real numbers.
In [tex] f(x)=(x-2)^2+4[/tex], [tex]a=1\ \textgreater \ 0[/tex], so the range is [tex]y\geq4[/tex]
So it's D.
The range of [tex]f(x)=a(x-h)^2+k[/tex] is
[tex]y\leq k[/tex] for [tex]a<0[/tex]
[tex]y\geq k[/tex] for [tex]a>0[/tex]
The domain of any quadratic function is all real numbers.
In [tex] f(x)=(x-2)^2+4[/tex], [tex]a=1\ \textgreater \ 0[/tex], so the range is [tex]y\geq4[/tex]
So it's D.
Answer:
The answer is D. The vertex is (2,4), the domain is all real numbers, and the range is y is greater than or equal to 4
Step-by-step explanation: