ANSWER
The correct answer is D.
EXPLANATION
The function given to us to graph is
[tex]y=(x-1)^2+3[/tex]
This function is written in the form
[tex]y=a(x-h)^2+k[/tex]
Where [tex](h,k)[/tex] is the vertex.
By comparing to the given function,
[tex]y=(x-1)^2+3[/tex]
The coordinates of the vertex are
[tex](1,3)[/tex].
Also [tex]a=1\:>\:0[/tex], the graph opens up.
We also need to determine the x and y intercepts.
At x-intercept [tex]y=0[/tex]
[tex]0=(x-1)^2+3[/tex]
[tex]-3=(x-1)^2[/tex]
We need to take the square of both sides. But square root of negative 3 gives an imaginary number. This means the graph will be hanging. It won't touch the x-axis.
At y-intercept [tex]x=0[/tex]
[tex]y=(0-1)^2+3[/tex]
[tex]y=1+3=4[/tex]
The y-intercept is [tex](0,4)[/tex].
We can now use the above information to graph the function as shown in the attachment.
We can see from the graph that [tex]y=(x-1)^2+3[/tex] is obtained when the graph of [tex]f(x)=x^2[/tex] is shifted up 3 units and to the right 1 unit.
