Answer:
[tex]g(x)=1(x-0)^{2}+4[/tex]
Step-by-step explanation:
Given:
The function,[tex]f(x)=x^{2}[/tex]
A standard parabola with vertex at origin is represented as:
[tex]f(x)=ax^{2}[/tex]
The above function is a standard parabola with vertex at origin and opening upward.
The vertex of the above function is at the origin [tex](0,0)[/tex] and the value of [tex]a[/tex] is 1.
Now, the function is translated 4 units up.
So, from the rule of function transformations, if a graph is moved up by [tex]c[/tex] units, then [tex]c[/tex] units is added to the function.
Therefore,
[tex]g(x)=f(x) + 4\\g(x)=x^{2}+4[/tex].
Also, the vertex of [tex]f(x)[/tex] will be translated 4 units up. So, the co-ordinates of the vertex of [tex]g(x)[/tex] will be [tex](0,0+4)=(0,4)[/tex]
Now, express the above function in the vertex form [tex]g(x)=a(x-h)^{2}+k[/tex].
We have [tex]a=1,h=0,k=4[/tex]
This gives,
[tex]g(x)=1(x-0)^{2}+4[/tex]