To find the range, you need to locate the vertex:
equation for vertex = ax^2 + bx +c
in x^2 + 3, a = 1, b = 0 and c = 3
now vertex form = a(x+d)^2 +e
d = b/2a = 0/2*1 = 0
e = c-b^2/4a = 3-0^2/4*1 = 3
The equation becomes: (x+0)^2 +3.
The range is Y ≥ 3 ( greater than or equal to 3)
Answer:
Range: [tex][3,\infty)[/tex]
Step-by-step explanation:
We have been given a function [tex]f(x)=x^2+3[/tex]. We are asked to find the range of our given function.
We know that range of a function is values of dependent variable (y) for which function is defined.
We can see that our given parabola is in vertex form [tex]y=a(x-h)^2+K[/tex] as [tex]y=1(x-0)^2+3[/tex] with a vertex at point [tex](0,3)[/tex].
Since our given parabola is an upward opening parabola, so point [tex](0,3)[/tex] is the minimum point.
We know that the range of an exponential function is form [tex]ax^2+bx+c[/tex] with a vertex at (h,k) is:
If [tex]a>0[/tex], then range is [tex]f(x)\geq k[/tex]
If [tex]a<0[/tex], then range is [tex]f(x)\leq k[/tex]
Since the value of a is positive and [tex]k=3[/tex], therefore, the range of our given function would be [tex]f(x)\geq 3[/tex] that is [tex][3,\infty)[/tex].
