Question

Find the range of the function. f(x) = x2 + 3

Answer 1

Answer:

Range: $[3,\infty)$

Step-by-step explanation:

We have been given a function $f(x)=x^2+3$. 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 $y=a(x-h)^2+K$ as $y=1(x-0)^2+3$ with a vertex at point $(0,3)$.

Since our given parabola is an upward opening parabola, so point $(0,3)$ is the minimum point.

We know that the range of an exponential function is form $ax^2+bx+c$ with a vertex at (h,k) is:

If $a>0$, then range is $f(x)\geq k$

If $a$, then range is $f(x)\leq k$

Since the value of a is positive and $k=3$, therefore, the range of our given function would be $f(x)\geq 3$ that is $[3,\infty)$.

Answer 2

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)