Question

How do you find the range of `f(x) = -x^2 + 3`?

Answer 1

`{y|y<=3}`

Explanation:

Since this function is quadratic, its graph is a parabola and it either has a minimum or a maximum value for `y`. To solve for the minimum/maximum value, we convert our equation to the vertex form `y=a(x-h)^2+k` by "completing the square".

The number `a` determines whether the parabola opens upward or downward. This is important because it will tell us whether we're looking for the minimum or maximum value of `y`. If it is positive, then we are looking for the minimum value. If it is negative, we are looking for the maximum value. The number then `k` tells us the minimum/maximum value of `y`.

Luckily, the function `f(x)=-x^2+3` is already in vertex form. You can look at it this way:
`f(x)=-(x-0)^2+3`

First, let's look at `a`. In this equation, `a=-1`. Since it is negative, it means that we are looking for the maximum value of `y`.

Next, we look at `k`. In this equation, `k=3`, meaning 3 is the maximum value for `y`.

The range will then be `{y|y<=3}`. You may also write it in set interval notation as `(-oo,3]`.

Answer 2

`(-oo,3]`

Explanation:

The given function represents a parabola opening downwards with vertex at (0,3). Hence its range would be `(-oo, 3]`

Answer 3

I found: `-oo<` `y` `<=3`

Explanation:

This function is represented graphically by a downwards parabola; this is because the `-1` in front of the `x^2` term.
To find the range (= possible `y` values) we need to find the highest point reached by our parabola, the vertex.
The `x` coordinate of the vertex is given as: `x_v=-b/(2a)`
where the coefficients `b and a` are found writing the function in general form as:
`f(x)=ax^2+bx+c=-1x^2+0x+3`
so:
`x_v=-(0)/(-1*2)=0`
giving for `y` (substituting `x=0` into your function):
`y_v=f(0)=3`

So the range (= possible `y` values) is:
`-oo<` `y` `<=3`

Graphically:
graph{-x^2+3 [-10, 10, -5, 5]}