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

Sep 21, 2015

$\left\{y | y \le 3\right\}$

#### 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 $y = a {\left(x - h\right)}^{2} + k$ 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 $f \left(x\right) = - {x}^{2} + 3$ is already in vertex form. You can look at it this way:
$f \left(x\right) = - {\left(x - 0\right)}^{2} + 3$

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 $\left\{y | y \le 3\right\}$. You may also write it in set interval notation as $\left(- \infty , 3\right]$.

Sep 21, 2015

$\left(- \infty , 3\right]$

#### Explanation:

The given function represents a parabola opening downwards with vertex at (0,3). Hence its range would be $\left(- \infty , 3\right]$

Sep 21, 2015

I found: $- \infty <$$y$$\le 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: ${x}_{v} = - \frac{b}{2 a}$
where the coefficients $b \mathmr{and} a$ are found writing the function in general form as:
$f \left(x\right) = a {x}^{2} + b x + c = - 1 {x}^{2} + 0 x + 3$
so:
${x}_{v} = - \frac{0}{- 1 \cdot 2} = 0$
giving for $y$ (substituting $x = 0$ into your function):
${y}_{v} = f \left(0\right) = 3$

So the range (= possible $y$ values) is:
$- \infty <$$y$$\le 3$

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