Let's begin by rewriting our function in the general expression of a quadratic function,

y = a x^{2} + b x + c .

f (x) = - x^{2} - 12 \Leftrightarrow f (x) = (- 1) x^{2} + 0 x + (- 12)

We can see that

a

=

- 1,

b

=

0,

and

c

=

- 12 .

Determining Maximum or Minimum Value

Before we determine the maximum or minimum recall that, if

a > 0,

the parabola has a minimum value. Conversely, if

a < 0,

the parabola has a maximum value.

In the given function, we have $a = - 1,$ which is less than $0 .$ Thus, the parabola will have a maximum value.

Finding the Maximum or Minimum Value

The minimum or maximum value of a parabola is always the

y -

coordinate of the vertex. We can find it by first looking for the

x -

coordinate of the vertex,

- \frac{b}{2 a} .

- \frac{b}{2 a}

SubstituteII

$a = - 1$ , $b = 0$

- \frac{0}{2 ( - 1 )}

ZeroPropMult

Zero Property of Multiplication

0

Now, we can find the

y -

coordinate by evaluating the function for

x = 0 .

f (x) = - x^{2} - 12

Substitute

$x = 0$

f (0) = - (0)^{2} - 12

ZeroPropMult

Zero Property of Multiplication

f (0) = - 12

The maximum value of our function is

- 12 .

Domain and Range

Unless there is a specific restriction given in the context of the problem, the domain of a quadratic function is all real numbers. In this case, there is no restriction on the value of

x .

Since the maximum value of the function is

- 12,

the range is all real numbers less than or equal to

- 12 .

Domain : Range : {all real numbers} {f (x) ∣ f (x) \leq - 12}

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