For the quadratic function $f (x) = a x^{2} + b x + c,$ the $y -$ coordinate of the vertex is the maximum value of the function when $a < 0 .$

Let's identify the values of

a,

b,

and

c

in the given quadratic function.

y = - x^{2} - 4 x - 2 ⇕ y = - 1 x^{2} + (- 4) x + (- 2)

We can see above that $a = - 1,$ $b = - 4,$ and $c = - 2 .$ We will now use these values to find the desired information.

We need to think of

y

as a function of

x, y = f (x) .

Since

a = - 1

is less than

0,

the parabola will open downwards. This means it will have a maximum value, which is given by

f (- \frac{b}{2 a}) .

Before we find the value of the function at this point, we need to substitute

a = - 1

and

b = - 4

in

- \frac{b}{2 a} .

- \frac{b}{2 a}

▼

Substitute values and evaluate

$a = - 1$ , $b = - 4$

- \frac{- 4}{2 ( - 1 )}

$a (- b) = - a \cdot b$

- \frac{- 4}{- 2}

$- \frac{a}{- b} = \frac{a}{b}$

\frac{- 4}{2}

Calculate quotient

- 2

Now we have to calculate

f (- 2) .

To do so, we will substitute

- 2

for

x

in the given function.

f (x) = - x^{2} - 4 x - 2

$x = - 2$

f (- 2) = - (- 2)^{2} - 4 (- 2) - 2

▼

Simplify right-hand side

Calculate power

f (- 2) = - 4 - 4 (- 2) - 2

f (- 2) = - 4 + 8 - 2

Add and subtract terms

f (- 2) = 2

This tells us that the maximum value of the function is

2 .

Unless there are any specified restrictions on the

x -

values, the domain of a quadratic function is all real numbers. Therefore, the domain of this function is all real numbers. Furthermore, since

a = - 1

is less than

0,

the range is all values less than or equal to the maximum value,

2 .

Domain : Range : All real numbers y \leq 2

Since

a = - 1

is less than

0,

the function increases to the left of the maximum value and decreases to the right of the maximum value, which we know occurs at

- 2 .

Increasing Interval : Decreasing Interval : To the left of - 2 To the right of - 2

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