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

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

f (x) = 2 x^{2} - 16 x - 42 \Leftrightarrow f (x) = 2 x^{2} + (- 16) x + (- 42)

We can see that

a

=

2,

b

=

- 16,

and

c

=

- 42 .

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 $a = 2,$ which is greater than $0 .$ Thus, the parabola will have a minimum 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

$a = 2$ , $b = - 16$

- \frac{- 1 6}{2 ( 2 )}

▼

Simplify

Multiply

- \frac{- 1 6}{4}

RemoveNegFracAndNum

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

\frac{1 6}{4}

CalcQuot

Calculate quotient

4

Now, we can find the

y -

coordinate by evaluating the function for

x = 4 .

f (x) = 2 x^{2} - 16 x - 42

Substitute

$x = 4$

f (4) = 2 (4)^{2} - 16 (4) - 42

▼

Simplify

CalcPow

Calculate power

f (4) = 2 (16) - 16 (4) - 42

Multiply

f (4) = 32 - 64 - 42

SubTerms

Subtract terms

f (4) = - 74

The minimum value of our function is

- 74 .

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 minimum value of the function is

- 74,

the range is all real numbers greater than or equal to

- 74 .

Domain : Range : {all real numbers} {f (x) ∣ f (x) \geq - 74}

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