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} - 6 x + 9 \Leftrightarrow f (x) = 1 x^{2} + (- 6) x + 9

We can see that

a

=

1,

b

=

- 6,

and

c

=

9 .

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 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

$a = 1$ , $b = - 6$

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

▼

Simplify

Multiply

- \frac{- 6}{2}

RemoveNegFracAndNum

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

\frac{6}{2}

CalcQuot

Calculate quotient

3

Now, we can find the

y -

coordinate by evaluating the function for

x = 3 .

f (x) = x^{2} - 6 x + 9

Substitute

$x = 3$

f (3) = 3^{2} - 6 (3) + 9

▼

Simplify

CalcPow

Calculate power

f (3) = 9 - 6 (3) + 9

Multiply

f (3) = 9 - 18 + 9

AddSubTerms

Add and subtract terms

f (3) = 0

The minimum value of our function is

0 .

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

0,

the range is all real numbers greater than or equal to

0 .

Domain : Range : {all real numbers} {f (x) ∣ f (x) \geq 0}

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