# What is the domain and range of f(x)= x^2 - 6x + 8?

Aug 9, 2015

Domain: $x \in R$ or $\left\{x : - \infty \le x \le \infty\right\}$. $x$ can take up any real values.

Range: $\left\{f \left(x\right) : - 1 \le f \left(x\right) \le \infty\right\}$

#### Explanation:

Domain:

$f \left(x\right)$ is a quadratic equation and any values of $x$ will give a real value of $f \left(x\right)$.

The function does not converge to a certain value ie: $f \left(x\right) = 0$ when $x \to \infty$

Your domain is $\left\{x : - \infty \le x \le \infty\right\}$.

Range:

Method 1-
Use completing the square method:
${x}^{2} - 6 x + 8 = {\left(x - 3\right)}^{2} - 1$
Hence you minimum point is $\left(3 , - 1\right)$. It is a minimum point because the graph is a "u" shape (coefficient of ${x}^{2}$ is positive).

Method 2-
Differentiate :
$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 2 x - 6$.

Let$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 0$

Therefore, $x = 3$ and $f \left(3\right) = - 1$
Minimum point is $\left(3 , - 1\right)$.
It is a minimum point because the graph is a "u" shape (coefficient of ${x}^{2}$ is positive).

Your range takes values between $- 1 \mathmr{and} \infty$

Aug 9, 2015

Domain $\left(- \infty , + \infty\right)$
Range $\left[- 1 , + \infty\right)$

#### Explanation:

It is a polynomial function, its domain is all real numbers. In interval notation this can be expressed as $\left(- \infty , + \infty\right)$
For finding its range, we can solve the equation y= ${x}^{2} - 6 x + 8$ for x first as follows:

$y = {\left(x - 3\right)}^{2} - 1$,
${\left(x - 3\right)}^{2} = y + 1$
x-3= $\pm \sqrt{y + 1}$
x= 3$\pm \sqrt{y + 1}$. It is obvious from this that y$\ge - 1$

Hence range is $y \ge - 1$. In interval notation this can be expressed as$\left[- 1 , + \infty\right)$