# How do you graph f(x)=3(x-4)^2+2 and identify the vertex, axis of symmetry, domain, range, max or min values, increasing and decreasing intervals?

Oct 31, 2016

The vertex is $\left(4 , 2\right)$
The axis of symmetry is $x = 4$
The domain is $\mathbb{R}$
The range is $\left(4 \le f \left(x\right) < + \infty\right)$
The min value is $\left(4 , 2\right)$

#### Explanation:

The vertex is when $x = 4$$\implies$$f \left(x\right) = 3 \cdot 0 + 2 = 2$
So the vertex is at $\left(4 , 2\right)$
The axis of symmetry is $x = 4$
For a polynomial function, the domain is $\mathbb{R}$
We need to calculate the derivative
$f ' \left(x\right) = 6 \left(x - 4\right)$
We have min or max when $f ' \left(x\right) = 0$
So, $6 \left(x - 4\right) = 0$ $\implies$ $x = 4$
We can do a chart
$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$4$$\textcolor{w h i t e}{a a a a a}$$+ \infty$
$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$
$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$\downarrow$$\textcolor{w h i t e}{a a a a a}$$\uparrow$

So we have a min at $x = 4$
So the range is $\left(4 \le f \left(x\right) < + \infty\right)$
The decreasing values are $- \infty < x < = 4$
The increasing values are $4 \le x < + \infty$

graph{3(x-4)^2+2 [-9.94, 12.56, -2.92, 8.33]}