Question

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?

Answer 1

The vertex is `(4,2)`
The axis of symmetry is `x=4`
The domain is `RR`
The range is `(4<=f(x)<+oo)`
The min value is `(4,2)`

Explanation:

The vertex is when `x=4` `=>` `f(x)=3*0+2=2`
So the vertex is at `(4,2)`
The axis of symmetry is `x=4`
For a polynomial function, the domain is `RR`
We need to calculate the derivative
`f'(x)=6(x-4)`
We have min or max when `f'(x)=0`
So, `6(x-4)=0` `=>` `x=4`
We can do a chart
`x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaa)` `4` `color(white)(aaaaa)` `+oo`
`f'(x)` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `+`
`f(x)` `color(white)(aaaaaa)` `darr` `color(white)(aaaaa)` `uarr`

So we have a min at `x=4`
So the range is `(4<=f(x)<+oo)`
The decreasing values are `-oo < x < =4`
The increasing values are `4<=x<+oo`

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