Question

How do you find the range of `f(x)=abs(x^2-8x+7)` for the domain `3 <= x <= 8`?

Answer 1

There are several approaches that will work quite well for this. Here's the one that I used:

Graph the parabola `y = x^2-8x+7`, the flip the negative part across the `x` axis to get the absolute value.

I chose this method, because it's relatively easy to find the `x` intercepts:
`x^2-8x+7 = 0`

`(x-7)(x-1) = 0`, the intercepts are `1,7` and the parabola opens upward:

The vertex has `x` coordinate equal to the midpoint between the intercepts, or `x=4` and `y` value
`y=(4)^2-8(4)+7 = 16-32+7 = -11`

graph{y = x^2-8x+7 [-18.32, 27.29, -11.85, 10.94]}

No graph the absolute values:

graph{y =abs( x^2-8x+7) [-7.87, 20.61, -1.88, 12.35]}

Now: `f(3) = 8` and `f(8) = 7`, but between `3` and `8` we will hit f(7)=0, so the range runs from `0` up to somewhere.

The maximum value of the absolute value on that domain will be the absolute value of the minimum of the parabola or `11`

The range for that domain is `0 <= y <= 11`