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

Apr 26, 2015

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

Graph the parabola $y = {x}^{2} - 8 x + 7$, 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:
${x}^{2} - 8 x + 7 = 0$

$\left(x - 7\right) \left(x - 1\right) = 0$, 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
$y = {\left(4\right)}^{2} - 8 \left(4\right) + 7 = 16 - 32 + 7 = - 11$

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: $f \left(3\right) = 8$ and $f \left(8\right) = 7$, 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 $0 \le y \le 11$