# What is the range of the function f(x)=|x^2-8x+7| ?

Aug 24, 2017

The range is: $0 \le f \left(x\right) < \infty$

#### Explanation:

The quadratic ${x}^{2} - 8 x + 7$ has zeros:

${x}^{2} - 8 x + 7 = 0$

$\left(x - 1\right) \left(x - 7\right) = 0$

$x = 1 \mathmr{and} x = 7$

Between 1 and 7 the quadratic is negative but the absolute value function will make these values positive, therefore, 0 is the minimum value of $f \left(x\right)$.

Because the value of the quadratic approaches $\infty$ as x approaches $\pm \infty$, the upper limit for f(x) does the same.

The range is $0 \le f \left(x\right) < \infty$

Here is a graph of f(x):

graph{|x^2 - 8x + 7| [-15.04, 13.43, -5.14, 9.1]}