# How do you find the range of the function y=f(x)=x/(x^2-5x+9)?

Apr 22, 2015

The answer is : $- \frac{1}{11} \le y \le 1$

Solution
$y = \frac{x}{{x}^{2} - 5 x + 9}$

$\implies y {x}^{2} - 5 y x + 9 y = x$

Send all terms to the left

$\implies y {x}^{2} + \left(- 1 - 5 y\right) x + 9 y = 0$

For real roots, ${b}^{2} - 4 a c \ge 0$

$\implies {\left(- 1 - 5 y\right)}^{2} - 4 \left(y\right) \left(9 y\right) \ge 0$

$\implies 1 + 10 y - 11 {y}^{2} \ge 0$

Now, all you've got to do is to factorize and find the range :)

$\implies \left(- 1 - 11 y\right) \left(- 1 + y\right) \ge 0$

Or $\left(11 y + 1\right) \left(y - 1\right) \le 0$

Hence $- \frac{1}{11} \le y \le 1$