Question

Find the range in which the value of the function $\frac{x^2+34x-71}{x^2+2x-7}$ lies for all real values.

Answer 1

Now,

$\begin{array}{c}y=\frac{x^2+34x-71}{x^2+2x-7}\\ y{x}^2+2xy-7y=x^2+34x-71\\ x^2\left(1-y\right)+2x\left(17-y\right)+7y-71=0x\in R\\ 4{\left(17-y\right)}^2-4\left(1-y\right)\left(7y-71\right)\ge 0\\ 4\left({17}^2+y^2-34y\right)-4\left(-7y^2-71+78y\right)\ge 0\\ \Rightarrow 8y^2-112y+360\ge 0\\ \Rightarrow y^2-14y+45\ge 0\\ \Rightarrow \left(y-9\right)\left(y-5\right)\ge 0\\ y\in \left(-\infty ,5\right)\cup \left(9,\infty \right)\end{array}$