+122 Solution: \(\mathrm{Range\:of\:}\frac{5x^2+20x+23}{x^2+8x+5}:\quad\begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le\frac{-\sqrt{421}+16}{11}\quad\mathrm{or}\quad\:f\left(x\right)\ge\frac{\sqrt{421}+16}{11}\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty\:,\:\frac{-\sqrt{421}+16}{11}]\cup\:[\frac{\sqrt{421}+16}{11},\:\infty\:)\end{bmatrix}\)
Steps: Rewrite \(\frac{5x^2+20x+23}{x^2+8x+5}\)as \(\frac{5x^2+20x+23}{x^2+8x+5}=y\)
\(=\frac{5x^2+20x+23}{x^2+8x+5}=y\)
Multiply both sides by \(x^2+8x+5\).
\(=\frac{5x^2+20x+23}{x^2+8x+5}\left(x^2+8x+5\right)=y\left(x^2+8x+5\right)\)
Simplify \(\frac{5x^2+20x+23}{x^2+8x+5}\left(x^2+8x+5\right):5x^2+20x+23\)
\(=5x^2+20x+23=y\left(x^2+8x+5\right)\)
The range is the set of y for which the discriminant is greater or equal to zero
Discriminant \(5x^2+20x+23=y\left(x^2+8x+5\right):\quad44y^2-128y-60\)
\(44y^2-128y-60\ge\:0\quad:\quad y\le\frac{-\sqrt{421}+16}{11}\quad\) or \(\quad\:y\ge\frac{\sqrt{421}+16}{11}\)
Check if the range interval endpoints are included
\(y=\frac{-\sqrt{421}+16}{11}\quad:\quad\)Included
\(y=\frac{\sqrt{421}+16}{11}\quad:\quad\)Included
Therefore, the range is \(f\left(x\right)\le\frac{-\sqrt{421}+16}{11}\quad\) or \(\quad\:f\left(x\right)\ge\frac{\sqrt{421}+16}{11}\).
