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help me find range

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Find the range of the function h(x) = (5x^2 + 20x + 23)/(x^2 + 8x + 5).

Dec 28, 2020

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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 $$\quad\:f\left(x\right)\ge\frac{\sqrt{421}+16}{11}$$.

Dec 29, 2020