 Kinsley Moon

2022-01-31

Finding the range of $y=\frac{{x}^{2}+2x+4}{2{x}^{2}+4x+9}$ (and $y=\frac{quadratic}{quadratic}$ in general) Sean Becker

Expert

The question can be easily solved by this technique: As $y=\frac{{x}^{2}+2x+4}{2{x}^{2}+4x+9}⇒2y=\frac{2{x}^{2}+4x+9-1}{2{x}^{2}+4x+9}$. Thus, $2y=1-\frac{1}{2\left(x+1{\right)}^{2}+7}$ Squares can never be less than zero so the minmum value of the function: $2\left(x+1{\right)}^{2}+7$ would be 7, or Maximum value of . This tells that miminum value of y will be $\frac{3}{7}$. From here you can easily tell the maximum and minimum values: $y\in \left[\frac{3}{7},\frac{1}{2}\right)$ ebbonxah

Expert

$y=\frac{{x}^{2}+2x+4}{2{x}^{2}+4x+9}=\frac{2{x}^{2}+4x+8}{2\left(2{x}^{2}+4x+9\right)}=\frac{2{x}^{2}+4x+9-1}{2\left(2{x}^{2}+4x+9\right)}=\frac{1}{2}\left(1-\frac{1}{2{x}^{2}+4x+9}\right)⇒2y=1-\frac{1}{2{x}^{2}+4x+9}$.
Now find the extremes of the range of the expression in the RHS of the above equation (which I believe you can; if not someone else or I myself shall try and add it) and divide them by 2 to get the required extremes(taking half since we get values for 2y and not y).

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