We have an answer to "Finding the range of y=x2+2x+42x2+4x+9 (and y=quadraticquadratic in..."

High School
Algebra
Algebra I
Quadratic function and equation

Kinsley Moon

Answered question

2022-01-31

Finding the range of

y = \frac{x^{2} + 2 x + 4}{2 x^{2} + 4 x + 9}

(and

y = \frac{q u a d r a t i c}{q u a d r a t i c}

in general)

Answer & Explanation

Sean Becker

Beginner2022-02-01Added 16 answers

The question can be easily solved by this technique: As

y = \frac{x^{2} + 2 x + 4}{2 x^{2} + 4 x + 9} \Rightarrow 2 y = \frac{2 x^{2} + 4 x + 9 - 1}{2 x^{2} + 4 x + 9}

. Thus,

2 y = 1 - \frac{1}{2 (x + 1)^{2} + 7}

Squares can never be less than zero so the minmum value of the function:

2 (x + 1)^{2} + 7

would be 7, or Maximum value of

\frac{1}{2 (x + 1)^{2} + 7} is \frac{1}{7}

. 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 [\frac{3}{7}, \frac{1}{2})

ebbonxah

Beginner2022-02-02Added 15 answers

y = \frac{x^{2} + 2 x + 4}{2 x^{2} + 4 x + 9} = \frac{2 x^{2} + 4 x + 8}{2 (2 x^{2} + 4 x + 9)} = \frac{2 x^{2} + 4 x + 9 - 1}{2 (2 x^{2} + 4 x + 9)} = \frac{1}{2} (1 - \frac{1}{2 x^{2} + 4 x + 9}) \Rightarrow 2 y = 1 - \frac{1}{2 x^{2} + 4 x + 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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