If x is real then the range of `(x^(2)-2x+9)/(x^(2)+2x+9)` is

Options

A. `(-oo, 0] uu (1, oo)`
B. `[(1)/(2), 2]`
C. `(-oo, -(2)/(9)]uu(1, oo)`
D. `(-oo, -6]uu[-2, oo)`

Answer

To find the range of the function \( y = \frac{x^2 - 2x + 9}{x^2 + 2x + 9} \), we will follow these steps:

### Step 1: Set up the equation
Let \( y = \frac{x^2 - 2x + 9}{x^2 + 2x + 9} \).

### Step 2: Rearranging the equation
We can rearrange this equation to eliminate the fraction:
\[ y(x^2 + 2x + 9) = x^2 - 2x + 9 \]
This simplifies to:
\[ yx^2 + 2yx + 9y = x^2 - 2x + 9 \]

### Step 3: Collecting like terms
Rearranging gives us:
\[ (y - 1)x^2 + (2y + 2)x + (9y - 9) = 0 \]
This is a quadratic equation in \( x \).

### Step 4: Finding the discriminant
For \( x \) to be real, the discriminant of this quadratic must be non-negative:
\[ D = b^2 - 4ac \]
Here, \( a = y - 1 \), \( b = 2y + 2 \), and \( c = 9y - 9 \). Thus, the discriminant is:
\[ D = (2y + 2)^2 - 4(y - 1)(9y - 9) \]

### Step 5: Expanding the discriminant
Calculating \( D \):
\[ D = (2y + 2)^2 - 4(y - 1)(9y - 9) \]
Expanding \( (2y + 2)^2 \):
\[ D = 4y^2 + 8y + 4 \]
Now expanding \( 4(y - 1)(9y - 9) \):
\[ 4(y - 1)(9y - 9) = 36y^2 - 36y - 36 \]
Thus, the discriminant becomes:
\[ D = 4y^2 + 8y + 4 - (36y^2 - 36y - 36) \]
\[ D = 4y^2 + 8y + 4 - 36y^2 + 36y + 36 \]
Combining like terms:
\[ D = -32y^2 + 44y + 40 \]

### Step 6: Setting the discriminant greater than or equal to zero
We need:
\[ -32y^2 + 44y + 40 \geq 0 \]
Dividing through by -4 (and reversing the inequality):
\[ 8y^2 - 11y - 10 \leq 0 \]

### Step 7: Finding the roots of the quadratic
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 8 \cdot (-10)}}{2 \cdot 8} \]
Calculating the discriminant:
\[ 121 + 320 = 441 \]
Thus:
\[ y = \frac{11 \pm 21}{16} \]
Calculating the two roots:
1. \( y = \frac{32}{16} = 2 \)
2. \( y = \frac{-10}{16} = -\frac{5}{8} \)

### Step 8: Testing intervals
We need to test the intervals determined by the roots \( y = -\frac{5}{8} \) and \( y = 2 \):
- For \( y < -\frac{5}{8} \): \( 8y^2 - 11y - 10 > 0 \)
- For \( -\frac{5}{8} < y < 2 \): \( 8y^2 - 11y - 10 \leq 0 \)
- For \( y > 2 \): \( 8y^2 - 11y - 10 > 0 \)

### Conclusion: Range of the function
The range of the function is:
\[ y \in \left[-\frac{5}{8}, 2\right] \]