The range of the function `y=(x+2)/(x^2-8x-4)`

Options

A. `(-oo,-(1)/(4)] cup [-(1)/(20),oo)`
B. `(-oo,-(1)/(4)) cup (-(1)/(20),oo)`
C. `(-oo,-(1)/(4)] cup (-(1)/(20),oo)`
D. none of these

Answer

To find the range of the function \( y = \frac{x + 2}{x^2 - 8x - 4} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ y = \frac{x + 2}{x^2 - 8x - 4} \]
We can rearrange it to express it in a standard form:
\[ y(x^2 - 8x - 4) = x + 2 \]
This simplifies to:
\[ yx^2 - 8yx - 4y = x + 2 \]
Rearranging gives us:
\[ yx^2 - (8y + 1)x - (4y + 2) = 0 \]

### Step 2: Identify coefficients
From the quadratic equation \( ax^2 + bx + c = 0 \), we identify:
- \( a = y \)
- \( b = -(8y + 1) \)
- \( c = -(4y + 2) \)

### Step 3: Calculate the discriminant
For the quadratic equation to have real roots, the discriminant \( D \) must be greater than or equal to zero:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-(8y + 1))^2 - 4(y)(-(4y + 2)) \]
This simplifies to:
\[ D = (8y + 1)^2 + 4y(4y + 2) \]

### Step 4: Expand and simplify the discriminant
Expanding the discriminant:
\[ D = (8y + 1)^2 + 16y^2 + 8y \]
Calculating \( (8y + 1)^2 \):
\[ D = 64y^2 + 16y + 1 + 16y^2 + 8y \]
Combining like terms:
\[ D = 80y^2 + 24y + 1 \]

### Step 5: Set the discriminant greater than or equal to zero
To find the range, we need:
\[ 80y^2 + 24y + 1 \geq 0 \]

### Step 6: Find the roots of the quadratic
Using the quadratic formula \( y = \frac{-b \pm \sqrt{D}}{2a} \):
\[ y = \frac{-24 \pm \sqrt{(24)^2 - 4 \cdot 80 \cdot 1}}{2 \cdot 80} \]
Calculating the discriminant:
\[ D = 576 - 320 = 256 \]
Thus:
\[ y = \frac{-24 \pm 16}{160} \]
Calculating the two roots:
1. \( y_1 = \frac{-8}{160} = -\frac{1}{20} \)
2. \( y_2 = \frac{-40}{160} = -\frac{1}{4} \)

### Step 7: Analyze the intervals
The roots are \( y = -\frac{1}{20} \) and \( y = -\frac{1}{4} \). We will test intervals around these points:
1. For \( y < -\frac{1}{4} \)
2. For \( -\frac{1}{4} < y < -\frac{1}{20} \)
3. For \( y > -\frac{1}{20} \)

### Step 8: Determine sign of the quadratic
Using test points:
- For \( y = -1 \) (in the first interval), \( 80(-1)^2 + 24(-1) + 1 = 80 - 24 + 1 = 57 > 0 \)
- For \( y = 0 \) (in the second interval), \( 80(0)^2 + 24(0) + 1 = 1 > 0 \)
- For \( y = 1 \) (in the third interval), \( 80(1)^2 + 24(1) + 1 = 105 > 0 \)

### Step 9: Conclusion on the range
The quadratic \( 80y^2 + 24y + 1 \) is positive for:
- \( y \leq -\frac{1}{4} \) or \( y \geq -\frac{1}{20} \)

Thus, the range of the function is:
\[ (-\infty, -\frac{1}{4}] \cup [-\frac{1}{20}, \infty) \]