Question

Based on the graph of j(x)=x2+16x+48
, write the domain and range.(2 points)
The domain is
<x<
. The range is y≥
.

Answer 1

To analyze the function \( j(x) = x^2 + 16x + 48 \), we can start by determining its domain and range.

### Domain
The domain of a polynomial function is always all real numbers. Therefore, the domain is:

\[
\text{Domain: } -\infty < x < \infty
\]

### Range
To find the range, we can identify the vertex of the parabola represented by the quadratic function. The general form of a quadratic function is \( ax^2 + bx + c \).

For \( j(x) \):
- \( a = 1 \) (which is positive, indicating the parabola opens upwards)
- \( b = 16 \)
- \( c = 48 \)

The x-coordinate of the vertex is given by the formula \( x = -\frac{b}{2a} \):

\[
x = -\frac{16}{2(1)} = -8
\]

Now, we can find the corresponding y-coordinate by substituting \( x = -8 \) back into the function:

\[
j(-8) = (-8)^2 + 16(-8) + 48 = 64 - 128 + 48 = -16
\]

The vertex of the parabola is at \( (-8, -16) \). Since the parabola opens upwards, the minimum value of \( j(x) \) is at the y-coordinate of the vertex.

Thus, the range of the function is all values of \( y \) greater than or equal to \(-16\):

\[
\text{Range: } y \ge -16
\]

In summary:

- **Domain:** \( -\infty < x < \infty \)
- **Range:** \( y \ge -16 \)