What’s a function equivalent to n(x)=1/2(x+4)^2+12

Answer 1

To find a function equivalent to \( n(x) = \frac{1}{2}(x + 4)^2 + 12 \), we can expand it.

1. Start with the expression:

   \[ n(x) = \frac{1}{2}(x + 4)^2 + 12 \]

2. Expand \( (x + 4)^2 \):

   \[ (x + 4)^2 = x^2 + 8x + 16 \]

3. Substitute this expansion back into \( n(x) \):

   \[ n(x) = \frac{1}{2}(x^2 + 8x + 16) + 12 \]

4. Distribute \( \frac{1}{2} \):

   \[ n(x) = \frac{1}{2}x^2 + \frac{8}{2}x + \frac{16}{2} + 12 \]

   This simplifies to:

   \[ n(x) = \frac{1}{2}x^2 + 4x + 8 + 12 \]

5. Combine like terms:

   \[ n(x) = \frac{1}{2}x^2 + 4x + 20 \]

Thus, an equivalent function is:

\[ n(x) = \frac{1}{2}x^2 + 4x + 20 \]