Rewrite the function by completing the square.

Question

\[h(x)= x^{2} +3 x -18\]
\[h(x)=\]
\[(x+\]
\[)^2+\]

Answer 1

To complete the square for the function \( h(x) = x^2 + 3x - 18 \), follow these steps:

1. Start with the quadratic expression \( x^2 + 3x \).
2. To complete the square, take the coefficient of \( x \) (which is 3), divide it by 2, and then square it: \[ \left(\frac{3}{2}\right)^2 = \frac{9}{4}. \]
3. Add and subtract this squared term inside the function: \[ h(x) = x^2 + 3x + \frac{9}{4} - \frac{9}{4} - 18. \]
4. Now, rewrite the quadratic part as a square: \[ h(x) = \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} - 18. \]
5. Combine the constant terms: \[ -\frac{9}{4} - 18 = -\frac{9}{4} - \frac{72}{4} = -\frac{81}{4}. \]

Putting it all together, we have: \[ h(x) = \left(x + \frac{3}{2}\right)^2 - \frac{81}{4}. \]

Thus, the completed square form of \( h(x) \) is: \[ h(x) = \left(x + \frac{3}{2}\right)^2 - \frac{81}{4}. \]