Question

Rewrite the function by completing the square.
\[h(x)= x^{2} +3 x -18\]
\[h(x)=\]
\[(x+\]
\[)^2+\]

Answer 1

To rewrite the function \( h(x) = x^2 + 3x - 18 \) by completing the square, we follow these steps:

1. Start with the quadratic expression:
\[
h(x) = x^2 + 3x - 18
\]

2. Focus on the quadratic and linear terms, \( x^2 + 3x \). To complete the square, we take half of the coefficient of \( x \), square it, and then add and subtract that square within the expression.

The coefficient of \( x \) is 3. Half of 3 is \( \frac{3}{2} \), and squaring it gives:
\[
\left(\frac{3}{2}\right)^2 = \frac{9}{4}
\]

3. We rewrite \( h(x) \) as follows:
\[
h(x) = \left(x^2 + 3x + \frac{9}{4}\right) - \frac{9}{4} - 18
\]

4. Now simplify the expression:
\[
h(x) = \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{72}{4}
\]
\[
h(x) = \left(x + \frac{3}{2}\right)^2 - \frac{81}{4}
\]

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

Putting it in the requested format:
\[
h(x) = \left(x + \frac{3}{2}\right)^2 - \frac{81}{4}
\]
So, the completed square form is:
\[
h(x) = \left(x + \frac{3}{2}\right)^2 - \frac{81}{4}
\]