Rewrite the function by completing the square.

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

1 answer

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: \[ h(x) = \left(x + \frac{3}{2}\right)^2 - \frac{81}{4} \]