Question

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

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