Question

How do you complete the square of f(x)=-1/2x^2+4x

Answer 1

First factor out the \(-1/2\). This gets you: \[
-2 \cdot f(x)=x^2-8x
\]Notice how: \[
(x+a)^2 = x^2+2ax+a^2
\]In this case \(2a=-8\) thus \(a=-4\). So we want: \[
x^2-8x+16
\]We do so by adding to both sides:\[
-2 \cdot + 16 f(x)=x^2-8x+16 = (x-4)^2
\]

Answer 2

\[
-2 \cdot f(x)+16=x^2-8x+16 = (x-4)^2
\]Then we solve for \(f(x)\), keeping our square but undoing other the stuff we did.\[ \begin{array}{rcl}
-2\cdot f(x) &=& (x-4)^2-16 \\
f(x) &=& -\frac{1}{2}(x-4)^2+8
\end{array}
\]

Answer 3

Thank you!!!