Completing the square of three parabolas

Text

Complete the square, rewriting the following parabolas in a form that makes the vertex coordinates explicit:

• (a) $y=x^2+6x$
• (b) $y=-2x^2+x-1$
• (c) $y=\dfrac{x^2}{4}-\dfrac{x}{8}+3$

Solution

(a) $y=x^2+6x=(x+3)^2-9$. Vertex $V\left(-3;-9\right)$.

(b) Factor out $-2$: $y=-2\left(x^2-\tfrac{x}{2}\right)-1=-2\left(x-\tfrac14\right)^2+2\cdot\tfrac{1}{16}-1=-2\left(x-\tfrac14\right)^2-\tfrac78$. Vertex $V\left(\tfrac14;-\tfrac78\right)$.

(c) Factor out $\tfrac14$: $y=\tfrac14\left(x^2-\tfrac{x}{2}\right)+3=\tfrac14\left(x-\tfrac14\right)^2-\tfrac14\cdot\tfrac{1}{16}+3=\tfrac14\left(x-\tfrac14\right)^2+\tfrac{191}{64}$. Vertex $V\left(\tfrac14;\tfrac{191}{64}\right)$.

$\boxed{V_a(-3;-9),\quad V_b\left(\tfrac14;-\tfrac78\right),\quad V_c\left(\tfrac14;\tfrac{191}{64}\right)}$