Calculus (3rd Edition)

We have parabola $y=ax^2+bx+c$. The square completion is done as follows: Step 1: Take coefficient of $x^2$ common throughout. $y=a\left(x^2+\dfrac{b}{a}x+\dfrac{c}{a}\right)$ Step 2: Add and subtract $\left(\dfrac{\text{Coefficient of x}}{2}\right)^2=\left(\dfrac{b}{2a}\right)^2$ inside the parenthesis. $\Rightarrow y=a\left(x^2+\dfrac{b}{a}x+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2+\dfrac{c}{a}\right)$ Step 3: Multiply and divide term with $x$ by 2. $\Rightarrow y=a\left(x^2+2\cdot \dfrac{b}{2a}\cdot x+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2+\dfrac{c}{a}\right)$ Step 4: Complete square and simplify. $\Rightarrow y=a\left(\left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2+\dfrac{c}{a}\right)$ $\Rightarrow y=a\left(\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a^2}+\dfrac{c}{a}\right)$ $\Rightarrow y=a\left(\left(x+\dfrac{b}{2a}\right)^2+\dfrac{-b^2+4ac}{4a^2}\right)$ $\Rightarrow y=a\left(x+\dfrac{b}{2a}\right)^2+\dfrac{-b^2+4ac}{4a}$ Thus, equation is of the form $y=a(x+h)^2+k$, where $h=\dfrac{b}{2a}$ and $k=\dfrac{-b^2+4ac}{4a}$, which is congruent to $y=ax^2$ with horizontal translation by $|h|$ units and vertical translation by $|k|$ units. Hence, proved.