For quadratic function $y=ax^2+bx+c$, where $a$, $b$ and $c$ are real numbers, and $a\neq 0$.
- The graph of $y=a(x-h)^2+k$
Properties
- If $a>0$, then the graph opens upwards. If $a<0$, then the graph opens downwards.
- The equation of the axis of the symmetry of the graph is $x=h$.
- The coordinates of the vertex is $(h,k)$.
- For $a>0$, the minimum value of $y$ is $k$ and the corresponding value of $x$ is $h$. For $a<0$, the maximum value of $y$ is $k$ and the corresponding value of $x$ is $h$.
- Completing the square
-
$\begin{array}{cl}
& x^2 + 8x + 12 \\
= & x^2 + 8x + \left(\dfrac{8}{2}\right)^2 – \left(\dfrac{8}{2}\right)^2 + 12 \\
= & (x^2 + 8x + 4^2) – 4^2 + 12 \\
= & (x+4)^2 – 4
\end{array}$ -
$\begin{array}{cl}
& x^2 – 8x – 12 \\
= & x^2 – 8x + \left(\dfrac{8}{2}\right)^2 – \left(\dfrac{8}{2}\right)^2 – 12 \\
= & (x^2 – 8x + 4^2) – 4^2 – 12 \\
= & (x-4)^2 – 26
\end{array}$ -
$\begin{array}{cl}
& 2x^2 + 8x + 12 \\
= & 2(x^2 + 4x) + 12 \\
= & 2[x^2 + 4x + \left(\dfrac{4}{2}\right)^2 – \left(\dfrac{4}{2}\right)^2] + 12 \\
= & 2(x^2 + 4x + 2^2) – 2(2)^2 + 12 \\
= & 2(x+2)^2 – 4
\end{array}$ -
$\begin{array}{cl}
& \dfrac{1}{3}x^2 + 4x + 15 \\
= & \dfrac{1}{3}(x^2 + 12x) + 15 \\
= & \dfrac{1}{3}[x^2 + 12x + \left(\dfrac{12}{2}\right)^2 – \left(\dfrac{12}{2}\right)^2] + 15 \\
= & \dfrac{1}{3}(x^2 + 12x + 6^2) – 12 + 15 \\
= & \dfrac{1}{3}(x+6)^2 + 3
\end{array}$
-
- The coordinates of the vertex $(h,k)$
\begin{equation*}
h=\frac{-b}{2a} \mbox{, } k=\frac{4ac-b^2}{4a}
\end{equation*}
