# Proof for the Vertex Form of Quadratics Start with the standard form and complete the square: $ \begin{gather} y = ax^2+bx+c \\ y = a \left(x^2+\frac b a x \right) + c \end{gather} $ The square we're interested in is: $\left(x+\frac b {2a} \right)^2 = x^2 + \frac b a x + \frac {b^2}{4a^2}$ We now plug in the square into the bracket above to get: $ \begin{gather} y = a \left(x^2+\frac b a x \right) + c \\ y = a \left( \left(x+\frac b {2a} \right)^2 - \frac {b^2}{4a^2} \right) + c \\ y = a \left(x+\frac b {2a} \right)^2 - \frac {b^2}{4a} + c \\ y = a \left(x- \left( - \frac b {2a} \right) \right)^2 - \frac {b^2}{4a} + c \tag{1} \end{gather} $ At the vertex of the quadratic its first derivative is 0:[^1] $ \begin{gather} \frac d {dx} \left [ ax^2 + bx + c \right] = 2ax + b \\ 2ax_v+b = 0 \\ x_v = - \frac b {2a} \end{gather} $ This is exactly what we have inside the brackets in the expression (1). We get $y_v$ by plugging in $x_v$: $ \begin{gather} y_v = a \left ( - \frac b {2a} \right)^2 + b\left( - \frac b {2a} \right) + c \\ y_v = a\frac {b^2}{4a^2} - \frac {b^2} {2a} + c \\ y_v = \frac {b^2}{4a} - \frac {2b^2} {4a} + c \\ y_v = -\frac {b^2}{4a} + c \\ \end{gather} $ $y_v$ also makes an appearance in expression (1), which we can now simplify into: $ \begin{gather} y = a \left(x- \left( - \frac b {2a} \right) \right)^2 - \frac {b^2}{4a} + c \\ y = a \left(x- x_v \right)^2 +y_v \end{gather} $ Which is known as the vertex form of a quadratic. [^1]: We can figure this out without calculus by applying [[Vieta's Formulas]], which tell us that the sum of the two roots is $r_1 + r_2 = \frac b a$. The $x$ of the vertex is the mid point: $\frac {r_1 + r_2} 2$. Similar reasoning is used to prove the simpler version of the quadratic formula: [[The Quadratic Formula#Proof for the Simpler Version]].