Remember that a quadratic function can be defined by equivalent expressions in different forms, which enable us to see different features of its graph. For example, these expressions define the same function:

$\begin{aligned} (x-3)(x-7) \qquad &\text {factored form}\\ x^2-10x+21 \qquad &\text {standard form}\\ (x-5)^2-4 \qquad &\text {vertex form}\end{aligned}$

• From factored form, we can tell that the $x$-intercepts are $(3,0)$ and $(7,0)$.
• From standard form, we can tell that the $y$-intercept is $(0,21)$.
• From vertex form, we can tell that the vertex is $(5,\text-4)$.

Graph of a quadratic function on a grid. X axis from 0 to 9, by 1’s. Y axis from negative 8 to 20, by 4’s. Origin O. Graph opens upward with points 0 comma 21, 3 comma 0, and 7 comma 0. Vertex at 5 comma negative 4.

Recall that a function expressed in vertex form is written as $a(x-h)^2 + k$. The values of $h$ and $k$ reveal the vertex of the graph: $(h, k)$ are the coordinates of the vertex. In this example, $a$ is 1, $h$ is 5, and $k$ is -4.

• If we have an expression in vertex form, we can rewrite it in standard form by using the distributive property and combining like terms.

  Let’s say we want to rewrite $(x-1)^2-4$ in standard form.

  $\displaystyle \begin{aligned} &(x-1)^2-4 \\ &(x-1)(x-1)-4 \\& x^2-2x+1-4 \\&x^2-2x-3 \end{aligned}$

• If we have an expression in standard form, we can rewrite it in vertex form by completing the square.

  Let’s rewrite $x^2 + 10x +24$ in vertex form.

  A perfect square would be $x^2 + 10x +25$, so we need to add 1. Adding 1, however, would change the expression. To keep the new expression equivalent to the original one, we will need to both add 1 and subtract 1.

  $\displaystyle \begin{aligned} &x^2 + 10x +24 \\& x^2 + 10x +24 + 1 - 1 \\ &x^2 + 10x + 25 - 1 \\ &(x+5)^2-1 \end{aligned}$

• Let’s rewrite another expression in vertex form: $\text-2x^2 + 12x - 30$.

  To make it easier to complete the square, we can use the distributive property to rewrite the expression with -2 as a factor, which gives $\text-2 (x^2 -6x + 15)$.

  For the expression in the parentheses to be a perfect square, we need $x^2 -6x + 9$. We have 15 in the expression, so we can subtract 6 from it to get 9, and then add 6 again to keep the value of the expression unchanged. Then, we can rewrite $x^2-6x+9$ in factored form.

  $\displaystyle \begin{aligned} &\text-2x^2 + 12x -30 \\&\text-2(x^2 -6x +15) \\& \text-2(x^2 -6x +15 -6 +6) \\ &\text-2(x^2 -6x +9 +6) \\ &\text-2\left((x-3)^2 + 6\right) \end{aligned}$

  This expression is not yet in vertex form, however. To finish up, we need to apply the distributive property again so that the expression is of the form $a(x-h)^2 + k$:

  $\displaystyle \begin{aligned} &\text-2\left((x-3)^2 + 6\right) \\&\text-2(x-3)^2 -12 \end{aligned}$

  When written in this form, we can see that the vertex of the graph representing $\text-2(x-3)^2 -12$ is $(3, \text-12)$.