A quadratic equation can be written in a way that makes the vertex easy to see. The vertex is the turning point of a parabola (or the point of the “V” on an absolute value graph). It’s where the graph reaches a maximum or minimum value.

This way of writing a quadratic is called vertex form:

$$y=a(x-h{)}^2+k$$

A similar form works for absolute value functions:

$$y=a|x-h|+k$$

Here’s what each parameter means:

• $h$ is the horizontal shift (how far the graph moves left or right from $x=0$).
• $k$ is the vertical shift (how far the graph moves up or down from $y=0$).
• $a$ is a constant that controls the shape:
  • $|a|>1$ makes the graph narrower (a vertical stretch).
  • $0<|a|<1$ makes the graph wider (a vertical shrink).
  • If $a<0$, the graph is reflected over the $x$-axis (flipped upside down).

The key benefit of vertex form is that the vertex is immediately visible: the vertex is at the point $(h,k)$. This also helps you picture the graph as a transformation of the parent function (like $y=x^2$ or $y=|x|$) without graphing from scratch.

Find the equation from a graph - quadratic

You may be given a graph and asked to find the equation that matches it. The main idea is:

• Find the vertex from the graph.
• Use the vertex to fill in $h$ and $k$ in vertex form.
• Use another point on the graph to solve for $a$.

We will find the equation associated with the following graph:

Start with the vertex. The vertex is at $(-2,-3)$, so:

$$\begin{gathered} h=-2 \\ k=-3 \end{gathered}$$

Substitute these values into the vertex form of a quadratic:

$$\begin{array}{rl}y& =a(x-h{)}^2+k\\ y& =a(x+2{)}^2-3\end{array}$$

Now solve for $a$ using any other point on the graph. From the graph, when $x=0$, $y=1$. Substitute $(0,1)$ into the equation:

$$\begin{array}{rl}y& =a(0+2{)}^2-3\\ 1& =a(2{)}^2-3\\ 1& =4a-3\\ 4& =4a\\ a& =1\end{array}$$

Now substitute $a=1$ to write the final equation:

$$y=(x+2{)}^2-3$$

So the process is: find the vertex, write the vertex-form equation, use another point to solve for $a$, then write the final equation.

Find the equation from a graph - absolute value

Using the same approach, we will find the equation associated with the following absolute value graph:

Start with the vertex:

$$\begin{gathered} h=-1 \\ k=1 \end{gathered}$$

Substitute into the absolute value vertex form:

$$y=a|x+1|+1$$

Now solve for $a$ using another point on the graph. Use the point $(0,3)$:

$$\begin{array}{rl}3& =a|0+1|+1\\ 3& =a+1\\ a& =2\end{array}$$

Form the final equation using this value of $a$:

$$y=2|x+1|+1$$

Find the graph from an equation

This is the reverse of what you did above: you’re given an equation and asked to identify the graph.

In vertex form, the vertex is always $(h,k)$, so you can locate the vertex immediately. Then use $a$ to determine the shape:

• If $a>1$, the graph is narrower (vertical stretch).
• If $0<a<1$, the graph is wider (vertical shrink).
• If $a<0$, the graph is reflected over the $x$-axis (flipped upside down).

Sometimes you won’t be asked to draw the graph, but instead to describe the transformations from the parent function. The parent function is the simplest version of the function:

• Quadratic parent function: $y=x^2$
• Absolute value parent function: $y=|x|$

For example:

What transformations exist for the function given by $y=2\ast (x+3{)}^2+4$?

Identify $a$, $h$, and $k$ by matching the equation to $y=a(x-h{)}^2+k$:

$$\begin{array}{rl}a& =2\\ h& =-3\\ k& =4\end{array}$$

Given this, you can describe the transformations:

• Since $a=2$ (positive and greater than $1$), the parabola has a vertical stretch by a factor of $2.
• Since $h=-3$, the graph is shifted left 3 (a horizontal translation).
• Since $k=4$, the graph is shifted up 4 (a vertical translation).