# Vertex Form of a Parabola

## (finding the equation of a parabola)

We learn how to find the equation of a parabola by writing it in vertex form

In the previous section, we learnt how to write a parabola in its vertex form and saw that a parabola's equation: $y = ax^2+bx+c$ could be re-written in vertex form: $y = a\begin{pmatrix}x - h \end{pmatrix}^2+k$ where:

• $$h$$: is the horizontal coordinate of the vertex
• $$k$$: is the vertical coordinate of the vertex.
This is illustrated here:

We can use the vertex form to find a parabola's equation. The idea is to use the coordinates of its vertex (maximum point, or minimum point) to write its equation in the form $$y=a\begin{pmatrix}x-h\end{pmatrix}^2+k$$ (assuming we can read the coordinates $$\begin{pmatrix}h,k\end{pmatrix}$$ from the graph) and then to find the value of the coefficient $$a$$.

This is explained in the step-by-stem mathod, below, as well as in the tutorials.

## How to find a parabola's equation using its Vertex Form

Given the graph of a parabola for which we're given, or can clearly see:

• the coordinates of the vertex, $$\begin{pmatrix}h,k\end{pmatrix}$$, and:
• the coordinates another point $$P$$ through which the parabola passes.
we can find the parabola's equation in vertex form following two steps:
• Step 1: use the (known) coordinates of the vertex, $$\begin{pmatrix}h,k\end{pmatrix}$$, to write the parabola's equation in the form: $y = a\begin{pmatrix}x-h \end{pmatrix}^2+k$ the problem now only consists of having to find the value of the coefficient $$a$$.
• Step 2: find the value of the coefficient $$a$$ by substituting the coordinates of point $$P$$ into the equation written in step 1 and solving for $$a$$.

This two-step method is better/further explained in the following tutorial, take the time to watch it now.

## Tutorial 1

In the following tutorial we learn how to find a parabola's using the coordinates of its vertex as well as the coordinates of its $$y$$-intercept.

## Exercise 1

Using the two-step method we've just learned, find the equation of each of the following parabola in the following two forms:

1. $$y=a\begin{pmatrix}x-h\end{pmatrix}^2+k$$
2. $$y = ax^2+bx+c$$

Note: this exercise can be downloaded as a worksheet to practice with

• Parabola 1:
1. $$y = 4 \begin{pmatrix}x-2\end{pmatrix}^2-7$$
2. $$y = 4x^2 - 16x + 9$$
• Parabola 2:
1. $$y = \begin{pmatrix}x-2\end{pmatrix}^2+ 3$$
2. $$y = x^2 - 4x +7$$
• Parabola 3:
1. $$y = -3 \begin{pmatrix}x+3 \end{pmatrix}^2+1$$
2. $$y=-3x^2 - 18x - 26$$
• Parabola 4:
1. $$y = 12 \begin{pmatrix}x-1\end{pmatrix}^2-4$$
2. $$y = 12x^2 - 24x + 8$$
• Parabola 5:
1. $$y = 2\begin{pmatrix}x+1 \end{pmatrix}^2+1$$
2. $$y = 2x^2+4x+3$$

## Tutorial 2

In the following tutorial we learn how to find a parabola's using the coordinates of its vertex as well as the coordinates of another point along its length (that isn't the $$y$$-intercept).

## Exercise 2

For each of the parabola shown here:

1. Find its equation in vertex form $$y=a\begin{pmatrix}x - h \end{pmatrix}^2+k$$
2. Find its equation in standard form $$y=ax^2+bx+c$$

Note: this exercise can be downloaded as a worksheet to practice with:

• Parabola 1:
1. $$y = 3 \begin{pmatrix}x-2\end{pmatrix}^2+1$$
2. $$y = 3x^2 - 12x + 13$$
• Parabola 2:
1. $$y = -2\begin{pmatrix}x-3\end{pmatrix}^2+ 5$$
2. $$y = -2x^2+12x-13$$
• Parabola 3:
1. $$y = 2 \begin{pmatrix}x+4 \end{pmatrix}^2-3$$
2. $$y=2x^2+16x + 29$$
• Parabola 4:
1. $$y = - \begin{pmatrix}x+1\end{pmatrix}^2-3$$
2. $$y = -x^2-2x-4$$
• Parabola 5:
1. $$y = 4\begin{pmatrix}x-3 \end{pmatrix}^2-5$$
2. $$y = 4x^2 - 12x + 31$$