# Find Equation of a Parabola from a Graph

Several methods are used to find equations of parabolas given their graphs. Examples are presented along with their detailed solutions and exercises.

## Examples with Detailed Solutions

Example 1 Graph of parabola given x and y intercepts

Find the equation of the parabola whose graph is shown below.

__Solution to Example 1__

The graph has two x intercepts at \( x = - 1 \) and \( x = 2 \). Hence the equation of the parabola may be written as

\( y = a(x + 1)(x - 2) \)

We now need to find the coefficient \( a \) using the y intercept at \( (0,-2) \)

\( -2 = a(0 + 1)(0 - 2) \)

Solve the above equation for \( a \) to obtain

\( a = 1 \)

The equation of the parabola whose graph is given above is

\( y = (x + 1)(x - 2) = x^2 - x - 2\)

Example 2 Graph of parabola given vertex and a point

Find the equation of the parabola whose graph is shown below.

__Solution to Example 2__

The graph has a vertex at \( (2,3) \). Hence the equation of the parabola in vertex form may be written as

\( y = a(x - 2)^2 + 3 \)

We now use the y intercept at \( (0,- 1) \) to find coefficient \( a \).

\( - 1 = a(0 - 2) + 3\)

Solve the above for \( a \) to obtain

\( a = 2 \)

The equation of the parabola whose graph is shown above is

\( y = 2(x - 2)^2 + 3\)

Example 3 Graph of parabola given three points

Find the equation of the parabola whose graph is shown below.

__Solution to Example 3__

The equation of a parabola with vertical axis may be written as

\( y = a x^2 + b x + c \)

Three points on the given graph of the parabola have coordinates \( (-1,3), (0,-2) \) and \( (2,6) \). Use these points to write the system of equations

\(
\begin{array}{lcl} a (-1)^2 + b (-1) + c & = & 3 \\ a (0)^2 + b (0) + c & = & -2 \\ a (2)^2 + b (2) + c & = & 6 \end{array}
\)

Simplify and rewrite as

\(
\begin{array}{lcl} a - b + c & = & 3 \\ c & = & -2 \\ 4 a + 2 b + c & = & 6 \end{array}
\)

Solve the above 3 by 3 system of linear equations to obtain the solution

\( a = 3 , b=-2 \) and \(c=-2 \)

The equation of the parabola is given by

\( y = 3 x^2 - 2 x - 2 \)

Example 4 Graph of parabola given diameter and depth

Find the equation of the parabolic reflector with diameter D = 2.3 meters and depth d = 0.35 meters and the coordinates of its focus.

__Solution to Example 4__

The parabolic reflector has a vertex at the origin \( (0,0) \), hence its equation is given by

\( y = \dfrac{1}{4p} x^2 \)

The diameter and depth given may be interpreted as a point of coordinates \( (D/2 , d) = (1.15 , 0.35) \) on the graph of the parabolic reflector. Hence the equation

\( 0.35 = \dfrac{1}{4p} (1.15)^2 \)

Solve the above equation for \( p \) to find

\(
p = 0.94
\)

The equation of the parabola is given by

\( y = 0.26 x^2 \)

The focus of the parabolic reflector is at the point

\( (p , 0) = (0.94 , 0 ) \)

## Exercises with Answers

Find the equation of the parabola in each of the graphs below

__Answers to the Above Exercises__

- \( y = x^2-3x-3 \)
- \( y = - (x + 2)^2 - 1 = - x^2 -4x -5 \)
- \( y = (x-2)(x+6) = x^2 + 4x - 12 \)

## More References and Links to Parabola

Equation of a parabola .Three Points Parabola Calculator.

Tutorial on How Parabolic Dish Antennas work?

Tutorial on how to Find The Focus of Parabolic Dish Antennas .

Use of parabolic shapes as Parabolic Reflectors and Antannas .