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

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

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

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

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

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.