# 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
$$(p , 0) = (0.94 , 0 )$$

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

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

## 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 .