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

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