## How To Write The Equation Of A Parabola

### Equation Of A Parabola

A parabola is a quadratic function or second degree polynomial that resembles a horse shoe and is symmetric with respect to either the x or y axis. The vertex of a parabola is the peak or valley of the function where the curve changes direction.

A parabola, just like a circle, ellipse and hyperbola, is a cross sectional cut of a conical section. For a parabola, similar to an ellipse, the cross sectional cut is taken at an angle with respect to the cone that gives it more of a rounded, u shape. The vertex of a parabola is the functions starting point and the location where it is commonly drawn from. The parabola equation of quadratic function can be used to create a data table of coordinates for points along the curve. A parabola symmetric about its y-axis may have one y-intercept (where it crosses the y-axis) and will have either two x-intercepts (where it crosses the x-axis) or none at all. Conversely, a parabola symmetric about its x-axis will have a single x-intercept and will have either two or zero y-intercepts.

The standard equation of a parabola is represented by the following.

Parabola, symmetric with respect to the x-axis:

x = ay2 + by + c

Parabola, symmetric with respect to the y-axis:

y = ax2 + bx + c

The vertex of a parabola can be found by the following formulations.

x-coordinate of the vertex: x = -b / 2a

y-coordinate of the vertex: y = (4ac â€“ b2) / 4a

A couple of examples are presented here.

1.) Find the vertex of the following function: x = 2y2 â€“ 8y + 3. Also find the coordinate of the x-intercept.

We recognize that this quadratic function is symmetric about the x-axis and the vertex can be found as follows.

x-coordinate = -(-8) / [2(2)] = 2

y-coordinate = [(4)(2)(3) â€“ (-8)2] / (4)(2) = -40 / 8 = -5

Vertex: (2, -5)

The x-intercept is the location where the curve crosses the x-axis, so to determine this point we set y = 0.

x = 2(0)2 â€“8(0) + 3

x = 3

x-intercept: (3, 0)

2.) Write the equation of the parabola with vertex (3/2, 1/2) and with the y-intercept of

(0, 5).

Since we have a single y-intercept, we know the function crosses the y-axis once and we know that the parabola is symmetric about the y-axis. The parabola takes the following form.

y = ax2 + bx + c

Since at x = 0, y = 5, we know that c = 5. We will look at the formulations for the x and y coordinates of the vertex.

y = [4ac â€“ b2] / 4a

1/2 = [4a(5) â€“ b2] / 4a

1/2 = (20a â€“ b2) / 4a

2a = 20a â€“ b2

-18a = -b2

a = b2 / 18

We will plug in this value for a into x = -b / 2a.

x = -b / 2a

3/2 = -b / [(2)(b2 / 18)]

3/2 = -b / [2b2 / 18)]

3/2 = -18b / 2b2

3/2 = -9 / b

(3/2)b = -9

b = -6

Now we can plug this value back into a = b2 / 18.

a = (-6)2 / 18

a = 36 / 18 = 2

So the equation of our parabola is y = 2x2 â€“ 6x + 5.

REFERENCE:

Bittinger, Marvin L., Ellenbogen, David, Keedy, Mervin L. (1994). Intermediate Algebra: Concepts and Application (4th ed.). USA: Addison-Wesley Publishing Company, Inc.