Finding the equation of a parabola

[Flatland]

How do you find the equation of a parabola if you are given it's vertex and 1 point? For example, find the quadratic equation of a parabola that has a vertex of (-2,-2) and goes through the point (-1,0)

 

[stalefish]

General quadratic equation: y=ax^2 + bx + c

Derivative: y' = 2ax + b

At the vertex, the derivate equals to zero. Use this fact and simultaneous equations to arrive at the equation.

 

[HallsofIvy]

More simply, since you are given the vertex of the parabola, you can write the equation y= a(x-x₀)²+ y₀ where x₀ and y₀ are the coordinates of the vertex. Choose a to force the parabola to go through the second point.

y= a(x-(-2))²- 2= a(x+2)²- 2. Setting x= -1, y= 0,
0= a(-1+2)²-2= a- 2 so a= 2.

By the way, this is assuming the parabola has a vertical line of symmetry. Otherwise there are an infinite number of parabolas satisfying these conditions.

 

[ivybond]

Know your parabolas

... assuming the parabola has a vertical line of symmetry...

Graph first, and you may find a shortcut for a given specific data.
If a point on a parabola is 1 to the right and 2 up from its vertex, it must be parabola
$$y = 2x^{2}$$
shifted horizontally and vertically, so its vertex (0,0) moves into (-2,-2), i.e. 2 to the left and 2 down:
$$y = 2(x+2)^{2} - 2$$