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)
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.
More simply, since you are given the vertex of the parabola, you can write the equation y= a(x-x_{0})^{2}+ y_{0} where x_{0} and y_{0} 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.
Know your parabolas 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 [tex]y = 2x^{2}[/tex] shifted horizontally and vertically, so its vertex (0,0) moves into (-2,-2), i.e. 2 to the left and 2 down: [tex]y = 2(x+2)^{2} - 2[/tex]