Question 4633: How does one find the particular equation of the quadratic function that passes through a vertex of (5, -1) and the point (8, -28)?
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! depends how much maths you know :-)
General quadratic is  . Put in both sets of (x,y) to get 2 equations in a and b. Solve these and ten re-write the general equation.
Alternatively, if you know calculus, the gradient, dy/dx of the general equation is 2ax+b. At the vertex, the gradient is zero, so put that into the equation and the x-value, and you get that 10a+b=0. Then put the values of the other point into the QUADRATIC general equation and use the fact that b=-10a, to find a.
Both methods are much the same. Take your pick.
jon.
|
|
|
