![]() ![]() Now we've reduced the problem to a two-equations, two-unknowns one. If the graph of a quadratic function opens up and the vertex is below the x-axis or if the graph opens down and the vertex is above the x-axis, then there will be two x-intercepts. Now set two pairs of equations equal to one another, eliminating C: equations 1 and 2, 2 and 3: If we solve all three equations for C, we get: We first use two sets of two equations to eliminate one of the variables ( C is usually easiest in these problems), then use those two equations to solve for the other two, and finally use the results to find the third variable. A quadratic function is a polynomial function with one or more variables. These three equations can be solved simultaneously by methods you already know. Notice that none of these appears to be the vertex of the parabola, but that won't matter. It's probably best at this point to work an example or two, so let's find the equation of the parabola above, which passes through (-4, 8), (1, -4) and (3, -2). Now that's three equations and three unknowns ( A, B, C), so we should have enough information to determine what they are. If each of out three points satisfies this equation for a certain set of A, B and C (the things we're really looking for), we have three equations: Let's say that a parabola passes through three known points, $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3).$ Now we know that the general form of the equation we're looking for is (a) Plot and recognise quadratic, cubic, reciprocal, exponential and circular functions. We'd like to be able to find the equation of that parabola. It's a very specific kind of curve, the graph of a quadratic function. That may be a bit hard to come to terms with at first, but remember that a parabola isn't just any curve. Through any three points, only one unique parabola can be drawn. ![]()
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