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Parabola vertex coordinates

When we mark several ordered pairs of a 2nd degree function, the graph we obtain corresponds to a parabola. The vertex is nothing more than a point of the function at which it changes direction.

In this way, the vertex is associated with concavity of the parabola, which can be the minimum point or the maximum point:

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  • When the parabola is concave upwards, then the vertex is the minimum point of the function.
  • When the parabola is concave downwards, then the vertex is the maximum point of the function.

If the vertex is a point on the parabola, then it has coordinates. But what are the coordinates of the vertex? Is there a formula to find these coordinates?

Yes. There are a few ways to find the coordinates of the vertex of a parabola. Next, we will show one of them.

How to calculate the coordinates of the vertex of the parabola

Considering a function of the 2nd degree, \dpi{120} \mathrm{f (x) ax^2 + bx + c}, the vertex of the parabola is a point \dpi{120} \mathrm{V(x_v, y_v)}, with coordinates given by:

\dpi{120} \mathrm{x_v \frac{-b}{2.a}} \: \: e\: \: \mathrm{y_v \frac{-\Delta }{4.a}} On what \dpi{120} \Delta \mathrm{ b^2 - 4.a.c} it's called discriminating and corresponds to the same value that we calculated to apply in the bhaskara's formula and find the roots of a 2nd degree equation.

Example of vertex of the parabola.
Example of vertex of the parabola.

Example: Determine the vertex of the function f(x) = x² + 3x – 28.

In this function, we have a = 1, b= 3 and c = -28.

Applying these values ​​in the formulas, we have:

\dpi{120} \mathrm{x_v \frac{-b}{2.a} \frac{-3}{2}} -1.5

It is

\dpi{120} \mathrm{y_v \frac{-\Delta }{4.a} \frac{-121}{4}} -30.25

\dpi{120} \Delta \mathrm{ b^2 - 4.a.c 3^2 - 4.1.(-28)} 9 + 112 121.

Therefore, the vertex of the function is the point V(-1,5; -30,25).

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