How is the vertex calculated

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The vertex of quadratic functions

On this page the symmetry of quadratic functions is proven and a formula for the coordinates of the vertex is derived. There is also a form → below for calculating the vertex, the vertex shape and the points of intersection with the coordinate axes.

All quadratic functions can be written in the normal form f (x) = ax² + bx + c, where a, b and c represent constant numbers called coefficients or parameters. In addition, the following should apply: a ≠ 0, because for a = 0 the quadratic summand is omitted and it is a linear or even constant function (even if b = 0).

The graph of all quadratic functions f (x) = ax² + bx + c (with a ≠ 0) is a parabola. All parabolas have in common, among other things, that they are open either upwards or downwards, have a lowest or highest point, which is called a vertex, and are obviously axially symmetrical to a vertical straight line that goes perpendicularly through this vertex.

symmetry

We now want to see whether the function f (x) = ax² + bx + c is really symmetrical. Symmetry means that the same function values ​​always occur to the left and right of the center of symmetry with the same distance of the x-values ​​from the center.

xS. be the x-value of this center. If d is the distance of an x-value from this center, then x = xS.+ d. The point x 'symmetrical to it has the value x' = xS.-d.

Assuming xS. is really the center of symmetry, then for all d ≠ 0 it must apply:

f (xS.-d) = f (xS.+ d)

This equation is always correct for d = 0, i.e. also for all positions that do not represent a center of symmetry. Therefore, in the following we can and must exclude this case and assume that d ≠ 0.

So let d ≠ 0. Now we write f (xS.-d) = f (xS.+ d) with f (x): = ax² + bx + c

f (xS.-d) = f (xS.+ d) a (xS.-d) ² + b (xS.-d) + c = a (xS.+ d) ² + b (xS.+ d) + c a (xS.² - 2dxS. + d²) + bxS. - bd + c = a (xS.² + 2dxS. + d²) + bxS. + bd + c axS.² - 2adxS. + ad² + bxS. - bd + c = axS.² + 2adxS. + ad² + bxS. + bd + c -2adxS. - bd = 2adxS. + bd | -2adxS. + bd -4adxS. = 2bd | : (- 4ad) (possible because d ≠ 0 and a ≠ 0) -b xS. = ——— 2a

Obviously, such a center of symmetry exists independently of d if d ≠ 0. Since the result only depends on a and b, we already have a formula for xS. won. The formula for yS., i.e. the y-coordinate of the vertex, is obtained by calculating f (xS.):

-b b² -b b² b² b² yS. = f (xS.) = f (——) = a ——— + b —— + c = ——— - ——— + c = c - —— 2a 4a² 2a 4a 2a 4a

The coordinates of the vertex S (xs| ys) a quadratic function f (x) = ax² + bx + c can thus be calculated by:

-b b² xS. = ——— yS. = c - --—— 2a 4a

If x² appears without a factor in the quadratic equation, i.e. a = 1, one writes instead of mostly. The vertex formulas are then as follows:

p p² xS. = - ——— yS. = q - ——— 2 4

In the following script, the vertex can be calculated for any quadratic function. The functions are even automatically brought into normal form first if they are quadratic or linear functions. As an alternative to the direct calculation of the vertex coordinates with the formulas, the vertex shape can optionally also be generated by adding a square.

The functions can be drawn on the plotter side. Extreme values ​​for non-square functions can also be calculated there.


Applications

  • Vertex shape of the function equation
    The vertex shape of f (x) = ax² + bx + c is f (x) = a · (x - xs) ² + ys. From it both the coordinates of the vertex and the parameter a can be read off immediately, with which the parabola can be drawn easily. This shape is also advantageous if the parabola is to be moved, because the parabola can be moved by moving the vertex.

    In order to write down a quadratic function in its vertex form, the quadratic addition can be carried out or the coordinates of the vertex can be determined with the formula in order to insert it into the above form. See also the next example.

    The formulas obtained above for the coordinates of the vertex can also be obtained by adding the square of general coefficients:f (x) = ax² + bx + c | : a f (x) / a = x² + b / a * x + c / a | quadratic addition: (b / (2a)) ² f (x) / a = x² + b / a · x + (b / (2a)) ² + c / a - (b / (2a)) ² f (x ) / a = (x + b / (2a)) ² + c / a - (b / (2a)) ² | A f (x) = a (x + b / (2a)) ² + c - b² / (4a) = a (x - (-b / (2a) )² + c-b² / (4a)
  • Moving a parabola
    In order to move a parabola to the left or right parallel to the x-axis, you usually need the vertex form of the function equation. For this purpose, one determines the coordinates of the vertex S (xS.| yS.) and puts it together with the parameter a from the function equation in the form f (x) = a (x - xS.)2 + yS. a. The negative sign in front of the x must be particularly importantS. in the vertex shape in conjunction with the sign of xS. be observed yourself. Because of the minus in the form there is always the opposite sign of xS. self.

    Once you have calculated the vertex, you only need to shift its coordinates, write down the result in vertex form and, if necessary, bring it back to normal form.

    Shift to the right: xS. increase by the distance; Shift to the left: xS. decrease by the distance; up: yS. increase by the distance; down: yS. decrease by the distance.

    Example:
    The parabola f (x) = -2x² - 3x + 7 should be shifted by 4 to the right and by 15 downwards.
    First its vertex is calculated: xS. = -b / (2a) = - (- 3) / (2 * (-2)) = -3/4 = -0.75
    yS. = c - b² / (4a) = 7 - (-3) ² / (4 * (-2)) = 7 - 9 / (- 8) = 7 + 9/8 = 65/8 = 8.125
    Vertex shape: f (x) = a (x - xS.) ² + yS. = -2 (x + 0.75) 2 + 8.125.
    Shifting 4 to the right shifts the vertex to x 'S. = -0.75 + 4 = 3.25. The shift down by 15 to y 'S. = 8,125 - 15 = -6,875.
    The vertex shape of the parabola shifted in this way is thus f '(x) = -2 (x - 3.25) ² - 6.875. Note the signs!
    Multiplying out gives the standard form: f '(x) = -2 (x - 3.25) ² - 6.875 = -2 (x² - 6.5x + 10.5625) - 6.875 = -2x² + 13x - 28

  • Extreme value problems
    Often there are problems in which one would like to know when a certain function assumes its greatest or also its smallest value. If the function is a quadratic one, this point can be calculated using the vertex formula. (Otherwise you need differential calculus for this.)

    Example:
    With a 5m long flywire net you want to limit a rectangular rabbit run on a house wall. The outlet has the length x and the width y. Since the house wall already gives a length limit, the following results for the division of the 5m on the three remaining sides: x + 2y = 5.
    It should now be considered at what length the surface area of ​​the run-out becomes maximally large.
    For the area F, F = x · y applies. This is the main condition. We are looking for the maximum value of F. Unfortunately, this “formula for F” still contains two variables (x and y). We can, however, eliminate a variable by solving the equation x + 2y = 5 ("secondary condition") for y and thus replacing y in the area equation:
    y = 2.5-0.5x
    F = x x y = x x (2.5-0.5x) = 2.5x - 0.5x²

    This is a quadratic function, i.e. you can calculate its extremum (the smallest / largest value) using its vertex. We read off: a = -0.5; b = 2.5 and c = 0. Because a <0 the parabola is open downwards; its vertex is therefore a maximum. Now one computes xMax as the x-coordinate of the vertex: xMax = xs = -b / (2a) = -2.5 / (- 1) = 2.5m. The second coordinate (“ys“) Of the vertex is in this case the maximum area FMax (not the maximum width that we called y, because the functional equation is F = -0.5x² + 2.5x and not y = -0.5x² + 2.5x): FMax = c - b² / (4a) = 0 - (2.5m) ² / (4 * (-0.5)) = 6.25m² / 2 = 3.125m². The width of the maximum run-out is given by y = 2.5m - 0.5 xs = 1.25m.


© Arndt Brünner, March 27, 2003
Version: 1.2.2004
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