The relationship between vertex coordinates and the number of zeros is understandable ...

Number of zeros in quadratic functions

• The number of zeros in a quadratic function can be zero, one or two. In addition, these are related to the vertex coordinates during the calculation.

• With a parabola that opens upwards, the vertex is at the lowest point and with a parabola that opens downwards at the highest point. Own Parabolas a zero, this is to be equated with the vertex coordinates.

• On the other hand, if the number of zeros is two, the vertex is exactly in the middle of these two points. For example, if they are at x₁ = 4 and x₂ = 6, just calculate 4 + 6 and then divide 10 by 2. The x-coordinate is equal to 5. You can get the y-value by plugging x = 5 into the given function.

Relationship between vertex coordinates and zeros

• The relationship between vertex coordinates and zeros can be explained with various display options. In addition to the normal form, there is also the linear factor form and the vertex form.

• The function f (x) = (x -4) (x -2) is an example of the linear factor form. It has the advantage that you can read off zeros 4 and 2 directly.


Calculate extrema - this is how it is done with polynomials

Calculate the extrema of the polynomial and give the relative maximum and minimum ...

• The transformation into the normal form is done by opening the brackets: f (x) = x²- 6x + 8.

• When reshaping from the normal form f (x) = x²- 6x + 8 in the vertex form you first have to remove the power of 2 from the first x, the second x and the +8 so that (x - 6) remains. Using the binomial formula (x - 3)² and the subsequent expansion of this you get (x² - 6x + 9). Finally, the +8 has to be taken into account. At +9 and +8 you get the difference 1. From the vertex form f (x) = ((x -3)² -1) the vertex coordinates (3 / -1) can be read off.

Excursus - Calculations of zeros

• Zeros can be determined in various ways. There is the linear factorization (the factoring out), the substitution method and the polynomial division.

• If there is no absolute term in the function, the linear factorization is used. This would be e.g. B. for the function f (x) = x³ + 110 x² - 102600x the case. In the first step, an x ​​can be factored out, so that x₁ = 0 is: f (x) = x (x² + 110 x - 102600). With the help of pq formula you can then use the other digits x₂ = -270 and for x₃ = 380 can be determined.

• If your function only has even exponents, you can use the so-called substitution method. Make sure that the function is first brought into normal form. Divide at f (x) = 2x⁴ - 18x² so first by 2. Your obtained function f (x) = x⁴ - 9x² must then be converted so that you can apply the pq formula. If you z. B. assume that u = x² is, in the next calculation step f (x) = u² - 9u the pq formula with u can be applied. At the end, don't forget to take the root and convert the u back to x. Your zeros are here at the positions x₁= 3, x₂ = -3 and x₃; 4 = 0 (read: double zero at the 0 position).

• at Functions of the form f (x) = x³ - x² - 3x + 72 you will get the first zero at x by trying it out₁ = 3. You can calculate this if you (x³ - x² - 3x + 72) divide by (x - 3). The result is x² - 2x -24. Then the pq formula can be used. The results x₂ = 6 and x₃ = -4 are correct.