The quadratic funtion

The quadratic funtion

I)History and Definition

The Babylonians:

Tablets of this time( 1 800-1 500 BC) keep a crowd of information, in particular they reveal us an already very developed algebra and testify of the control of the Babylonians to solve quadratic equations. The Babylonian clay tablet N 13901 of British Museum ( London), was qualified as " real small textbook of algebra, dedicated to the quadratic equation and to the systems of equations, and giving the fundamental resolutory procedures ".

Diophante in 4th century.

Diophante pursues the searches for the Babylonians. He will have an algebraic approach of the problem.

In 8th century, the Indian mathematician Sridhar Acharya proposes a method to calculate both real roots.

Towards 820-830, Al-Khwarizmi, member of the scientific community gathered around the caliph al Mamoun, describes, in its treaty of algebra, algebraic transformations allowing to solve equations of the 2nd degree.

The negative roots are ignored until the 16th.

According to the ideas developed by Stevin in 1585, Girard in 1629 gives examples of equations with negative roots. 

" The negative in geometry indicates a regression, while the positive corresponds to a progress(promotion). ". He does not moreover have more scruples with the complex roots.

A quadratic function (or trinomial function) is written : f(x)=ax²+bx+c where a,b,c are reals and a is different to zero. It is defined sur R and represented graphically by a parabola.

II)Vertex form/ completed square form

Any trinomial f(x)= a(x-α)²+ß  where α = - b/2a and ß =-b²-4ac/4a.

This expression is the vertex form of the trinomial ax²+bx+c.

Demonstration:

f(x)= ax²+bx+

f(x)=a(x²+(b/a)x+c/a)

f(x)=a((x+b/2a)²-b²/4a²+c/a)

f(x)=a((x+b/2a)²-b²/4a²+4ac/4a²)

f(x)=a(x-(-b/2a))²-b²-4ac/4a

III)Changes

The summit S of the parabola a has for coordinates (α, ß ).

when a > 0 when a<0

IV)Solve a quadratic equation

We call « roots » of the trinomial, the solutions of the equation ax²+bx+c=0. We call discriminant, the number Δ=b²-4ac

When Δ is negative(Δ<0),the equation ax²+bx+c=0 has no /any solution.

When Δ is equal tp 0(Δ=0), the equation ax²+bx+c=0 has only one solution : -b/2a

When Δ is positive(Δ>0), the equation ax²+bx+c=0 has two solution:the first is x1=-b-racine Δ/2a and the second solution is x2=-b+racine de Δ/2

This

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