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There will be many polynomials of degree 2 that pass through the points (1, 5) and (3, 7). The situation can be described by a system of two linear equations in three variables that has many solutions. Find an equation (involving a parameter r) that represents this family of polynomials. (Let the coefficient of the x2 term in the equation be r.) y(x) = Determine the polynomials that open up and those that open down.

Respuesta :

Answer:

Step-by-step explanation:

Given that a quadratic function(polynomial of degree 2) passes through two points (1,5) and (3,7)

The function would be of the form

[tex]y=ax^2+bx+c[/tex]

with if a>0 open up and if a<0 open down

Since passes through(1,5) and (3,7) these points satisfy the equation

[tex]5= a+b+c\\7 = 9a+3b+c[/tex]

Let us eliminate c easily

2 = 8a +2b or 4a+b =1

b =1-4a

Thus parametric equation we can write as

[tex]y=ax^2+bx+c[/tex]

[tex]y = ax^2+(1-4a)x+c[/tex]

where c is arbitrary

If a>0 this will be open up otherwise open down.

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