There will be many polynomials of degree 2 that pass through the points (1, 7) and (3, 9). 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.)

Any polynomial of the form f(x)=rx²+(1-4r)x+(3r+6) passes through these points.

Step-by-step explanation:

Let f(x)=rx²+bx+c be a polynomial that passes through (1,7) and (3,9). This means that f(1)=7 and f(3)=9, that is,

f(1)=r(1)²+b(1)+c=r+b+c=7, and

f(3)=r(3)²+b(3)+c=9r+3b+c=9

We have the equations r+b+c=7 and 9r+3b+c=9. Substract the second equation minus 9 times the first to obtain -6b-8c=-54, then 6b+8c=54 or equivalently b=9-4c/3.

Substract the second equation minus 3 times the first to obtain 6r-2c=-12. Then 2c=6r+12, thus c=3r+6. We have written c in terms of r. Substitute in b=9-4c/3 to obtain b=9-4(r+2)=-4r+1. The polynomial acquires the form

f(x)=rx²+(1-4r)x+(3r+6).

1,2,4,6 are the correct options.