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).

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.

a 270° rotation about point p says on e d g e n u it y (had to space it out because it says it's a misspelling so i couldn't submit it