Quadratic equation and Radical function

[vt33]

Quadratic equation and Radical function!

Please help with these two problems!

1) a) Find a radical function where f(4)=0 and f(16)=6
b) Find a radical function where f(4)=0 and f(2)=2

2) a) find a quadratic function where the vertex is (3,-1) and has a vertical intercept of (0,8)
b)find a quadratic function where the X- intercepts are (3,0) and (-5,0) and has a vertical intercept of (0,-30)
c) Find a quadratic function where f(1) = 4,f(2)=6, and f(4)=22
I know it's a lot. I'm not looking for the answers, I just need to know how to set it up. Thanks in advance for anyone who chooses to assist me.

 

[robert Ihnot]

Well, anyway, in c) where you have the information to determine three constants, we set F(x) =Ax^2+Bx+C. F(1)=A+B+C=4; F(2) = 4A+2B+C=6; F(4)=16A+4B+C=22.

 

[tacman]

1) a) One possible radical function could be f(x) = √(6/12) * √(x-4) where x≥4. This function will have a value of 0 when x=4 and a value of 6 when x=16.
b) Another possible radical function could be f(x) = √(x-2) where x≥2. This function will have a value of 0 when x=4 and a value of 2 when x=2.

2) a) One possible quadratic function could be f(x) = -2(x-3)^2 + 8. This function will have a vertex at (3,-1) and a vertical intercept at (0,8).
b) Another possible quadratic function could be f(x) = -3/5(x-3)(x+5). This function will have x-intercepts at (3,0) and (-5,0) and a vertical intercept at (0,-30).
c) One possible quadratic function could be f(x) = (3/2)x^2 - (13/2)x + 4. This function will have a value of 4 when x=1, a value of 6 when x=2, and a value of 22 when x=4.