Help finding points

[savusavu]

I am stuck with a problem that gives me several points and the standard curve equation only. I know how to find the equation and other points when I have one point in which y=0 and some other point on the curve, but I am not sure how to do it without this.

A curve, y=ax^2 + bx+ c, passes through several points (-1,22); (1,8); (3,10); (-2,p); (q,17). Give possible answers for p and q. I have tried rearranging and solving 22= a(-1 + n) (-1 +m); 8 = a(1 +n) (1+m) and 3= a(10+n) (10+m) and solving together. Could you please help? I am sure it is something quite easy, but I am just not seeing it!!! Thank you so much!!!

 

[pappus]

I am stuck with a problem that gives me several points and the standard curve equation only. I know how to find the equation and other points when I have one point in which y=0 and some other point on the curve, but I am not sure how to do it without this.

A curve, y=ax^2 + bx+ c, passes through several points (-1,22); (1,8); (3,10); (-2,p); (q,17). Give possible answers for p and q. I have tried rearranging and solving 22= a(-1 + n) (-1 +m); 8 = a(1 +n) (1+m) and 3= a(10+n) (10+m) and solving together. Could you please help? I am sure it is something quite easy, but I am just not seeing it!!! Thank you so much!!!

1. With the coordinates of the first 3 points you can determine the values of a, b and c. Plug in the x- and y-values. You'll get a system of simultaneous equations:
\(\displaystyle \left|\begin{array}{rcl}a(-1)^2+b(-1)+c&=&22 \\ a(1)^2+b(1)+c&=&8 \\ a(3)^2+b(3)+c&=&10 \end{array}\right.\)

Solve for a, b and c.

2. Now plug in into the brand-new equation x = -2 to determine p.

3. Plug in x = q and y = 17. Solve this quadratic equation for q.

4. For confirmation only: All values a, b, c and p are integers. Only for q you'll get a (simple) rational number.