I've tried to figure it out by substituting the x and y values into an equation (y=ax2+bx+c) and writing a system of equations, but I got confused.
How do I find the equation in standard form of the parabola passing through the points (2, 20), (-2,-4), and (0,8)?
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You have three unknowns: a, b, and c, as parabola coefficients. You have three points, which lie on the parabola. Plug in their coordinates to obtain three equations containing a,b, and c.
1) point (2,20)
2) point (-2, -4)
3) point (0,8)
So the final system looks as follows:
Since c is determined by the third equation already, plug its value into the first and the second equation.
Add two equations together, to obtain 8a=0; a=0; then 2b=12 or b=6.
Answer: a=0; b=6; c=8
Equation of parabola: y=6x+8.
So parabola in your case is degenerate and all three points lie on a straight line.
Always begin with the zero...(0,8)...Let's plug in...
Let's plug-in one set of coordinates...I randomly select (2,20)...
Let's plug-in the other set of coordinates...(-2,-4)
Let's add the two above equations together...
Let's subtract the two equations from each other...
y = ax2+bx+c
Plug in (0, 8),
8 = c
Plug in (2, 20), (-2,-4),
20 = 4a+2b+c ......(1)
-4 = 4a-2b+c ......(2)
(1)-(2): 24 = 4b
b = 6
a = (1/4)(2b-c-4) = (1/4)(2*6-8-4) = 0
Answer: y = 6x+8 (This is a straight line.)