I've tried to figure it out by substituting the x and y values into an equation (y=ax^{2}+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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3 Answers
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)
a*2^{2}+b*2+c=20 or
4a+2b+c=20
2) point (-2, -4)
a*(-2)^{2}-2b+c=-4 or
4a-2b+c=-4
3) point (0,8)
a*0^{2}+b*0+c=8 or
c=8
So the final system looks as follows:
4a+2b+c=20
4a-2b+c=-4
c=8
Since c is determined by the third equation already, plug its value into the first and the second equation.
We obtain:
4a+2b+8=20
4a-2b+8=-4
4a+2b=12
4a-2b=-12
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.
Hi Lillian;
y=ax^{2}+bx+c
Always begin with the zero...(0,8)...Let's plug in...
8=a(0)^{2}+b(0)+c
8=c
y=ax^{2}+bx+8
Let's plug-in one set of coordinates...I randomly select (2,20)...
20=a(2)^{2}+b(2)+8
20=4a+2b+8
12=4a+2b
Let's plug-in the other set of coordinates...(-2,-4)
-4=a(-2)^{2}-2b+8
-12=4a-2b
Let's add the two above equations together...
0=8a
0=a
Let's subtract the two equations from each other...
24=0+4b
24=4b
6=b
y=6x+8
y = ax^{2}+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
From (2),
a = (1/4)(2b-c-4) = (1/4)(2*6-8-4) = 0
Answer: y = 6x+8 (This is a straight line.)