How do I find the equation in standard form of the parabola passing through the points (2, 20), (-2,-4), and (0,8)?

Question

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.

Answer 1

Hi Lillian;

 

y=ax²+bx+c

Always begin with the zero...(0,8)...Let's plug in...

8=a(0)²+b(0)+c

8=c

y=ax²+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

Answer 2

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*2²+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*0²+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.

Answer 3

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

From (2),

a = (1/4)(2b-c-4) = (1/4)(2*6-8-4) = 0

Answer: y = 6x+8 (This is a straight line.)