^{2}+bx+c) and writing a system of equations, but I got confused.

^{2}+b*2+c=20 or

^{2}-2b+c=-4 or

^{2}+b*0+c=8 or

**a=0; b=6; c=8**

**Equation of parabola: y=6x+8.**

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.

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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)

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**

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

Let's plug-in the other set of coordinates...(-2,-4)

-4=a(-2)^{2}-2b+8

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.)

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