y = ax^2 + bx + c
0 = a(0^2) + b(0) + c ... 0 = 0 + 0 + c
-2 = a(1^2) + b(1) + c ... -2 = a + b + c
-4 = a(-1^2) + b(-1) + c ... -4 = a - b + c
solve the system for a, b, and c
What is the equation of a parabola containing the points (0,0),(1,-2)and (-1,-4)?
4 answers
y =a x^2 + b x +c
0 = a*0 + b*0 + c so c = 0
y = a x^2 + b x
-2 = a + b
y = a x^2 + b x
-4 = a - b
add the two
-6 = 2 a
a = -3
then b = 1
so
y = -3 x^2 + x
0 = a*0 + b*0 + c so c = 0
y = a x^2 + b x
-2 = a + b
y = a x^2 + b x
-4 = a - b
add the two
-6 = 2 a
a = -3
then b = 1
so
y = -3 x^2 + x
let the equation be
y = ax^2 + bx + c
since (0,0) is a point on it
0 = 0+ 0 + c, so c = 0, and we can go with
y = ax^2+ bx
If (1, -2) lies on it ---> -2 = a+b
if (-1,-4) lies on it ---> -4 = a - 4b
subtract them:
2 = 5b
b = 2/5
sub back in to get a and write your equation
y = ax^2 + bx + c
since (0,0) is a point on it
0 = 0+ 0 + c, so c = 0, and we can go with
y = ax^2+ bx
If (1, -2) lies on it ---> -2 = a+b
if (-1,-4) lies on it ---> -4 = a - 4b
subtract them:
2 = 5b
b = 2/5
sub back in to get a and write your equation
made an error, go with Damon and Scott