Asked by Patricia
find the equation of the parabola passing through the origin with the turning point (2;-8).help mi.
Answers
Answered by
Reiny
the equation must look like:
y = a(x-2)^2 - 8
and it must pass through (0,0)
0 = a(-2)^2 - 8
8 = 4a
a = 2
equation: y = 2(x-2)^2 - 8
y = a(x-2)^2 - 8
and it must pass through (0,0)
0 = a(-2)^2 - 8
8 = 4a
a = 2
equation: y = 2(x-2)^2 - 8
Answered by
Steve
or, you know that one root is at (0,0) and since the vertex is at x=2, the other root must be at (4,0).
So, y = ax(x-4)
y(2) = -8, so
-8 = a*2(-2)
a = 2
y = 2x(x-4)
This agrees with Reiny's solution above.
So, y = ax(x-4)
y(2) = -8, so
-8 = a*2(-2)
a = 2
y = 2x(x-4)
This agrees with Reiny's solution above.
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