find the equation of the parabola passing through the origin with the turning point (2;-8).help mi.
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
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.
To find the equation of a parabola, we need to use the standard form of the equation:
y = a(x - h)^2 + k
Where (h, k) represents the coordinates of the vertex (turning point) of the parabola, and "a" is a coefficient that determines the shape of the parabola.
In this case, we know the turning point is (2, -8). So we can substitute these values into the equation:
y = a(x - 2)^2 - 8
Since the parabola passes through the origin (0, 0), we can substitute these values as well:
0 = a(0 - 2)^2 - 8
Simplifying:
0 = 4a - 8
To solve for "a", we can isolate it by adding 8 to both sides of the equation:
8 = 4a
Then, divide both sides by 4 to solve for "a":
2 = a
Now we have the value of "a". We can substitute it back into the equation to get the final equation of the parabola:
y = 2(x - 2)^2 - 8
So, the equation of the parabola passing through the origin with the turning point (2, -8) is y = 2(x - 2)^2 - 8.