Find the equation of a parabola where the vertix is at origin and the parabola opens to the left and passes through (-1,2).
vertex at (0,0) means
y = ax^2
So, plug in (-1,2) and you have
2 = a(-1)^2
a = 2
so,
y = 2x^2
My bad. opens to the left:
x = ay^2
-1 = a*2^2
a = -1/4
x = -1/4 y^2
or
y^2 = -4x
http://www.wolframalpha.com/input/?i=y^2+%3D+-4x
To find the equation of the parabola with a vertex at the origin and opens to the left, we can use the standard form for a parabola equation. The equation of a parabola with a vertical axis of symmetry can be written as:
y = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola. In this case, since the vertex is at the origin, the equation simplifies to:
y = ax^2
Now, we need to find the value of 'a'. To do that, we can use the fact that the parabola passes through the point (-1, 2). Substituting these values into the equation gives us:
2 = a(-1)^2
2 = a(1)
2 = a
Therefore, the value of 'a' is 2. Plugging this value back into the equation, we get the equation of the parabola:
y = 2x^2
So the equation of the parabola with a vertex at the origin, opens to the left, and passes through (-1,2) is y = 2x^2.