Suppose a parabola has its vertex at (0, 1) and its zeros at x = -2 and x = 2. Then equals _____.
A. 1-1/4x^2
B. 4-x^2
C. X^2-1
D. X^2-4
The vertex is at (0,4) sorry, I copied it wrong.
since the vertex is above the x-axis, we will have (with a > 0)
y = -a(x-h)^2 + k
We know (h,k) = (0,4), so
y = -ax^2 + 4
Since y(2) = 0, 0=-4a+4, so a=1
y = -x^2 + 4, so (B)
To determine the equation of a parabola given its vertex and zeros, we can use the vertex form of a parabola equation:
y = a(x - h)^2 + k,
where (h, k) represents the vertex of the parabola. In this case, the vertex is (0, 1), so we have:
y = a(x - 0)^2 + 1,
y = ax^2 + 1.
Now, we need to find the value of 'a' to complete the equation. Since the zeros of the parabola are x = -2 and x = 2, we can write the equation as:
y = a(x + 2)(x - 2).
To find 'a', we can substitute one of the points (either (-2, 0) or (2, 0)) into the equation and solve for 'a'. Let's use the point (-2, 0):
0 = a(-2 + 2)(-2 - 2),
0 = a(-4)(-4),
0 = 16a,
a = 0.
So, the equation of the parabola is:
y = 0(x + 2)(x - 2),
y = 0.
Therefore, none of the provided options (A, B, C, D) are correct, as the equation of the parabola is y = 0.