To change the function to one with two real zeros and a vertex of (3, 2), we need to adjust the values of h and k in the standard form of a quadratic function, y = a(x – h)2 + k.
Given that the vertex is (3, 2), we have h = 3 and k = 2.
The new function will be y = a(x – 3)2 + 2.
To find the value of a, we can use the fact that the vertex form of a quadratic function is given by y = a(x – h)2 + k, and the coefficient a determines the direction of opening and the stretch/compression of the parabola. If a > 0, the parabola opens upwards, if a < 0, the parabola opens downwards.
From the original function y = -2(x + 4)2 - 3, we see that a = -2. Since we want the parabola to have two real zeros, we need to make a > 0. Therefore, we need to change the sign of a to positive.
So, the new function is y = 2(x – 3)2 + 2.
To find the zeros of the new function, we set y = 0 and solve for x:
0 = 2(x – 3)2 + 2
-2 = 2(x – 3)2
-1 = (x – 3)2
±√(-1) = x – 3
x = 3 ± i
Therefore, the new function has two complex zeros at x = 3 + i and x = 3 - i.
The quadratic function y = –2(x + 4)2 – 3 with no real zeros is graphed.
Determine the values of h and k that will change the function to one with two real zeros and a vertex of (3, 2). What are the zeros of the new function?
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