Question - Finding the Equation of an Upward-Opening Parabola
Solution:
The question in the image is asking for an equation of a parabola that opens upward, has a minimum value of 3, and has an axis of symmetry at x = -3.A parabola that opens upward will have a positive coefficient for the squared term (x^2). The vertex form of a parabola's equation is:f(x) = a(x - h)^2 + kwhere (h, k) is the vertex of the parabola, and a determines the direction and width of the parabola.Here, we're looking for a parabola with a minimum value, which means its vertex will be at the bottom of the parabola (since it opens upward). The minimum value of the function is given by the value of k in the vertex form, which is 3 in this case.The axis of symmetry is x = h. For this problem, the axis of symmetry is given as x = -3, which means h must be -3.So we're looking for a parabola with a vertex at (-3, 3). Therefore, the equation would be in the form:f(x) = a(x + 3)^2 + 3Since a must be positive (because the parabola opens upward), and none of the options in the image display a negative a, we can simply look for an equation that fits the vertex form with the correct h and k values.Option A matches the given requirements:A. f(x) = (x + 3)^2 + 3Thus, the equation representing the parabola with the described properties is given by Option A.
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