Question - Identifying a Parabolic Equation with Given Characteristics
Solution:
The image contains a question asking which equation represents a parabola that opens upward, has a minimum value of 3, and has an axis of symmetry at $$ x = -3 $$.The options provided are four quadratic equations:A. $$ f(x) = -(x + 3)^2 + 3 $$B. $$ f(x) = (x - 3)^2 + 6 $$C. $$ f(x) = (x + 3)^2 - 6 $$D. $$ f(x) = (x + 3)^2 + 3 $$A parabola that opens upward will have the square term with a positive coefficient. Options B, C, and D all have the square term with a positive coefficient, and thus they represent parabolas that open upward. Option A is incorrect because the negative sign in front would make the parabola open downward.Regarding the axis of symmetry, it is determined by the $$ h $$ in the vertex form of a parabola $$ f(x) = a(x - h)^2 + k $$, where $$ (h, k) $$ is the vertex of the parabola. The axis of symmetry is x = h. So in this case, we need $$ h = -3 $$. Options C and D both have $$ x + 3 $$, which can be written as $$ x - (-3) $$, indicating that the axis of symmetry is at $$ x = -3 $$, which meets the axis of symmetry criteria.Lastly, the parabola has a minimum value of 3, which means that the vertex is at $$ (h, k) = (-3, 3) $$. So we want the constant term (after completing the square) to be 3. Only option D has a constant term of +3.The correct answer based on all these conditions is:D. $$ f(x) = (x + 3)^2 + 3 $$ This equation represents a parabola that opens upward, has its vertex and therefore its minimum value at 3, and has an axis of symmetry at $$ x = -3 $$.
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