Identifying a Parabolic Equation with Given Characteristics
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 \( x = -3 \).
