Question - Finding the Equation of a Parabola with Given Parameters
Solution:
To find the equation of a parabola that opens upwards, has a minimum value of -3, and an axis of symmetry at x = -3, you should look for a quadratic equation in the vertex form:$$ f(x) = a(x-h)^2 + k $$where (h, k) is the vertex of the parabola. Since the axis of symmetry is at x = -3, h must be -3. The minimum value of -3 indicates that k = -3, and since the parabola opens upwards, the coefficient 'a' must be positive.Looking at the options:A. $$ f(x) = -(x+3)^2 - 3 $$ - This option has both a negative 'a' (which would open downwards) and the wrong sign in front of 3 (it would result in an axis of symmetry at x = -(-3) = 3, not x = -3).B. $$ f(x) = (x+3)^2 - 6 $$ - This equation has the right sign for 'a' (which would open upwards) but its minimum value would be -6, not -3.C. $$ f(x) = (x-3)^2 + 6 $$ - This equation has the axis of symmetry at x = 3, not x = -3. Additionally, its vertex would be at (3, 6), which does not match the given minimum value of -3.D. $$ f(x) = (x+3)^2 - 3 $$ - This is the correct equation. It has a positive 'a', indicating that the parabola opens upwards, and the vertex form matches the given vertex of the parabola: (h, k) = (-3, -3).Therefore, the correct equation is:D. $$ f(x) = (x+3)^2 - 3 $$
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