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The vertex form of a quadratic function is y=a(x-h)^2+k, where (h,k) is the vertex of the parabola.
Axis of Symmetry: x=6
Minimum Value: 7
Domain: All real numbers
Range: y≥ 7
General Formula:y=& a(x- h)^2 + k Equation:y=& 1/2(x- 6)^2+ 7 We can see that a= 12, h= 6, and k= 7. Note that from the vertex form it follows that the coordinates of the vertex are ( 6, 7).
The axis of symmetry is a vertical line that divides the parabola into two mirror images. This line passes through the vertex and follows a specific format, x= h. For this exercise, we have h= 7. Therefore, the axis of symmetry is the line x= 7.
Before we determine the maximum or minimum, recall that if a>0, the parabola opens upwards. Conversely, if a<0, the parabola opens downwards.
In the given function, we have a= 12, which is greater than 0. Thus, the parabola opens upwards and we will have a minimum value. The minimum or maximum value of a parabola is always the y-coordinate of the vertex, k. For this function, it is k= 7.
Unless there is a specific restriction given in the context of the problem, the domain of a quadratic function is all real numbers. In this case, there is no restriction on the value of x. Since the minimum value of the function is 7, the range is all real numbers greater than or equal to 7. Range:& y≥ 7