In the present problem, it is presented a figure of a graph of a parabola with concavity facing up, what implies in that the graph will not have a Max., but a Min. value, and it is in the vertex.
Let us start with the A.O.S. (Axis Of Symmetry). For a parabola, it is the line that passes in the minimum point, that can be identified in the present figure as being in the x = - 1 point. And for this reason, we are able to answer the first part as:
A.O.S.: x = -1Now, as mentioned above, the vertex lies in the minimum point, which is located in the x = -1 and with height y = 2. This allow us to answer the second part as:
Vertex: (-1, 2)About the y-interception, it happens when the graph passes through the y-axis. It happens when x = 0, and the answer is just the value of Y where the graph assumes x=0, which is at:
Y-intercept: y = 4
Because the parabola has concavity facing up, the graph presents a value of minimum, but no maximum. And this is possible to be identified as the point where x = -1 and y = 2, which lead us to the following answer for the third part of the question:
Min.: (-1 , 2)The domain for any second-order polynomial is the entire real number because there is no restriction for the value of x. And for this reason, the answer of the fourth part of the answer is:
Domain: All real numbersAbout the range, for a parabola with the concavity facing up, it can be determined as all the real numbers bigger or equal to the minimum value of the function, which is y = 2. From this reason, we are able to answer the last part of the question as:
Range: All real numbers ≥ 2
<-- volume formula.
