Let f(t)=6−2t2. Evaluate f(t+1).
1 answer:
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The <em>correct answers</em> are:
C) No: we would need to know if the vertex is a minimum or a maximum; and
C)( 0.25, 5.875).
Explanation:
The domain of any quadratic function is all real numbers.
The range, however, would depend on whether the quadratic was open upward or downward. If the vertex is a maximum, then the quadratic opens down and the range is all values of y less than or equal to the y-coordinate of the vertex.
If the vertex is a minimum, then the quadratic opens up and the range is all values of y greater than or equal to the y-coordinate of the vertex.
There is no way to identify from the coordinates of the vertex whether it is a maximum or a minimum, so we cannot tell what the range is.
The graph of the quadratic function is shown in the attachment. Tracing it, the vertex is at approximately (0.25, 0.5875).
