Question

HAPPY APRIL FOOLS- CAN ANYBODY ANSWER MY QUESTION---------

Answer 1

This is an interesting question.  Wish there were more questions of this kind.

This question helps us recognize the use of the vertex form of a quadratic expression/function.

The vertex form is in the form
$f(x)=y=a(x-h)^2+k$

The extreme value of the function occurs when x=h, i.e. when the first term vanishes, which leaves the value of the function equal to k.  I.e. the vertex of the function is at (h,k).

For example, when
f(x)=5(x-4)^2+7
at x=4, f(4)=5(4-4)^2+7=5(0)+7=7,
in other words, f(x) is at its minimum when x=4, with a value of 7,
even simpler, the vertex of the function is at (4,7).

How do we know if it is a maximum or minimum?

If the first parameter "a" is positive, then at any other value than x=h, the function has a greater value than the vertex, hence a>0 => minimum.
Similarly, if the first parameter "a" is negative, then whenever x does not equal h, the value of the function is smaller, hence a<0 => maximum.

For example, with f(x)=5(x-4)^2+7, a=+5 >0, so (4,7) is a minimum.
Check: f(0)=5(0-4)^2+7=16+7=23 > 7, or (0,23) verifies that (4,7) is a minimum.

For given situations A, B, C, & D, we see that only one function is in the vertex form, i.e. y=a(x-h)+k, namely situation B
B. y=-3(x-2)^2+5, where a=-3, h=2 and k=5.
This means that the function has a maximum at the vertex (2,5).  It has a maximum because a=-3 < 0, as discussed above.

Oh yes, enjoy the rest of April Fools Day!  lol