Question

A) If y=x^2+kx+3, determine the value(s) of k for which the min value of the function is an integer?

Answer 1

`k` must be any even integer.

Method: Find an expression, in terms of `k`, for the minimum value of the function and determine what sorts of `k` will yield an integer value.

Because this question is asked in "Algebra", I will not use calculus, (Which is not needed to answer.)

I will assume that you know that the minimum value of a quadratic function occurs at the vertex of the parabola.

Two possibilities:

First: If you know that the vertex of the parabola: `y=ax^2+bx+c` occurs at `x=-b/(2a)`, then you know that the minimum value for this function occurs at `x=-k/2`.
Putting this value in for `x`, we get minimum value: `y=((-k)/2)^2+k((-k)/2)+3=k^2/4-k^2/2+3=(-k^2)/4+3=3-k^2/4`.

Second possibility: If you know how to complete the square to write the expression in standard form, do so
`y=x^2+kx+3=x^2+kx+k^2/4-k^2/4+3`
`y=(x+k/2)^2+3-k^2/4`
The vertex is at `(-k/2, 3-k^2/4)`

However you found it, the minimum value, `3-k^2/4` will be an integer exactly when `k^2/4` is an integer. And that will be an integer exactly when `k` is an even integer.