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

1 Answer
Mar 6, 2015

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.