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.