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

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: $y = a {x}^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$, then you know that the minimum value for this function occurs at $x = - \frac{k}{2}$.
Putting this value in for $x$, we get minimum value: $y = {\left(\frac{- k}{2}\right)}^{2} + k \left(\frac{- k}{2}\right) + 3 = {k}^{2} / 4 - {k}^{2} / 2 + 3 = \frac{- {k}^{2}}{4} + 3 = 3 - {k}^{2} / 4$.

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

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.