Find the smallest integer value of $c$ such that the function $f(x)=\frac{2x^2+x+5}{x^2+4x+c+12x+x^2}$ has a domain of all real numbers.

Answer 1

The function $f(x)$ is defined for all real numbers except for those that make the denominator equal to zero. Hence we need $x^2+4x+c+12x+x^2=0$. Simplifying gives $2x^2+16x+c=0$. This polynomial is a quadratic, which means the graph of $y=2x_{^2}+16x+c$ is a parabola.
We are looking for the smallest integer value of $c$ such that this parabola has no real roots. There are a few ways to see what values of $c$ this is true for. For example, we can complete the square: $2x^2 + 16x + c = 2(x^2+8x) + c = 2((x+4)^2-16)+c = 2(x+4)^2 - 32 + c$. Since the square of a real number is nonnegative, the smallest possible value of $2(x+4)^2$ is 0, and adding 0 to $-32$ will not change its value. Therefore, the smallest integer value of $c$ such that this parabola has no real roots is $c=\boxed{-32}$.