Infern sign of quadratic equation from coefficients

Question

Consider the quadratic function $f(x)= ax^2 + bx + c$ where $x \in \{1,2,\ldots\}$. Presume that $b^2 - 4ac < 0$ holds and $a \neq 0$. We know that there does not exist a $x \in \mathbb R$ such that $f(x) = 0$. So either $f(x) > 0$ or $f(x) < 0$ for all $x$. Is the following statement true: $f(x) > 0$ for all $x$ if $ a > 0$ and $f(x) < 0$ for all $x$ if $ a < 0$.

Answer 1

Yes, the idea being to compute $$ \lim_{x \to + \infty} a x^{2} + b x + c $$ and check that this is $$ \begin{cases} + \infty & \text{if $a > 0$,}\\ - \infty & \text{if $a < 0$.} \end{cases} $$

Hint

For $x > 0$, write $$a x^{2} + b x + c = x^{2} \left(a + \frac{b}{x} + \frac{c}{x^{2}}\right).$$

Answer 2

Just another way, $b^2-4ac < 0 \implies ac>0$, so $a$ and $c=f(0)$ share the same sign. As $f(x)$ has no zero, $f(x)$ never changes sign, so $a$ and $f(x)$ have the same sign...

Answer 3

By completing the square,

$$ax^2+bx+c=a(x-d)^2+e$$ where $d,e$ are some constants. As $(x-d)^2\ge0$ is unbounded, the expression has the sign of $a$.