($y=ax^2+bx$) finding range of values for $a$ and $b$ so that the parabola's height is always greater than the distance between its zeros

Question

So long story short, teaching myself algebraic and graphical modelling because of covid.

I can't seem to find anything online about my particular question.

Let's say an object is launched from the ground and it lands some distance away. This graphed would appear within the first quadrant (so $a<0$ and $b>0$). It's equation could be written as $y=ax^2+bx$.

What range of values must $a$ and $b$ be so that the height of the objects trajectory ($y$-coordinate of the vertex) is always greater than its total distance traveled.

How do you go about establishing or finding $a$ and $b$'s range if nothing is given to you other than it's graphed quadrant position?

Answer 1

An intuitive way

It is a well known fact that the vertex of the parabola is $\left(-\frac{b}{2a},-\frac{b^{2}-4ac}{4a}\right)$. For a projectile, the range is the intercept made on the axis. The direct form for $|\alpha - \beta|=\frac{\sqrt{b^2-4ac}}{|a|}$ where $\alpha$ and $\beta$ are the roots of the equation.

From this you can easily deduce $\frac{\sqrt{b^2-4ac}}{|a|} \leq -\frac{b^{2}-4ac}{4a}$ would be your required condition.

A graph for understanding

Answer 2

It is given that $y=ax^2+bx$. You can easily find the zeroes. Also, it is obvious that the extremum of the parabola occurs at $x=-b/2a$ (by calculus or completing square). From here, you can find the maximum height.