Finding $q$ such that the curve with equation $y = \frac{x^2 +qx + 1}{2x + 3}$ has the $x$-axis as a tangent

Question

The following question and official answer appeared in an A-levels Cambridge exam. Can anyone please argue rigorously as to why for $x$-axis to be the tangent all that is required is to make the discriminant equal to zero?

The curve $C$ has equation $$y = \frac{x^2 +qx + 1}{2x + 3}$$ where $q$ is a positive constant.

Find the value of $q$ for which the $x$-axis is a tangent to $C$. and sketch $C$ in this case.

Answer : All that was required was a statement to the effect that for tangency with the x-axis the discriminant of the quadratic form $x^2 + qx + 1$, that is $q^2 - 4$, must be zero. Since it is given that $q > 0$ then $q = 2$

Answer 1

Since this is a differentiable function, all you need to prove tangency to the $x$ axis is a root at a local minimum (or maximum). That happens when the quadratic in the numerator has a unique root.

Answer 2

When you set the curve equal to 0 (x-axis), the condition for tangency is to have root(s) with multiplicity greater than 1, therefore set $\Delta=b^2-4ac=0$ to get $q^2-4=0$ and then $q=\pm2$

Answer 3

If the axis $x$ is tangent to the curve, it means that there is a point in the curve where the ordinate of the point is zero. But the point in question is a point of max or min or inflection.

The tangent has equation $y=m x+k$.

Necessarily $k=0$, then the angular coefficient $m$ must be zero. But $m$ is nothing more than the first derivative calculated at the point of tangency, that is $P(x_{0},0)$:

$y’=\frac{2x^{2}+6x+3q-2}{(2x+3)^{2}}$,

$y’=0$ for $x_{0}=\frac{-3+\sqrt{13-6q}}{2}$, and $x_{1}=-\frac{+3+\sqrt{13-6q}}{2}$.

So that the values are real: $q<\frac{13}{6}$.

Substituting the values of $x_{0}$, where the function must be zero, we get:

$y=\frac{x^{2}+q x+1}{2x+3}$;

$0=\frac{x_{0}^{2}+q x_{0}+1}{2x_{0}+3}$,

Solving you get: $q=-2$ and $q=2$.

We are interested in $q>0$, so the function becomes: $y=\frac{x^2+2x+1}{2x+3}$.

This equation represents a curve that has a minimum at the point $P(-1,0)$, and the horizontal tangent at this point is the abscissa axis $x$, of equation $y=$0.