Show that the expression $\frac{px^2+3x-4}{p+3x-4x^2}$ will be capable of all values when $x$ is real, provided that p has any value between 1 and 7 .

Question

Show that the expression $\frac{px^2+3x-4}{p+3x-4x^2}$ will be capable of all values when $x$ is real, provided that $p$ has any value between 1 and 7 .

What I've tried :

Let $\frac{px^2+3x-4}{p+3x-4x^2} = y$. Then, $px^2 + 3x - 4 = py + 3xy - 4x^2y$
or, $(p+4y)x^2 + 3(1-y)x -4-py = 0$ Since, $x \in R$, $D ≥ 0$ implying, $9(1-y)^2-4(p+4y)(-4-py) ≥ 0$ or, $(9+16p)y^2+(46+4p^2)y+9+16p ≥ 0$ which further implies $9+16p ≥ 0$ .

I'm stuck here can't think further and there are solutions on the net but none of them explains well after this step ..

Answer 1

$$(9+16p)y^2+(46+4p^2)y+9+16p\ge0$$Which further implies $9+16p\ge0$.

That's not right. Notice that all three summands there could possibly be negative, for example (say) $1+3-2\ge0$ is a possibility or maybe $-5+3+3\ge0$ or ... I made those numbers up and when I say "possibility" I mean in the general sense, if you have $a+b+c\ge0$ it can be the case that any of $a,b,c$ are negative.

Also note that you want that quantity to be nonnegative for all $y$. So you're basically asking:

For which values of $p$ is the quadratic: $$y\mapsto(9+16p)y^2+(46+4p^2)y+(9+16p)$$Nonnegative?

And there's a standard way to answer that question, which is by showing the discriminant of the quadratic is nonnegative and the leading coefficient is positive. So try doing that!