Silhouette of a Fish

It's the same as asking for what's the smallest value of $a$ such that there exist a $y$ for all $x \geq b$ for the equation given.

If $\text{RHS} < 0$ , then there's no real solution for $y$ . So $\text{RHS} \geq 0$

Which means the minimum value of $f(x) = a + 15x - 9x^2 + x^3$ occurs at $f'(x) = 0 \Rightarrow 3x^2 - 18x + 15 = 0 \Rightarrow x = 1, 5$

Substitution of the values of $x$ gives the pairs $(x,16y^2) = (1,a+7), (5,a-25)$

So either $a = -7$ or $a = 25$ only (not both). Suppose $a = -7$ then the equation factors $16y^2 =(x-1)^2 (x-7)$ which implies that $x = 1$ is a solution but $1 <x<7$ is not a solution thus disobeying the condition. Hence, $a = \boxed{25}$ only.

Fun fact: the graph resembles a Tschirnhausen Cubic .

The RHS is a cubic polynomial and it can only have 2 roots, one of which is a double root, so that $y$ is real.

Let's call the solutions $t,s$ where $t$ is the double root. By vieta's, we have

$2t+s=9\longrightarrow s=9-2t$

$2ts+t^2=15\longrightarrow 2t(9-2t)+t^2=15$

$0=3t^2-18t+15\longrightarrow (3t-3)(t-5)=0\longrightarrow \boxed{t=1,5}$

Plugging in our values for $t$

$2+s=9\rightarrow s=7$

$10+s=9\rightarrow s=-1$

important $t=1$ is extraneous for one simple reason: while the graph is continuous after $x=7$ , I specifically stated that there must be no value of $x<b$ (b=7 here) that can satisfy this equation. However, the single point $(1,0)$ is a solution (hence the title). Thus $b\neq 7$

This means $t=5,~s=-1$ thus

$t^2s=-a\longrightarrow \boxed{ a=25}$

I'll leave it to someone else to post the calc solution by finding the minimum using derivatives