Given that $\displaystyle y = tx^2 + 8x + 10 - t$ find the range of values of $\displaystyle t $ for which $\displaystyle y$ is always positive.
The answer depends on both $\displaystyle t$ and $\displaystyle x$.
$\displaystyle tx^2 + 8x + 10 - t > 0$
Isolating the variable $\displaystyle t$ produces
$\displaystyle tx^2 - t > -8x - 10$
$\displaystyle \iff t(x^2 - 1) > -8x - 10$
$\displaystyle \iff t > \dfrac{-8x - 10}{x^2 - 1}$ if $\displaystyle x^2 - 1 > 0$ or $\displaystyle t < \dfrac{-8x - 10}{x^2 - 1}$ if $\displaystyle x^2 - 1 < 0$
For any $\displaystyle t\neq 0$ express
$\displaystyle y(x)=\ldots=t\left(x+\dfrac{4}{t}\right)^2-\dfrac{16}{t}+10-t$
We have to find only values of $\displaystyle t>0$ (why?). For $\displaystyle t>0$ , $\displaystyle y(x)> 0$ for all $\displaystyle x\in\mathbb{R}$ iff
$\displaystyle y(-4/t)=-\dfrac{16}{t}+10-t>0$
Now, solve the last inequality on $\displaystyle t$ .
First for y to always be positive the parabola must open up this means that
$\displaystyle y=ax^2+bx+c$
$\displaystyle a =t > 0$ so we can divide by $\displaystyle t$
Now we know that the vertex will be the lowest point and it occurs at
$\displaystyle x=-\frac{b}{2a}=\frac{-8}{2t}=\frac{-4}{t}$
Now if we put this back into the equation we want to solve
$\displaystyle \displaystyle y > 0 \iff t\left( \frac{-4}{t}\right)^2+8\left( \frac{-4}{t}\right)+10-t >0 \iff t^2-10t+16 <0$
We can reduce to this because we already know that $\displaystyle t > 0$
so the inequality is preserved.
Here is a fun animation