Given that $\displaystyle y = tx^2 + 8x + 10 - t$ find the range of values of $\displaystyle t$ for which $\displaystyle y$ is always positive.

2. 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 $\displaystyle x^2 - 1 < 0$

3. Originally Posted by Ilsa
Given that $\displaystyle y = tx^2 + 8x + 10 - t$ find the range of values of $\displaystyle t$ for which $\displaystyle y$ is always positive.

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 $\displaystyle t$ .

4. Originally Posted by Ilsa
Given that $\displaystyle y = tx^2 + 8x + 10 - t$ find the range of values of $\displaystyle t$ for which $\displaystyle y$ is always positive.

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

5. How dull of me not to realize $\displaystyle y > 0 \iff t > 0$ (or clever of you to realize so).

6. Originally Posted by NOX Andrew
The answer depends on both $\displaystyle t$ and $\displaystyle x$.
It seems clear that $\displaystyle t$ is a parameter and $\displaystyle x$ the variable i.e. the question should be:

Find all values of $\displaystyle t\in\mathbb{R}$ such that $\displaystyle y_t(x)=tx^2+8x+10-t> 0$ for all $\displaystyle x\in\mathbb{R}$ .