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

2. The answer depends on both $t$ and $x$.

$tx^2 + 8x + 10 - t > 0$

Isolating the variable $t$ produces

$tx^2 - t > -8x - 10$

$\iff t(x^2 - 1) > -8x - 10$

$\iff t > \dfrac{-8x - 10}{x^2 - 1}$ if $x^2 - 1 > 0$ or $t < \dfrac{-8x - 10}{x^2 - 1}$ if $x^2 - 1 < 0$

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

For any $t\neq 0$ express

$y(x)=\ldots=t\left(x+\dfrac{4}{t}\right)^2-\dfrac{16}{t}+10-t$

We have to find only values of $t>0$ (why?). For $t>0$ , $y(x)> 0$ for all $x\in\mathbb{R}$ iff

$y(-4/t)=-\dfrac{16}{t}+10-t>0$

Now, solve the last inequality on $t$ .

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

First for y to always be positive the parabola must open up this means that
$y=ax^2+bx+c$

$a =t > 0$ so we can divide by $t$

Now we know that the vertex will be the lowest point and it occurs at

$x=-\frac{b}{2a}=\frac{-8}{2t}=\frac{-4}{t}$

Now if we put this back into the equation we want to solve

$\displaystyle y > 0 \iff t\left( \frac{-4}{t}\right)^2+8\left( \frac{-4}{t}\right)+10-t >0 \iff t^2-10t+16 <0$

We can reduce to this because we already know that $t > 0$

so the inequality is preserved.

Here is a fun animation

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

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

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