Is there a general formula for a quadratic that is always positive?

Question

I encountered a problem of quadratic. It asks for a quadratic that is in the form $f(x)=ax^2+bx+c$. It is always positive and $b$ is greater than $a$. Than it asks me to find $f(17)$ based on $f(16)=20$. Is there a fast and easy way to do this? My idea is letting $16a+b=2\sqrt{5}$. But than I don't know how to do it since $a$ nor $b$ can be irrational. Any help will be appreciated!

Answer 1

Condition $f(x)\geq 0$ gives you that $a>0$ because, if $a<0$, then $\lim_{x\to\pm\infty}f(x) = -\infty$, so there will be some $x$ such that $f(x) < 0$.

We can write down $f(x) = a(x-\alpha)^2+\beta$.

Since $a(x-\alpha)^2 \geq 0$, it follows that $f(x)\geq \beta$ with $f(\alpha) = \beta$, so $\beta$ is the minimum value $f$ achieves. We have $f(x)\geq 0\iff \beta \geq 0$.

If we expand, we get that $f(x) = ax^2 - 2a\alpha x + a\alpha^2 + \beta$ (i.e. $b=-2a\alpha$) and condition $b>a$ gives us $-2a\alpha > a$. Since $a>0$, this means that $b>a\iff\alpha <-\frac 12$.

Finally, $f(16) = 20$ gives us two things: $\beta\leq 20$ and $a=\frac{20-\beta}{(16-\alpha)^2}$, for $\beta \neq 20$ (I will get back to case $\beta = 20$).

Thus, pick any $\alpha < -\frac 1 2$ and $0\leq \beta < 20$ and you have that $$f(x) = \frac{20-\beta}{(16-\alpha)^2}(x-\alpha)^2 + \beta$$ satisfies all the conditions. In particular, $$f(17) = \frac{20-\beta}{(16-\alpha)^2}(17-\alpha)^2 + \beta.$$

If $\beta = 20$, then we have that $a(16-\alpha)^2 = 0$, and since $a>0$, it means that $\alpha = 16$. But, this is contradiction with requirement that $\alpha <-\frac 12$.