Finding sign of leading coefficient of a quadratic equation In a given quadratic equation $f(x)=ax^2+bx+c$ if $f(-1)>-4, f(1)<0$ and $f(3)>5$, then how can I find the sign of $a$?
Answer in the textbook: $a>0$

If $a=0$, then you have a straight line. Then $$f(1)=\frac{f(-1)+f(3)}2>\frac{-4+5}2=\frac 12>0$$
Then for a parabola with $a>0$ you have $$f(\frac{x+y}2)<\frac{f(x)+f(y)}2$$ and for $a<0$ you have $$f(\frac{x+y}2)>\frac{f(x)+f(y)}2$$
Since $$\frac{f(-1)+f(3)}2>0$$ and $$f(1)<0$$ then $$f(\frac{-1+3}2)<\frac{f(-1)+f(3)}2$$
