Finding sign of leading coefficient of a quadratic equation

Question

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: $a>0$

Answer 1

You have that:

$$a - b + c > - 4 \tag{1}$$ $$-a - b - c > 0 \tag{2}$$ $$ 9a + 3b + c > 5 \tag{3}$$

Now here is an idea... let us try to find positive numbers $t, p, q$ such that when you multiply (1) by $t$, (2) by $p$ and (3) by $q$ and then you add the 3 inequalities, the $b$ and $c$ would be eliminated.

This requirement (for elimination of $b$ and $c$) gives us:

$$-t-p+3q = 0$$ and $$t-p+q = 0$$

Solving this system (for $t$ and $p$) gives us: $t=q$ and $p=2q$.

So we can use e.g. $q=1, p=2, t=1$

So we get:

$$a - b + c > - 4 \tag{1'}$$ $$-2a - 2b - 2c > 0 \tag{2'}$$ $$ 9a + 3b + c > 5 \tag{3'}$$

Now summing up these 3 inequalities we get:

$$8a \gt 1$$

Answer 2

Putting the values into the assumed quadratic form

$$a - b + c > - 4$$ $$-a - b - c > 0$$ $$ 9a + 3b + c > 5$$

On adding the first two inequalities, we have

$$-2b > -4 \implies b < 2$$

Now, if we add the second and third inequality, we get

$$8a + 2b > 5 \implies 4a + b > 5$$

But since $b < 2$, this means that$4a > 3$ for this to be true

$$\implies a > \frac{3}{4} > 0$$

Answer 3

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 $$f(\frac{-1+3}2)<\frac{f(-1)+f(3)}2$$