Can someone prove that "a" has to be positive if...?

Question

If f(x) = ax^2 + bx + c, where a, b, and c are real numbers, and f(x) > 0 for all x, then does 'a' have to be positive? Who can prove this? I have tried some arguments using concepts from calculus but have not, to my mind, been successful. It's driving me crazy, so I'd love it if one of you can produce a clear, cogent proof that 'a' has to be positive if f is everywhere positive.

After considering the first 6 or so responses, it seems that Dr. Ivan has the most rigorous response, but I feel that he makes an unwarranted assumption; namely, that b^2 - 4ac < 0. I do not see why this has to be the case. Can someone shed some light on this issue?

Answer 1

First, a can be zero, if b=0 and c>0. Then f(x)=c>0 for all x.

We can prove that a has to be nonnegative.

Suppose a<0 and f(x)>0 for all x.
f(x) > 0 implies that f(x)≠0 for all x, which in turn implies that the discriminant of the quadratic trinomial ax^2+bx+c is negative.
The discriminant = b^2-4ac.
Thus,
b^2-4ac<0 =>
4ac>b^2
The number on the right is nonnegative, therefore, c has to be negative too for this inequality to hold:
c<0.
Now, substitute x=0:
f(0)=c<0
Contradiction completes the proof.

Another way to prove a is nonnegative is to employ limits at infinity.

We know that if x approaches +infinity, a polynomial's behavior is determined by the dominant term. Here the dominant term is ax^2. Since the degree of f(x) is even, as x approaches +infinity f(x) approaches + infinity if and only if a>0 (unless a=0). Otherwise f(x) approaches -infinity and cannot be positive for all x.

Answer 2

ax^2 + bx + c > 0
x(ax+b) + c>0

Suppose a is negative. We see that as x gets larger, ax will be a greater magnitude in the negatives. Eventually it will negate b (if not already, ie b<0) so that (ax+b)<0. And as x gets larger, x(ax+b) will be an even greater magnitude in the negatives. Eventually it will negate c (if not already, ie c<0) so that x(ax+b)+c < 0. Therefore, if a<0, then it will go negative.

Therefore, a>=0 in order to be positive for all values of x.

Answer 3

f(x) is a parabola. a determines the direction of it, up or down. The statement f(x) > 0 for all x, means that the value at the vertex f(-b/2a) > 0 and it is turned up a > 0.

Dr. Ivan's proof is the way I would have done this. But I hate to see repetitions of the same thing in these threads. Here is just another, intuitive way to see it.

Answer 4

Ain't gonna prove sump'n that jest ain't so....

Let f(x) = ax^2 + bx + c,
where a, b, and c are real numbers,
and f(x) > 0 for all x.
Let a = -1, b = 1, and c =1.
Hence, f(0) = (-1)0 + 1(0) + 1 = 1.
Thus, a does not have to be positive to enforce f(x) > 0. QED

Answer 5

sure if f is a parabola , with a positive, the parabola is like a smiling mouth, with a negative like a crying mouth. since f > 0 for all x, the parabola has to be a smiling mouth thus a > 0.

Answer 6

You proved that there is not any greatest integer. You assumed the existence of the main important integer and built a variety larger than it.hence u proved there is not any greatest integer.because of the fact a greatest integer does not exist we can't teach that a million is the main important integer.