Determining c in Quadratic Function Turning Point

[Monoxdifly]

The graph's turning point of a quadratic function \(\displaystyle f(x)=ax^2+bx+c\) is over the X-axis. If the coordinate of the turning point is (p, q) and a > 0, the correct statement is ...
A. c is less than zero
B. c is more than zero
C. q is less than zero
D. q equals zero

Since the point (p, q) is over the X-axis that means q is more than zero, so the options C and D are out of question. What should I do to determine if c is positive or negative? I'm stuck on this. Thanks.

 

[skeeter]

hint ...

what does the sign of the discriminant say about a quadratic’s roots?

 

[Monoxdifly]

If it's positive, it has two real roots. Okay, so D > 0.

D > 0
\(\displaystyle b^2-4ac>0\)
\(\displaystyle b^2>4ac\)
\(\displaystyle 4ac<b^2\)
\(\displaystyle c<\frac{b^2}{4a}\)
A square number must be positive, so \(\displaystyle b^2\) is positive. Since a > 0 then 4a > 0, so \(\displaystyle \frac{b^2}{4a}\) is still positive.
There's still a possibility that c is either negative or positive. What should I do next?

 

[skeeter]

you were given $a>0$ which means the graph of the resulting parabola opens upward.
you were also given that the vertex of the parabola, $(p,q)$ was over the x-axis, hence $q > 0$.

If $q > 0$, the parabola has no x-intercepts ... the given quadratic has no real roots $\implies b^2-4ac <0$

 

[HOI]

"Over the x-axis" seems to me a strange way of saying "above the x-axis".

 

[skeeter]

"Over the x-axis" seems to me a strange way of saying "above the x-axis".

I took it to be one of the three possibilities ... over, on, or under.

 

[Monoxdifly]

@skeeter:
Ah, I see. Thanks for your help.

@Country Boy:
Sorry, I didn't know the proper term so I came up with what I had in mind at best, without realizing that the simple "above" is already the proper term.