Guys this app is being retarded. That equation would form a parabola

it would be in quadrants 1 and 2 unless there was a specific restriction on the domain of x stated in the problem.

>>8668015
You are falsely assuming the parabola is upight. A tilted parabola can certainly be within any quadrant.

Consider y = x^2 + 100. Now rotate the axes left by 45 degrees.

>>8668046
Ignore this totally wrong sorry.

>>8668046
>nitpicking because muh smart but being plain wrong because the given equation invalidates your argument
good job

>>8668059
>>8668055

>>8668055
This is an upright parabola and so must be at least in two quadrants. either 1+2 or 3+4.

But it need not cross the X axis: if the quadratic equation has no solutions it will not touch it.

>>8668015
The answer is 0. Consider the function [math]f(x) = (x + 1)^2 + 1 = x^2 + 2x + 2[/math]. Clearly, in its first form, you can see it's like the graph of [math]f(x) = x^2[/math], but shifted to the left 1 and up 1. No real roots, yet all positive coefficients.

>>8668138
That makes sense but what about the quadrant part of the answer?

>>8668015
They meant to say the vertex is completely within the first quadrant

If a, b, and c are all positive, it will never intercept the x-axis.

>>8668247
That's not true. Consider [math]f(x) = x^2 + 3x + 1[/math]. The discriminant is [math]3^2- 4(1)(1)=5[/math], which is positive, so there are real roots. Graph it if you don't believe me.

Forget about what quadrant it's in. A parabola can cross the x-axis 0 times. If the vertex is above the x-axis and it opens up, or if the vertex is below the x-axis and it opens down.

Consider a=b=c=1.

>>8668015

It's an error but A is still the right answer and you're still a brainlet for answering otherwise