So if the graph is above the x axis, it covers all x values but if the graph touches the x axis it doesn't cover all x values since it doesn't include the x intercept values because the graph is> 0 instead of it being greater than or equal to zero.
Is that what you were trying to convey?
Yes, the inequality states that the graph is
strictly positive for all real
x
So if the graph touches the x axis, the values between the 2 intercepts (f(x)) is negative so can't be true. Is that what you are saying? Sorry, for the constant questioning
Well 'touching' the
x-axis refers to the graph of the quadratic being tangent to the
x-axis at one point, which means the graph will consist of two equal real roots and so the value of the quadratic will equal
0 at those points (roots). This does not the satisfy the condition that the quadratic must be
strictly positive. On the other hand, if the graph crosses (which is what you're asking) the
x-axis, it (the quadratic) will consist of two distinct real roots. So, just like you said, if it crosses the x-axis then it will mean that the quadratic will take on negative values for some
x values, and (again) won't satisfy the condition that the quadratic is
strictly positive for
all real
x. Of course, the coefficient of the leading term must be positive for the graph to be above the
x-axis, in this case
, but this condition is already satisfied since we found
n to lie in the interval
.