As a polynomial function, the graph of  is continuous. By the Intermediate Value Theorem, if  or , then there must exist a value  such that . 

Set  and . It is not true that , so the Intermediate Value Theorem does not prove that there exists  such that . However, it does not disprove that such a value exists either. For example, observe the graphs below:

Both are polynomial graphs fitting the given conditions, but the only the equation graphed at right has a zero on .

The graph of a quadratic function  has an -intercept at any point  at which , so we set the quadratic expression equal to 0:

Since the question simply asks for the number of -intercepts, it suffices to find the discriminant of the equation and to use it to determine this number. The discriminant of the quadratic equation 

is 

.

Set , and evaluate:

The discriminant is positive, so the  has two real zeroes - and its graph has two -intercepts.

The graph of the quadratic function  is a parabola with its vertex at the point with coordinates

.

Set ; the -coordinate is 

Evaluate  by substitution:

The vertex has 0 as its -coordinate; it is therefore on an axis.

As a polynomial function, the graph of  is continuous. By the Intermediate Value Theorem (IVT), if  or , then there must exist a value  such that .

Setting ,  and examining the first condition, the above becomes:

if , then there must exist a value  such that  - or, restated,  must have a zero on the interval . Since , . the condition holds, and by the IVT, it follows that  has a zero on .

As a polynomial function, the graph of  is continuous. By the Intermediate Value Theorem, if  or , then there must exist a value  such that .

Setting  and , assuming for now that , and looking only at the second condition, this statement becomes: If , then there must exist a value  such that  - or, equivalently,  must have a zero on .

However, the conclusion of this statement is false:  has no zeroes at all. Therefore,  is false, and  has no negative values for any . By similar reasoning,  has no negative values for any . Therefore, by the IVT, by way of its contrapositive, we have proved that  is positive everywhere.

The zeros of the parabola are at  and , so when placed into the formula 

, 

each of their signs is reversed to end up with the correct sign in the answer. The coefficient can be found by plugging in any easily-identifiable, non-zero point to the above formula. For example, we can plug in  which gives 

The -intercept of the graph of a function  is the point at which it crosses the -axis; its -coordinate is 0, so its -coordinate is . 

,

so, by setting ,

,

making  the -intercept.

The parabola of an equation of the form  is vertical, and faces upward or downward depending entirely on the sign of , the coefficient of . This coefficient, , is negative; the parabola is concave downward.

As a polynomial function, the graph of  is continuous. By the Intermediate Value Theorem, if  or , then there must exist a value  such that .

Setting ,  and , this becomes:  If  or , then there must exist a value  such that  - that is,  must have a zero on .

However, the question is asking us to use the converse of this statement, which is not true in general. If  has a zero on , it does not necessarily follow that  or  - specifically, with , it does not necessarily follow that . A counterexample is the function shown below, which fits the conditions of the problem but does not have a negative value for :

The answer is false.

The graph of an equation of the form

is a horizontal parabola. Whether it is concave to the left or to the right depends on the sign of . Since , a negative number, the parabola is concave to the left.