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.

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

The number of -intercepts of the graph is equal to the number of real zeroes of the above equation, which can be determined by evaluating the discriminant of the equation, . Set , and evaluate:

The discriminant is equal to zero, so the quadratic equation has one real zero, and the graph of  has exactly one -intercept.

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 a positive -coordinate and a positive -coordinate, putting it in the upper right quadrant, or Quadrant I.

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 a negative -coordinate and a negative -coordinate, putting it in the lower left quadrant, or Quadrant III.

First, we need to get this function into a more standard form.

Now we can see that while the function is being horizontally compressed by a factor of 2, it's being translated 3 units to the right, not 6. (It's also being vertically stretched by a factor of 4, of course.)

By definition, , so