By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

1 is a zero of  if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is a zero.

 is a zero of  if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0:

 is not a zero.

1 is the only rational zero.

The Intermediate value Theorem states that if  is a continuous function, , and  and  are of unlike sign, it must hold that  for some . Equivalently, if  has values on the endpoints of an interval that differ in sign, then  has a zero on that interval. Since all polynomial functions are continuous, we may apply this theorem by evaluating  

for , and finding two consecutive values such that  differs in sign.

By substitution, we can find that:

The change in sign occurs on the interval ; therefore, a zero of  is located on this interval.

By the Rational Zeroes Theorem, any rational zero of a polynomial with integer coefficients must be equal to a factor of the constant divided by a factor of the leading coefficient, taking both positive and negative numbers into account.

The constant 17, being prime, has only two factors, 1 and 17; the leading coefficient is 1, which only has 1 as a factor. Thus, the only possible rational zeroes of the given polynomial are given in the set

,

a set of four elements. This makes 4 the correct choice.

By Descartes' Rule of Signs, the number of positive real zeroes of a polynomial with real coefficients is, at maximum, the number of its sign changes, but equal to some number of the same odd-even parity. This number can be seen to be two:

This means that this polynomial must have either zero or two such zeroes.

The function  shifts a function f(x) units to the left. Conversely,  shifts a function f(x) units to the right. In this question, we are translating the graph two units to the left.

To translate along the y-axis, we use the function  or .

To begin, we analyze the equation given: the base equation,  is shifted left one unit and vertically stretched by a factor of 2. The graph of the equation  is:

To solve the inequality, we need to take a test point and plug it in to see if it matches the inequality. The only points that cannot be used are those directly on our parabola, so let's use the origin . If plugging this point in makes the inequality true, then we shade the area containing that point (in this case, outside the parabola); if it makes the inequality untrue, then the opposite side is shaded (in this case, the inside of the parabola). Plugging the numbers in shows:

Simplified as:

Which is not true, so the area inside of the parabola should be shaded, resulting in the following graph: