In order to determine the end behavior of a polynomial function, it must first be rewritten in standard form. Standard form means that the function begins with the variable with the largest exponent and then ends with the constant or variable with the smallest exponent. 

For f(x) in this case, it would be rewritten in this way: 

When this is done, we can see that the function is an Even (degree, 4) Negative (leading coefficent, -3) which means that both sides of the graph go down infinitely. 

In order to answer questions of this nature, one must remember the four ways that all polynomial graphs can look: 

Even Positive: 

Even Negative: 

Odd Positive: 

Odd Negative: 

The way to figure out this problem is by understanding behavior of polynomials.

The sign that occurs before the  is positive and therefore it is understood that the function will open upwards. the "8" on the function is an even number which means that the function is going to be u-shaped. The only answer choice that fits both these criteria is:

Starting with

moves the parabola by  units to the right.

Similarly moves the parabola by  units to the left.

Hence the correct answer is option .

When we look at the function we see that the highest power of the function is a 3 which means it is an "odd degree" function. This means that the right and left side of the function will approach opposite directions. *Remember O for Odd and O for opposite. 

In this case we also have a negative sign associated with the highest power portion of the function - this means that the function is flipped. 

Both of these combine to make this an "odd negative" function. 

Odd negative functions always have the right side of the function approaching down and the left side approaching up. 

We represent this mathematically by saying that as x approaches negative infinity (left side), the function will approach positive infinity: 

...and as x approaches positive infinity (right side) the function will approach negative infinity:

Then set each factor equal to zero, if any of the ( ) equal zero, then the whole thing will equal zero because of the zero product rule. 

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.