The highest exponent of the variable term is two (). This tells that this function is quadratic, meaning that it is a parabola.

The graph below will be the answer, as it shows a parabolic curve.

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 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:

To simplify, we must first simplify the absolute values.

Now, combine like terms:

Define . Then, if , it follows that .

By the Intermediate Value Theorem (IVT), if is a continuous function, and and are of unlike sign, then for some . As a polynomial, is a continuous function, so the IVT applies here.

Evaluate for each of the following values:

Only in the case of does it hold that assumes a different sign at both endpoints - . By the IVT, , and , for some .

crosses the axis when equals 0. So, substitute in 0 for :

We have the following answer choices.

The first equation is a cubic function, which produces a function similar to the graph. The second equation is quadratic and thus, a parabola. The graph does not look like a prabola, so the 2nd equation will be incorrect. The third equation describes a line, but the graph is not linear; the third equation is incorrect. The fourth equation is incorrect because it is an exponential, and the graph is not an exponential. So that leaves the first equation as the best possible choice.