The functionshifts 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 functionor.

To begin, we analyze the equation given: the base equation,is shiftedleftone unit andvertically stretchedby a factor of 2. The graph of the equationis:

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 inequalitytrue, then we shade the area containing that point (in this case,outsidethe parabola); if it makes the inequalityuntrue, then the opposite side is shaded (in this case, theinsideof the parabola). Plugging the numbers in shows:

Simplified as:

Which isnottrue, so the area inside of the parabola should be shaded, resulting in the following graph:

is a degree 3 polynomial, so we don't have any easy formulas for calculating possible roots--we just have to check individual values to see if they work. We can use therational root testto narrow the options down. Remember, if we have a polynomial of the formthen any rational root will be of the form p/q where p is a factor ofand q is a factor of. Fortunately in this case,so we only need to check the factors of, which is -15. Let's start with the easiest one: 1.

It doesn't work.

If we try the next number up, 3, we get this:

It worked! So we know that a factor of our polynomial is. We can divide this factor out:

and now we need to see ifhas any roots. We can actually solve quadratics so this is easier.

There aren't any real numbers that square to get -5 so this has no roots. Thus,only has one root.

As a polynomial function, the graph ofis continuous. By the Intermediate Value Theorem, ifor, then there must exist a valuesuch that.

Setand. It is not true that, so the Intermediate Value Theorem does not prove that there existssuch that. However, it does notdisprovethat 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.

One way to answer this question is as follows:

Let. By a corollary of the Factor Theorem,is divisible byif and only if the sum of its coefficients (accounting for minus symbols) is 0.has

as its coefficient sum, sois indeed divisible by.

Let. By a corollary of the Factor Theorem,is divisible byif and only if the alternating sum of its coefficients (accounting for minus symbols) is 0.

To find this alternating sum, it is necessary to reverse the symbol before all terms of odd degree. In, there is one such terms, theterm, so the alternating coefficient sum is

,

sois not divisible by.

As a polynomial function, the graph ofis continuous. By the Intermediate Value Theorem, ifor, then there must exist a valuesuch that.

Setting, and looking at the second condition alone, this becomes: If, then there must exist a valuesuch that- that is,must have a zero on. The conditions of this statement are met , since- and- sodoes have a zero on this interval.

By the Factor Theorem,is a factor of a polynomialif and only if. It is given thatis a factor of, so it follows that.

is an even function, so, by definition, for allin its domain,. Setting,; by substitution,. It follows thatis also a factor of, making the statement true.

Start by visualizing the graph associated with the function:

相关条款在括号内quared x-variable will shift the parabola horizontally, while terms outside of the parentheses will shift the parabola vertically. In the provided equation, 2 is located outside of the parentheses and is subtracted from the terms located within the parentheses; therefore, the parabola in the graph will shift down by 2 units. A simplified graph oflooks like this:

Remember that there is also a term within the parentheses. Within the parentheses, 1 is subtracted from the x-variable; thus, the parabola in the graph will shift to the right by 1 unit. As a result, the following graph matches the given function: