Polynomial function/inequality

This article is filled with information that will assist you to learn polynomials of degrees up to 5:

A polynomial f(x) with real coefficients and of degree n has n zeros (not essentially all different). A few or all are real zeros and appear as x-intercepts when the graph of f(x) is plotted.

f(x) = ax5 + bx4 + cx3 + dx2 + ex + f

By altering the values of the coefficients: a, b, c, d, e and f, it is not simple to draw any conclusion when you change all 5 coefficients at the same time. You can constantly reduce the degree (highest power) by setting a few parameters to zero. For instance, if you set parameter a to zero and b to a non zero value, you get a polynomial of degree 4.

Leading Coefficient Test for polynomial function

1. Set limit a (leading coefficient) to a positive value (polynomial of degree 5) and set b, c, d, e and f to a few values.

As x increases without bounds, check if the right side of the graph rise or fall?

As x decreases without limits, check if the right side of the graph rise or fall?

Alter the value for b, c, d, e and f and observe if the behavior you observed earlier alters.

2. Position “a” to zero and b (leading coefficient) to a positive value (polynomial of degree 4) and do the same thing you did above in 1 above and 2.

3. Position a and b to zero and c (leading coefficient) to a positive value (polynomial of degree 3) and carry out the same operation as in 1 above and 2 above.

4. Position a, b and c to zero and d (leading coefficient) to a positive value (polynomial of degree 2) and carry out the same operation as in number 1 and 2 above.

5. Position a, b, c and d to zero and e (leading coefficient) to a positive value (polynomial of degree 1) and carry out the same operation as in 1 above and 2 above.

From your observation, what can is your conclusions about the behavior of the graph of the polynomial f(x) with an even degree n and a positive leading coefficient as x increases without bounds? What have you got to say about the behavior of the same polynomial as x decreases without limits?

What is your conclusion about the behavior of the graph of the polynomial f(x) with an even degree n and a negative leading coefficient as x increases without limits? What do you observe about the behavior of the same polynomial as x decreases without limits?

What do you conclude about the behavior of the graph of the polynomial f(x) with a odd degree n and a positive leading coefficient as x increases limitlessly? What do you say about the behavior of the same polynomial as x decreases limitlessly?

What about the behavior of the graph of the polynomial f(x) with a odd degree n and a negative leading coefficient as x increases without limitlessly? What do you say about the behavior of the same polynomial as x decreases limitlessly?

Solving Inequalities:

Solving linear inequalities is equivalent to solving linear equations, apart from one small but significant detail: you turn over the inequality sign whenever you multiply or divide the inequality by a negative. The easiest way to illustrate this is with a few examples:

(1) x + 3 < 2 The only variation between the linear equation "x + 3 = 2" and this linear inequality is that inequality has a "less than" sign, rather than an "equals" sign. The solution method is precisely the same: subtract 3 from either side.

Observe that the solution to a "less than, but not equal to" inequality is graphed with a parentheses (or besides an open dot) at the endpoint, showing that the endpoint is not included within the solution.

(2) 2 - x > 0

The only variation between the linear equation "2 – x = 0" and this linear inequality is the "greater than" sign in place of an "equals" sign.

Observe that "x" in the solution does not "need" to be on the left. Nevertheless, it is frequently easier to picture what the solution means with the variable on the left. You are free to rearrange things to go with your taste.

(3) 4x + 6 ≤ 3x - 5

The only difference between the linear equation "4x + 6 = 3x – 5" and this inequality is the "less than or equal to" sign in place of a plain "equals" sign. The solution method is precisely the same.

Observe that the solution to a "less than or equal to" inequality is plotted with a square bracket (or rather a closed dot) at the endpoint, showing that the endpoint is included inside the solution.

(4) 2x > 4

The solution technique here is to divide the two sides by a positive two.

(5) -2x > 4

This is the particular case observed above. When you separate them by the negative two, (minus two, -2) you would have to turn over the inequality sign.

The rule for example 5 above frequently appears awkward to students the first time they come across it. However to understand it, you’ll consider inequalities with numbers in there, rather than variables. You’re aware that that the number four is larger than the number two: 4 > 2. Multiplying through the inequality with –1, you will obtain –4 < –2, which the number line illustration proves that it is true:

If you hadn't turned over the inequality, you would have gotten "–4 > –2", which evidently is not true.

The inequalities we treated earlier are referred to as "linear" inequalities. We were dealing with linear expressions like "x – 2" ("x > 2" is just "x – 2 > 0", earlier. When we have an inequality with "x2" as the highest-degree term, it is known as a "quadratic inequality". The solution for this type of inequality is more intricate.

Solve x2 – 3x + 2 > 0

Initially, you ought to obtain the x-intercepts of the related quadratic, due to the fact that the intercepts are where y = x2 – 3x + 2 is equal to zero. Graphically, an inequality like this requires you to find where the graph is above or below the x-axis. It is much simpler to find where it in reality crosses the x-axis, thus, it is better to begin from there.

Factorizing, you would get x2 – 3x + 2 = (x – 2)

(x – 1) = 0, so x = 1 or x = 2. In that case the graph crosses the x-axis at 1 and 2, and the number line is divided into the intervals :(negative infinity, 1), (1, 2), and (2, positive infinity). Between the x-intercepts, the graph is either above the axis (and therefore positive, or greater than zero), or else below the axis (and therefore negative, or less than zero).

There are two dissimilar algebraic ways of checking for this positivity or negativity on the intervals. These methods are shown below:

(1) Test-point method.

Through this method, the intervals between the x-intercepts are (negative infinity, 1), (1, 2), and (2, positive infinity). Choose a point (any point) inside each interval and estimate the value of y at that point. Whatever the sign on that value is, that is the sign for that the whole interval.

For (negative infinity, 1), choose x = 0 for an example; then y = 0 – 0 + 2 = 2, which is positive. This says that y is positive on the whole interval of (negative infinity, 1), and this interval is therefore part of the solution because I’m looking for a "greater than zero" solution.

For the interval (1, 2), select, for example, x = 1.5; then y = (1.5)2 – 3(1.5) + 2 = 2.25 – 4.5 + 2 = 4.25 – 4.5 = –0.25, which is negative. Then y is negative on this whole interval, and this interval is thus not part of the answer.

For the interval (2, positive infinity), Select, for example, x = 3; then y = (3)2 – 3(3) + 2 = 9 – 9 + 2 = 2, which is positive, and this interval is in this case part of the answer. Then the complete answer for the inequality is x < 1 and x > 2. This is represented different ways as:

(2) Factor method.

Factoring, I get y = x2 – 3x + 2 = (x – 2)(x – 1). Now I will consider each of these factors separately.

The factor x – 1 is positive for x > 1; likewise, x – 2 is positive for x > 2. Thinking back to when you first learned about negative numbers, you know that (plus)×(plus) = (plus), (minus)×(minus) = (plus), and (minus)×(plus) = (minus). Therefore, to compute the sign on y = x2 – 3x + 2, you just really require to know the signs on the factors. Then you can apply what you know about multiplying negatives.

Subsequently, you would observe that the solution of x2 – 3x + 2 > 0 are the two intervals with the "plus" signs:

(negative infinity, 1) and (2, positive infinity).

Another example: Solve –2x2 + 5x + 12 < 0.

The first thing you would do is to find the zeroes, which are the endpoints of the intervals: y = –2x2 + 5x + 12 = (–2x – 3) (x – 4) = 0 for x = –3/2 and x = 4. Therefore, the endpoints of the intervals will be at –3/2 and 4. The intervals are between the endpoints, thus, the intervals are (negative infinity, –3/2], [–3/2, 4], and [4, positive infinity). Observe that brackets are used for the endpoints in "or equal to" inequalities, rather than parentheses, due to the fact that the endpoints will be included in the last solution.)

To obtain the intervals where y is negative by the Test-Point Method, all you have to do is choose a point in each interval. You can use points like x = –2, x = 0, and x = 5.

To obtain the intervals where y is negative by the Factor Method, you merely solve each factor: –2x – 3 is positive for –2x – 3 > 0, –3 > 2x, –3/2 > x, or x < –3/2; and x – 4 is positive for x – 4 > 0, x > 4. View more on psalmfresh.blogspot.com for more educating and tricks article Thank you