Polynomial graphing and Solving

Polynomials, Graphing and Solving

Now we turn our attention to graphing and solving polynomials. To graph a polynomial we'll need to find the zeroes of the polynomial (the X intercepts), the Y intercept, and then using general observations we'll sketch its graph. The leading term of a polynomial is the term with the largest exponent. For example, in the polynomial 3x4 + 2x2 + x – 1, the leading term is 3x4. (The largest exponent is 4.) The reason we look at the leading term is to get an idea how the graph looks for very large positive and negative values of x. The leading term determines the overall shape of the graph. y = 3x4 + 2x2 + x – 1 In this equation as x becomes a very large positive number then the term 3x4 becomes much larger than the terms 2x2, x and -1. For example if x becomes 100,000 then x4 is 1,00...000 (20 zeroes) and x2 is 1,00...000 (10 zeroes). (20 -10 = 10 fewer trailing zeroes) When x becomes 100,000,000 the x4 is 1,00....000 (32 zeroes) and x2 is 1,00....000 (16 zeroes) (32 – 16 = 16 fewer trailing zeroes) As x continues to increase in powers of ten, the leading term will have more and more trailing zeroes then the squared term. We say that leading term dominates all other terms. As x continues to increase the other terms tend to have little effect on the graph. This behavior is true for negative x as x becomes more and more negative. Now for the graphs: The diagrams below show the extremes for large magnitudes of x (both positive and negative) for the leading term Axn. The dashed lines are shown only to indicate the rest of the graph which we are not interested in at this point. The X axis is shown with the usual positive values to the right and the negative values to the left. The Y axis is not shown only for generality, i.e., we only care about how the graph looks above and below the X axis. A > 0 and n even Here the coefficient A is greater than 0 (A>0) and the exponent n is even. Any number, positive or negative raised to an even power will be positive. Since A is positive then their product will be positive. For these conditions the y values become and remain positive as x becomes more positive and more negative. A<0 and n even Here A < 0 with n even. This case is the same as the previous but the negative A reflects the tails so the y values become and remain negative. a > 0 and n odd Here we have A > 0 and n is odd. With an odd power, a positive number to that power creates a positive result while a negative number to that odd power will create a negative result. (+1)3 = +1, (-1)3 = -1 So, when x becomes more and more positive the y values become more and more positive and remain positive. And as x becomes more and more negative the y values become more and more negative, and remain negative. A < 0 and n odd With A < 0 and n odd, the negative A causes the previous graph to be mirrored across the X axis. When x is negative and raised to an odd power the result is negative, and this negative result multiplied by negative A becomes positive. And as x becomes more negative, y becomes and remains positive as shown. When x is positive and raised to an odd power the result is positive, but when the result is multiplied by negative A we get a negative y. So, as x continues to increase positively then y becomes and remains negative. The result is the mirror image across the X axis of the previous diagram. In all of the previous examples it was stated that Y becomes and remains positive or negative. The meaning here is that eventually the graph will take on these shapes with no further bends since the leading term dominates at these extremes for x. Now we turn our attention to the shape of the polynomial function between the tails above. It turns out that if the degree of the polynomial is n, then at most there can be n x intercepts. Also, there can be no more than n-1 turning points. (The proof of these statements is usually covered in a college course on Algebraic Structures or Theory of Algebra.) This is a plot of y = x3 - 2x2 - 9x + 14 The leading term has the exponent n=3, which is odd and as you can see the tails of this plot follow our previous discussion. (The leading coefficient is 1 which is positive.) The blue points 1, 4, and 6 are the x intercepts; we have 3 and at most we can have is n = 3. (Determining these points will be shown later.) The green points 2 and 5 are the turning points; 2 at most, since n - 1 = 3 - 1 = 2. The Y intercept is the red point 3. Every polynomial has one and only one Y intercept (this is easily seen if you recall that the Y intercept occurs when x = 0, and when you substitute this x into any polynomial only the constant term remains while all the other terms become 0.) Now let's have a look at how the terms look individually and how they affect each other when plotting the graph. Odd Powers graphs of: x, x3, x5, x7                  Here we see a straight line for the exponent 1, then the curves for the odd exponents. The larger the exponent the closer that curve is to the X axis for x between -1 and 1, and the longer it stays near the X axis and the steeper it becomes beyond x = 1 and x = -1. For |x| < 1 as the exponent increases, x to that power decreases. for example: x =0.2 then x2 = 0.04, x3 = .008, and x4 = .0016 notice how the values get smaller. Even Powers graphs of x2, x4, x6 The larger the exponent, the flatter the graph is on along the X axis between x = -1 and x = 1 and the steeper it becomes as x increases beyond 1 or decreases beyond -1. Intercepts with even leading terms: The constant term of a polynomial has the effect of moving the graph vertically along the Y axis. Recall that this constant term is the Y intercept. n is even. Shown are the positions for the graphs: xn + 1, xn + 0, xn - 2 Intercepts with odd leading terms: The observations are the same as for even leading terms. n is odd. Shown are the positions for the graphs: xn + 1, xn + 0 , xn - 2 Adding the linear term with even leading term exponents: Here we have y = xn + x (In this graph below n = 2.) We'll move from left to right from x = -1 to x = -½ then to x = 0: y = x2 + x At x = -1, (-1)2 + (–1) = 0 and the graph crosses the X axis at x = -1. At x = 0, 02 + 0 = 0. At x = -½, the graph reaches its minimum value of (-½)2 – ½ = -¼ At this point x2 becomes dominant (larger) and the tail swings positive and remains positive. This point is an estimate for our purposes, the curve must swing up so that it crosses the X axis at x = 0. For negative x < -½, x2 is dominant and very soon the x term contributes very little. For larger magnitudes of x, the graph will become x2 and the x term will all but vanish. If the linear term had a negative slope then the effect would be to shift the graph along the positive X axis as shown next. y = x2 - x At x = 1, (1)2 – (1) = 0. The polynomial crosses the X axis at x = 1. At X = 0, (0)2 – (0) – 0. The polynomial crosses the X axis at x = 0. For 0 < x < 1, at x = ½, (½)2 - ½ = ¼ - ½ = - ½. For x > ½ the leading term dominates and the curve swings positively to the point x = 1. Adding the linear term with positive slope to odd leading term exponents: Here we have y = x3 + x At point 3, x = 0, so y = 0. At point 1 where x = -1 y = -1 and y3 = -1 so their sum is -2 at point 2. Likewise at point 4 where x = 1, y = 1 and y3 = 1 so their sum = 2 at point 5. A positive slope for the linear term tends to straighten out the odd degree polynomial into the cyan polynomial . Adding the linear term with negative slope to odd leading term exponents. y = x3 – x and y = x3 - 3x The red curve is y = x3, The blue line is y = -x and the magenta line is y = -3x. As the slope of the linear term becomes more negative (the line rotates clockwise from the blue line to magenta line (dark black arrow top left of graph.) The effect is to pull the turning points farther apart (vertically), as you can see in the sequence: the red graph to the blue graph to the magenta graph. The curved blue arrow shows the effect of -x on x3. The red curve is twisted clockwise to become the blue curve. The curved magenta arrow just below the curved blue arrow shows the effect -3x has on x3. The red curve twists further clockwise to become the magenta curve. Point 1 is x = 0 so y = 0 for all graphs. Point 2 (at x = -1 ) is the result of adding -1 (from y = x3 = (-1)3 = -1) to the +1 (from y = -x = -(-1) = 1) giving the result 0. Point 3 ( at x = -1) is the result of adding the -1 (from y = x3 ) to the +3 (from y = -3x) giving the result 2. Example: sketch y = x3 + x2 First sketch each term. (The light green and red curves in the next graph.) For positive x both terms are positive and y will increase positively eventually following the x3 term since that term is dominant for large x. For negative x between 0 and -1 the x2 term is larger so the x2 graph pulls up the x3 graph until x = -1. At this point their sum equals zero and as x becomes more negative the x3 graph is dominant and for large negative x the graph becomes x3. For x > 0, the values for both terms are positive so the graph remains positive. The squared term adds to the cubed term causing the final graph to increase more for lower values of x, but the cubed term takes over at x = 1. x3 > x2 when x > 1 (here: x3 > x2  x3 – x2 > 0  x2(x-1) > 0 Now x2 is always greater than 0 so we must have (x – 1) > 0, so x > 1 is when the leading term takes over.) Example: sketch y = x4 + x3 First sketch each term. (Red and green curves in the next graph.) For positive x both terms add to a positive y and the graph becomes x4 as x gets very large. This happens at x = 1, since this is the value of x where the terms are equal. For negative x between 0 and -1 x3 is larger then x4, so x3 dominates and pulls the x4 graph below the X axis until x becomes -1. At x = -1 their sum equals 0 ((-1)4 = 1) and ((-1)3 = -1) so 1 + -1 = 0. Then as x becomes more negative x4 dominates x3 and the graph takes on the shape of x4. Example: sketch y = 3x4 + 2x2 + x – 1 First sketch each term. ( Each curve is labeled in the next graph.) The leading term is even so sketch in the tails close to its graph. Also place a dot at the y intercept (where x = 0). Now as x decreases from 0 to -1, the negative portion of y = x and y = -1 act to pull the final curve down into the dotted region shown in the graph following this one. At x = -0.4, (dashed oval) the leading term and the squared term add very little, about 0.2, the linear term adds -0.4 and the constant term adds -1. Taken together we get about -1 - 0.4 + 0.2 = -1.2. This value is plotted in the next graph. We have enough points to sketch in the final curve. Polynomial curves are smooth which means it must return to the Y intercept y=-1. And from there we can proceed to the tail on the positive X and also the tail on the negative left. Here is a plot of the final curve. Notice how the linear term y = x distorts the shape of the curve. Before we saw a shift due to the linear term, but now we have two higher terms with linear exponents. The combined effect is to distort the graph somewhat. Finally, the constant term moves the curve down the Y axis. Example: sketch y = x5 + 2x2 - x + 1 First sketch each term. Recall that a linear term with negative slope tends to rotate and stretch the leading term x5 clockwise as shown next. The dashed curve is the result of the linear term. Sketch in the tails as well. Now we need to move the sketch to intersect the Y axis at the y intercept ( the constant term of the polynomial, 1.) The dashed curve moves vertically to the positive y = 1. All the terms between x = 0 and x = -1 contribute more than the leading term, x5, so the effect of this will be to pull the graph of x5 for negative x high. For positive x all terms give positive y values except for the linear term. Taken together the linear term is quickly overtaken and the final curve must almost immediately swing positive. The curve will not be symmetric about the Y axis. At x = -1 we can add each term labeled 1, 2, 3, 4 and 5 on the graph to get the y value 3. (-1 + 1 + 1 + 2 = 3). (this is one bend) As x increases toward zero from x = -1 all terms decrease as well so the graph must cross the Y axis at x = 0 and the intercept y = 1. At x = 0.5 The points 5, 6, and 7 add to y = 1 (-0.5 + 0.5 + 1 = 1). This means there is another bend between x = 0 and x = 0.5. If we look at x = 0.25 then the term 2x2 contributes 2(0.25)2 = 0.125 and the term -x contributes -0.25, added together the y value is -0.125 plus the constant term 1 gives 0.875 for y. Here is a plot of the final graph. As before the linear term distorts the graph over the negative X axis and has little effect over the positive X axis. Summary: plot each term of the polynomial ( different colors can be very helpful) The terms with smaller exponents than the leading term exponent are dominant on the interval -1 < x < 1. These lower terms act to distort the primary shape of the graph of the leading term. Apply each term to change the shape of the leading term's graph in succession. (not eh constant term yet) Move the final graph to the intercept which is the constant term. Calculate the values between turns of the polynomial, plot them and obtain a final shape for the plot to sketch the rest into place.   FINDING THE ZEROES OF A POLYNOMIAL   The Fundamental Theorem of Algebra states that every polynomial of degree 1 or larger has at least one zero in the system of complex numbers.  The Factor Theorem assures us that for a polynomial, if we substitute c for x in the polynomial and that polynomial sums to zero then (x - c) is a factor of the polynomial.                 For polynomial f(x), if f(c) = 0, then (x -c) is a factor.font> Solving polynomials refers to finding the value(s) of x that when substituted into the polynomial results in 0.  There can not be more than n such roots where n is the degree of the leading term of the polynomial. We are interested in real roots; it is possible for a polynomial to not have any real roots. These roots are also called the "zeroes" of the polynomial. Using the techniques for factoring quadratic and cubic equations will help us find these zeroes. Here I'll present a technique that will show all possible rational values of x that could be roots of the polynomial. Now the point here is they could be roots but are not necessarily roots. We'll do this by example. This example is detailed and shows a number of techniques, so move through it slowly. The factoring method has been covered in the section on factoring but the technique is explained in detail here again. Find the rational zeros for the polynomial 3x4 –11x3+ 10x – 4 We look for all possible values for the ratio p/q where pis a factor of -4 p is a factor of the constant term and q is a factor of 3. q is a factor of the leading term coefficient p can be -1, 4, 1, -4, 2, -2 (all factors of -4) q can be 1, 3, -1, -3 (all factors of 3) so these are all the possible values of p/q: -1/1, -1/3, -1/-1, -1/-3, 4/1, 4/3, 4/-1, 4/-3 1/1, 1/3, 1/-1, 1/-3 -4/1, -4/3, -4/-1, -4/-3 2/1, 2/3, 2/-1, 2/-3, -2/1, -2/3, -2/-1, -2/-3 eliminating the repeated ratios, this reduces to -1, -1/3, 1, 1/3, 4, 4/3, -4, -4/3, 2, 2/3, -2, -2/3 -1 and 1 are easy to check by direct substitution: 3(-1)4 – 11(-1)3 + 10(-1) - 4 = 3 + 11 – 10 – 4 = 0 (yes) 3(1)4 – 11(1)3 + 10(1) – 4 = 3 - 11 + 10 – 4 = -2 (no) Since x = -1 is a root of the polynomial that means (x + 1) is a factor. recall: x = -1 --> x + 1 = 0, so (x + 1) is a factor Now factor 3x4 – 11x3+ 10x – 4 (by inspection) (x + 1)( a x3 + b x2 + c x - 4 ) First we find a: The only way to get 3x4 is by multiplying terms x and x3, we cannot change the coefficient of x, so amust be 3. Now rewrite the equation: (x + 1)( 3 x3 + b x2 + c x - 4 ) = 3x4– 11x3+ 10x – 4 Next we find b: we need to create - 11 x3. x3 comes from the product of the x and x2 terms, factors 1 and b, together with the product of the terms 1 and 3x3. This last pair is already fixed, (1)( 3x3), all we can alter is b. So (1)( 3x3) + x(bx2) = 3x3 + bx3 = (b+3)x3 = -11x3 (b + 3) = -11, so b = -14 Now rewrite the equation: (x + 1)( 3 x3 -14 x2 + c x - 4 ) = 3x4– 11x3+ 10x – 4 Last we find c: We need to create 10x. The x2 and x3 terms are not involved to create the x term. Firstx and -4 are multiplied together and the 1 and c x are multiplied together and summed to get the x term. (x)(-4) + (1)(cx) = ( c – 4 )x = 10x So (c – 4) = 10, that is c = 14. So finally we have: (x + 1)( 3 x3 – 14 x2 + 14 x – 4) Three of the remaining eleven factors may be a root of the second term. (Three since the leading term has the power 4 which means there can only be 4 and we've already found one, x = -1. ) use synthetic division on the second term: ( 3x3 – 14 x2 + 14 x – 4) (x + 1/3) 3 -14 14 -4 -1/3 -1 5 -19/3 3 -15 19 -31/3 not 0 (x - 1/3) 3 -14 14 -4 1/3 1 -13/3 29/9 3 -13 29/3 -7/9 not 0 (x + 2) 3 -14 14 -4 -2 -6 40 -108 3 -20 54 -113 not 0 (x - 2) 3 -14 14 -4 2 6 -32 -56 3 -8 -28 -60 not 0 (x - 2/3) 3 -14 14 -4 2/3 2 -8 4 3 -12 6 0 yes Now, since 2/3 is a root we have x - 2/3 = 0 so x = 2/3 and 3x = 2 or 3x – 2 = 0, is a factor now factor: (3 x3 – 14 x2 + 14 x – 4) again by inspection (3x – 2)( A x2 +B x + C ) We need 3x2 from the terms 3x and AX2 (no other terms will give x3) A must be 1 since 3x(1x2) = 3x3 rewrite the equation (3x – 2)( x2 + Bx + C) = (3 x3 – 14 x2 + 14 x – 4) Now we need -14x2 from the terms 3x,Bx and -2, x2 3Bx2 -2x2 = -14x2 3Bx2 = -12x2 B = -4 Substitute and rewrite: (3x – 2)( x2 - 4x +C) = (3 x3 – 14 x2 + 14 x – 4) The terms that give -4 are -2,C. -4 = -2C, so C = 2 (3x – 2)( x2 - 4x + 2) now (x2 – 4x + 2) is prime (since b2 – 4ac = 16 – 4(1)(2) = 8, which is not a square,) so combining these two factors with the first (x + 1)( 3 x3 – 14 x2 + 14 x – 4) we get (x + 1)(3x - 2)(x2 - 4x + 2) So, this polynomial has two rational roots: x = -1 and x = 2/3. The quadratic term (x2 - 4x + 2) has two real roots: using the quadratic formula: (-(-4) ± √8)/2(1) = (4 ± 2√2)/2 = 2±√2 x = 2 + √2 x = 2 - √2 To completely factor the polynomial use these last factors like so: since x = 2 + √2 is a zero for the polynomial then x - (2 + √2 ) is a factor of the polynomial as is x - (2 - √2 ) (x + 1)(3x – 2)(x - (2 + √2 )) ( x - ( 2 - √2)) Finally we have for a complete factorization: (x + 1)(3x – 2)(x - 2 - √2)(x – 2 + √2) Now in retrospect using synthetic division for each possible rational root can be cumbersome. A spreadsheet with the possibilities listed down one column and the second column containing the polynomial evaluating the first column will quickly determine if the polynomial has any rational roots. For example in the snapshot below: The original equation was entered directly (a power function could have been used for the exponent terms but was not for clarity.) You can see it in the cell entry bar. A1 and A11 contain the rational roots (in bold). Another way to minimize the number of synthetic divisions to perform (without a spreadsheet,) is to sketch the graph and see where the possible zeros are. This time I'll start with the leading and secondary terms on the graph. On the left side of the graph (negative X axis) both terms are positive, the first for the even (4) exponent and the second because of the negative coefficient, -11. So the left tail of the two will become very steep (shown in light blue). Of course at 0 Y is zero so we pass through the origin. For x between zero and one, x4 is smaller so the graph will swing negative for awhile. At some point 3x4 will dominate -11x3 and swing the graph positive. When they equal they will add to 0, crossing the X axis. This happens when 3x4 equals 11x3, that is at x = 11/3 (about 3.66) (how? 3x4 – 11x3 = 0, solve for x) So, somewhere in between x = 0 and x = 3.66 the graph needs to turn positive again. (Note, if -11 where -2 instead, the graph would have a shallower dip). Just where this turn happens is not important for our purposes, techniques in Calculus will show us exactly. Polynomial curves are smooth and we know it must swing back to the X axis. Now let's add the effect of the linear term +10x. The previous curve is shown in red. Adding the linear term shown in green acts to rotate the previous curve counter clockwise as shown to the final light blue curve. Now let's add the constant term, recall that this will only move the graph along the vertical axis to that point. The previous curve is shown in red, and the constant term is shown in green. The final plot for this polynomial is shown in light blue. From this graph we can determine that the most likely candidates for rational roots are those nearby the crossings of the graph with the X axis, -1, 2/3, and 3.5. This observation narrows down the candidates to -1 and 2/3, both of which happen to be the rational roots of this polynomial. This is another FREE ALGEBRA PRINTABLE presented to you from the Algebra section of K12math.com

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