#4a.

The basic equation (mathematical model) is y = ax³ + bx² + cx + d, where x = gigawatt hours (i.e., millions of kwh) and y = cost in thousands of dollars. (Actually, you can choose any units you wish, but these units keep the numbers small and manageable.) We have four pieces of information: (1) when x=1, y=20; (2) when x=2, y=29; (3) when x=3, y=34; and (4) when x=4, y=41. Plugging these four facts in, one at a time, gives four linear equations in four unknowns:
a + b + c + d = 20
8a + 4b + 2c + d = 29
27a + 9b + 3c + d = 34
64a + 16b + 4c + d = 41
You can solve this system of equations using linear combinations, substitution, Cramer’s Rule, or a variety of other methods . . . but by far the easiest is to use your calculator’s augmented matrix capability. Here’s what you key in: MATRX EDIT [A]. The matrix is 4 x 5 (i.e., 4 rows by 5 columns) with entries equal to the coefficients seen above. In other words, you have 1, 1, 1, 1, and 20 in the first row (press ENTER after each one), 8, 4, 2, 1, and 29 in the second row, and so on until you have entered all four rows. Then press 2nd QUIT. To compute the solution, key in MATRX MATH rref( MATRX NAMES [A] ENTER ) ENTER. This will display as rref([A]). You can now read off the coefficients: a=1, b=–8, c=26, and d=1. Therefore, the mathematical model is y=x³ – 8x² + 26x + 1 (in thousands of dollars), which is the answer to part (a).

b.

Plug in x=1.5 to get y=25.375 (write your answer as $25,375). The fastest way is to use your calculator’s "Y=" button to define a function Y₁=x³ – 8x² + 26x + 1; then key in VARS Y-VARS FCN Y₁ followed by (1.5) ENTER. This will display as Y₁(1.5).

c.

Plug in x=0 to get y=1 (i.e., $1000); apparently $1,000 is a service charge.

d,e.

Note: There are two methods; choose whichever one you prefer. You may get errors if you enter values that are outside your graphing window.
Method 1: Define Y₂=35, and use 2nd CALC INTERSECT to find where Y₁ crosses this value.
Method 2: In your head, think of setting y equal to 35. In other words, 35 = x³ – 8x² + 26x + 1, from which simple algebra gives you 0 = x³ – 8x² + 26x – 34. That is, we want to find a root (i.e., zero) of a new function that is x³ – 8x² + 26x – 34. So, just edit Y1 slightly so that the last term is –34, and then use 2nd CALC ZERO to find the zero.

Which method do you like better? It doesn’t matter; either way, you get x=3.191 million kwh (answer).

R1.

Done in class. Be sure you know how to do this using the method of synthetic substitution. The answer is 19.

R2a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.

–i
i
1
3iÖ7
Done in class (in the complex plane, go 5 units to the left of the origin and 3 units up).
–17 – 13i
5+9i
66+26i
145
143+24i
3i
Ö137

R3a.

By the quadratic formula, the solution set is { –2/3 + i (Ö 26)/3, –2/3 – i (Ö 26)/3}.

b.

In class, we did a method that nobody liked, which went like this: Let U and V denote the two roots. Then the equation has a factored form of (x – U) (x – V) = 0. We’ve seen this many times with real roots U and V; all we’re saying is that it works for complex roots as well. OK, then since we know our complex roots are 3+4i and 3–4i, let’s plug those in for U and V and let the dust settle. After some messy algebra, you get x² – 6x + 25 = 0, which is the answer. You must include the "=0" since the problem asked for an equation. Note that any nonzero multiple of this equation (for example, –3x² + 18x – 75 = 0) would also be a correct answer.

Here’s another optional method (use it if you wish): Since the quadratic formula gives us two conjugate answers, the sum of the two roots would give us a nice cancellation of the Ö (b2–4ac)/2a part. In other words, we know that the sum of the two roots is –b/2a + mess + (–b/2a) – mess. The "mess" cancels, leaving –b/2a – b/2a, or simply –b/a. But . . . we also know that if 3+4i is a root, the other root must be 3–4i, from which you can see that the roots have a sum of 6. If the sum is –b/a, and the sum is also 6, then we know –b/a=6. The number a can basically be any nonzero number (think of parabolas: all a does is tell you whether the parabola opens up or down, and how fat it is; there are infinitely many parabolas you could find that have the same pair of roots). So, let’s make our lives easy and choose a=1. Since –b/a = 6, we now have –b=6, or b=–6. Our parabola is y = ax² + bx + c = x² – 6x + c, where c is the only unknown. Plug in the known root 3+4i to get 0 = (3+4i)² – 6(3+4i)+c and solve for c. You should get c=25. Therefore, y = x² – 6x + 25 is our parabola of interest. However, the problem asked for a quadratic equation (not a function), and the equation that is true for these roots is y = 0 = x² – 6x + 25, which we usually write as x² – 6x + 25 = 0. Note that any nonzero multiple of this equation (for example, –3x² + 18x – 75 = 0) would also be a correct answer.

R4a.
b.
c.

21 (note you must use synthetic substitution; calculator is only for double-check)
I assume you can do this. It’s a squiggle that starts at (–3, 35) and ends at (3, –37).
Factor out the real zero at x=2 to get P(x) = (x – 2) (–2x² – 5x – 4). The second factor has roots of –5/4 plus or minus i(Ö 7)/4. Overall solution set: {2, –5/4 + i(Ö 7)/4, –5/4 – i(Ö 7)/4}.

d.

Use synthetic division with a divisor of –2. (Don’t use +2; the divisor in synthetic division is the thing that is being subtracted from x. Also, remember to set up your synthetic division as 1 0 0 17 in the first row since you need placeholders for the quadratic and linear terms) You should get a remainder of 9. Here’s how to double-check your work: Plug x=–2 into the polynomial x³+17 and see what you get!

e.

This would be an "M"-shaped curve that sweeps back and forth across the x-axis. It should cross the x-axis twice on the left side of the y-axis and twice more on the right side.

R6a.

This is very similar to #4 from section 10–6 except without all the "word problem" stuff. See the solution above for #4 to see how to proceed. You should get the following augmented matrix:
First row: –1, 1, –1, 1, –10
Second row: 8, 4, 2, 1, –7
Third row: 27, 9, 3, 1, 2
Fourth row: 64, 16, 4, 1, 45
The rref([A]) solution gives coefficients of 3, –10, 2, and 5. Therefore, the answer is P(x) = 3x³ – 10x² + 2x + 5.

b.

By synthetic substitution (which is even faster than your calculator here!), you get –63.

c.

There cannot possibly be any integer roots less than –5 or greater than 5. (Why not? Hint: Remember the Rational Root Theorem.) So, if you’re in a hurry, just enter the function as Y₁ on your TI-83, and then press 2nd TBLSET TblStart=–5 D Tbl=1 Auto Auto 2nd TABLE. As you scroll down, you’ll see that the only integer root is at x=1.