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Algebra 2 9-12 / Week 02 / Tuesday
2/6
Week 02 · Polynomial Functions

Tuesday

Zeros and ends
// A box whose volume is a cubic
⏱ about 20 min

Tuesday: Zeros and Ends

Wren tapes graph paper across the Glass House chalkboard. "Rough sketches only. Two facts and a curve."

"Which two?" Comet asks.

"Where it crosses zero, and what it does far away," Wren says. "Try p(x) = (x + 2)(x - 1)(x - 3). What do you notice?"

"Each factor is zero somewhere," Comet says. "x = -2, x = 1, x = 3. Three dots on the axis."

Nova projects the expanded form, x³ - 2x² - 5x + 6. "Would you like a hint? Far away, only the x³ matters."

"Big negative x gives a big negative x³," Comet says. "So the left end falls. The right end rises."

She draws a curve down from the left, up through -2, down through 1, up through 3 and away.

"Rough, but right," Wren says. "Three zeros, two turns, and the ends point the right way."

Way one: zeros from factors

A product is zero when a factor is zero. So each factor x - a gives a zero at x = a.

For p(x) = (x + 2)(x - 1)(x - 3) the zeros are -2, 1, 3. The graph meets the x-axis at each one.

For p(x) = x³ - 4x, factor first: x(x - 2)(x + 2). Zeros -2, 0, 2.

Way two: end behavior from the leading term

Far from zero, the highest-power term is so large that the others hardly matter. Look only at the leading term.

Odd degree: the two ends point opposite ways. Even degree: both ends point the same way.

A positive leading coefficient sends the right end up. A negative one sends it down.

Leading termDegreeLeft endRight end
x³odddownup
-x³oddupdown
x⁴evenupup
-2x⁴evendowndown

The crew's sketch table

PolynomialZerosDegreeEnd behavior
x³ - 4x-2, 0, 23down on the left, up on the right
(x + 2)(x - 1)(x - 3)-2, 1, 33down on the left, up on the right
x⁴ - 5x² + 4-2, -1, 1, 24up on the left, up on the right
-x³ + x² + 6x-2, 0, 33up on the left, down on the right
The graph of y = (x + 2)(x - 1)(x - 3). It passes -2, 1 and 3, down on the left and up on the right.
The graph of y = x⁴ - 5x² + 4. Both ends point up. It crosses the x-axis at -2, -1, 1 and 2.

A degree-n polynomial has at most n zeros and at most n - 1 turning points. The sketch between the zeros is rough, and that is fine.

FIND THE ZEROS
  • Read the question.
  • Tap your answer.
What are the zeros of p(x) = (x + 2)(x - 1)(x - 3)?
What are the zeros of p(x) = x³ - 4x?
What are the zeros of p(x) = -x³ + x² + 6x? Factor out -x first.
READ THE ENDS
  • Read the question.
  • Tap your answer.
How does the graph of y = (x + 2)(x - 1)(x - 3) behave at the far left and far right?
How does the graph of y = x⁴ - 5x² + 4 behave at the far left and far right?
How does the graph of y = -x³ + x² + 6x behave at the far left and far right?
The crew's box function is V(x) = 4x³ - 84x² + 432x. How does its graph behave at the far left and far right?
SKETCH P(X) = (X + 2)(X - 1)(X - 3)
  • ?Odd degree and positive coefficient: down on the left, up on the right
  • ?Set each factor to zero: the zeros are -2, 1 and 3
  • ?Draw a smooth curve through the zeros that matches both ends
  • ?Expand or inspect to find the leading term, x³
  • ?Mark the three zeros on the x-axis
WHY THIS EXERCISETwo facts fix the shape of a rough sketch. Everything else is a smooth curve joining them.
StatementTrue or false?
The graph of a degree-3 polynomial can have four zeros.?
A positive leading coefficient sends the right end of the graph up.?
x⁴ - 5x² + 4 points the same way at both ends.?
(0 - 2 + 2) × (0 - 2 - 1) × (0 - 2 - 3) = 0?
1 - 2 - 5 + 6 = 0?
WHY THIS EXERCISEChecking a zero by substitution proves the sketch starts in the right places.
Try it
Write (x - 1)(x + 3)(x - 4) on a card. Find its zeros, its degree and its end behavior, then sketch it.
Now put a minus sign in front and sketch again. What flipped?
Sketch p(x) = (x - 1)(x + 3)(x - 4) from your card. Mark the zeros, then draw the curve with the right ends.

Good sketching. Tomorrow is Lab day: you build the box, fill it and find the cut that holds the most.

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