MAC 2311
GRAPHING POLYNOMIALS
Polynomials are differentiable on the set of real numbers (implying continuity on the set of real numbers).
So, the graph of a polynomial is a smooth curve (no corner points or cusps) on −∞,∞
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To sketch the graph of a polynomial function y = P x :
1) Find the y-intercept P(0) and if possible algebraically, any x-intercepts.
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2) Determine the end behavior of f (the ‘arms’ of the graph); that is, find lim P x and lim P x .
x→∞
x→−∞
3) Find all critical points, intervals of increasing and decreasing, and relative extrema. Since f is
differentiable everywhere, every critical point will be a stationary point; that is, a zero of f ′ .
4) Find all intervals of concave up, concave down, and any inflection points.
5) Use the information in the above four steps to sketch the graph. Do not forget to plot the intercepts,
relative extrema, and inflection points. Also, do not overlook step two-the end behavior must be
correct. A combined sign chart helps with step 4.
EXAMPLES:
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PROBLEM ONE: Sketch the graph of f x = x 3 − 6x 2 − 9x + 3
1) The y-intercept is f(0) = -3. The x-intercepts cannot be found algebraically.
2) End Behavior- ‘arms’
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lim f x = lim x 3 − 6x 2 − 9x + 3 = lim x 3 = ∞ and lim f x = lim x 3 − 6x 2 − 9x + 3 =
x→∞
x→∞
x→∞
x→−∞
x→−∞
lim x 3 = −∞
x→−∞
3)
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f ′ ( x ) = 0 ⇒ 3x
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f ′ x = 3x 2 − 12x + 9 and f ′′ x = 6x − 12
2
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So, f has two stationary points-----x=1 and x=3.
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− 12x + 9 = 0 ⇒ 3 x 2 − 4x + 3 = 0 ⇒ 3 x − 3 x − 1 = 0 ⇒ x = 1 or x = 3
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f ′′ x = 0 ⇒ 6x − 12 = 0 ⇒ 6 x − 2 = 0 ⇒ x = 2
f has a possible inflection point at x=2.
4)
f′
x−3
x −1
+
−
−
−
−
+
+
+
+
− − − − − − − − − − −− − − − − − − − − − − − − −− − − − − − − −
1
3



f ′′
−
+
x−2
−
+
6
+
+
− − − − − − − − − − − − − − − − − − −− − − − − − − − − − − − − − −
2
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The combined sign chart reveals that f is increasing and concave down on the interval −∞,1 ,
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decreasing and concave down on 1,2 , decreasing and concave up on 2,3 and increasing and
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point ( 2,f ( 2)) = ( 2,−1) is an inflection point of f.
concave up on 3,∞ . f has a relative maximum of f(1) = 1 and a relative minimum of f(3) = -3. The
5) Graph:
y
3
2
(1,1)
1
x
–1
1
2
3
–1
(2,-1)
–2
–3
(3,-3)
–4
–5
4
5
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PROBLEM TWO: Sketch the graph of f x = x 4 − 6x 2
1) The y-intercept is f(0) = 0. The x-intercepts are the solutions to x 4 − 6x 2 = 0
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x 4 − 6x 2 = 0 ⇒ x 2 x 2 − 6 = 0 ⇒ x 2 x + 6 x − 6 = 0 ⇒ x = 0 or x = − 6 or x = 6
2) End Behavior:
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lim f x = lim x 4 − 6x 2 = lim x 4 = ∞ and lim f x = lim x 4 − 6x 2 − 9x + 3 =
x→∞
x→∞
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x→∞
x→−∞
x→−∞
lim x 4 = ∞ +∞ for both limits since the degree is even and leading coefficient > 0
x→−∞
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3)
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f ′ ( x ) = 0 ⇒ 4x
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f ′ x = 4x 3 − 12x and f ′′ x = 12x 2 − 12
3
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− 12x = 0 ⇒ 4x x 2 − 3 = 0 ⇒ 4x x + 3 x − 3 = 0 ⇒ x = 0 or x = 3 or x = − 3
So, f has three stationary points------ x = 0, x = − 3 and x = 3 .
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f ′′ x = 0 ⇒ 12x 2 − 12 = 0 ⇒ 12 x 2 − 1 = 0 ⇒ 12 x + 1 x − 1 = 0 ⇒ x = −1 or x = 1
f has possible inflection points at x = −1 and x = 1 .
NOTE: f is an even function; that is, f(-x) = f(x) for all x (verify that). So, the graph of f is symmetric to the
y-axis. The y-axis symmetry allows for graphing on 0,∞ and then applying the symmetry to sketch over
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( −∞,0) . So, a sign analysis only on (0,∞ ) is required.
−
−
+
+
f′
x− 3
x+ 3
4x
+
+
+
+
− − − − − − − − − − − − − − − − −− − − − − − − − − − − − − −
0
3


4)
f ′′
x −1
x +1
12
−
−
+
+
+
+
+
+
− − − − − − − − − − − − − −− − − − − − − − − − − − − − −
0
1
5)
y
x
–4
(- 6 , 0)
–2
2
–2
–4
(-1, -5)
(1, -5)
–6
–8
(-
3 , 0)
– 10
– 12
(
3 , 0)
( 6 , 0)
4